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0.8 + 0.02
|
0.82
| |
The function $f(x)$ satisfies
\[f(xy) = \frac{f(x)}{y}\]for all positive real numbers $x$ and $y.$ If $f(30) = 20,$ find $f(40).$
|
15
| |
There exists a scalar $c$ so that
\[\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) + \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) + \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) = c \mathbf{v}\]for all vectors $\mathbf{v}.$ Find $c.$
|
2
| |
A) For a sample of size $n$ taken from a normal population with a known standard deviation $\sigma$, the sample mean $\bar{x}$ is found. At a significance level $\alpha$, it is required to find the power function of the test of the null hypothesis $H_{0}: a=a_{0}$ regarding the population mean $a$ with the hypothetical value $a_{0}$, under the competing hypothesis $H_{1}: a=a_{1} \neq a_{0}$.
B) For a sample of size $n=16$ taken from a normal population with a known standard deviation $\sigma=5$, at a significance level of 0.05, the null hypothesis $H_{0}: a=a_{0}=20$ regarding the population mean $a$ with the hypothetical value $a_{0}=20$ is tested against the competing hypothesis $H_{1}: a \neq 20$. Calculate the power of the two-sided test for the hypothesized value of the population mean $a_{1}=24$.
|
0.8925
| |
A company has calculated that investing x million yuan in project A will yield an economic benefit y that satisfies the relationship: $y=f(x)=-\frac{1}{4}x^{2}+2x+12$. Similarly, the economic benefit y from investing in project B satisfies the relationship: $y=h(x)=-\frac{1}{3}x^{2}+4x+1$.
(1) If the company has 10 million yuan available for investment, how should the funds be allocated to maximize the total investment return?
(2) If the marginal effect function for investment is defined as $F(x)=f(x+1)-f(x)$, and investment is not recommended when the marginal value is less than 0, how should the investment be allocated?
|
6.5
| |
Let \( z = \frac{1+\mathrm{i}}{\sqrt{2}} \). Then calculate the value of \( \left(\sum_{k=1}^{12} z^{k^{2}}\right)\left(\sum_{k=1}^{12} \frac{1}{z^{k^{2}}}\right) \).
|
36
| |
Calculate the value of the expression given by: $$2 + \cfrac{3}{4 + \cfrac{5}{6}}.$$
|
\frac{76}{29}
| |
A traffic light cycles as follows: green for 45 seconds, then yellow for 5 seconds, and then red for 50 seconds. Mark chooses a random five-second interval to observe the light. What is the probability that the color changes during his observation?
|
\frac{3}{20}
| |
Let \( p, q, r, s \) be distinct real numbers such that the roots of \( x^2 - 12px - 13q = 0 \) are \( r \) and \( s \), and the roots of \( x^2 - 12rx - 13s = 0 \) are \( p \) and \( q \). Find the value of \( p + q + r + s \).
|
-13
| |
Given two vectors in space, $\overrightarrow{a} = (x - 1, 1, -x)$ and $\overrightarrow{b} = (-x, 3, -1)$. If $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$, find the value of $x$.
|
-1
| |
Given that Ron incorrectly calculated the product of two positive integers $a$ and $b$ by reversing the digits of the three-digit number $a$, and that His wrong product totaled $468$, determine the correct value of the product of $a$ and $b$.
|
1116
| |
Given that $f(x)$ is an even function defined on $\mathbb{R}$ and satisfies $f(x+2)=- \frac{1}{f(x)}$. When $1 \leq x \leq 2$, $f(x)=x-2$. Find $f(6.5)$.
|
-0.5
| |
Find $\begin{pmatrix} 3 & 0 \\ 1 & 2 \end{pmatrix} + \begin{pmatrix} -5 & -7 \\ 4 & -9 \end{pmatrix}.$
|
\begin{pmatrix} -2 & -7 \\ 5 & -7 \end{pmatrix}
| |
A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$. Then the area of $S$ has the form $a\pi + \sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$.
|
29
|
We can describe the line parallel to the imaginary axis $x=\frac{1}{2}$ using polar coordinates as $r(\theta)=\dfrac{1}{2\cos{\theta}},$
which rearranges to $z=\left(\dfrac{1}{2\cos{\theta}}\right)(cis{\theta})\implies \frac{1}{z}=2\cos{\theta}cis(-\theta).$
Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral:
$[S] = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta.$
Thankfully, this is a routine computation:
$[S] = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta$
$[S] = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12\left[\frac{1}{2}\sin{2\theta}+\theta\right]_0^{\frac{\pi}{6}}=12\left(\frac{\sqrt{3}}{4}+\frac{\pi}{6}\right)=2\pi+3\sqrt{3}=2\pi + \sqrt{27}$
$a+b = \boxed{029}$.
|
Find the number of positive integers $n$ that satisfy
\[(n - 1)(n - 5)(n - 9) \dotsm (n - 101) < 0.\]
|
25
| |
Five volunteers participate in community service for two days, Saturday and Sunday. Each day, two people are selected to serve. Calculate the number of ways to select exactly one person to serve for both days.
|
60
| |
Given non-zero plane vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ satisfy $|\overrightarrow{a}|=2$, $|\overrightarrow{b}-\overrightarrow{c}|=1$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{π}{3}$, calculate the minimum value of $|\overrightarrow{a}-\overrightarrow{c}|$.
|
\sqrt{3} - 1
| |
It is known that \( b^{16} - 1 \) has four distinct prime factors. Determine the largest one, denoted by \( c \).
|
257
| |
How many distinct arrangements of the letters in the word "example" are there?
|
5040
| |
Vasya has:
a) 2 different volumes from the collected works of A.S. Pushkin, each volume is 30 cm high;
b) a set of works by E.V. Tarle in 4 volumes, each volume is 25 cm high;
c) a book of lyrical poems with a height of 40 cm, published by Vasya himself.
Vasya wants to arrange these books on a shelf so that his own work is in the center, and the books located at the same distance from it on both the left and the right have equal heights. In how many ways can this be done?
a) $3 \cdot 2! \cdot 4!$;
b) $2! \cdot 3!$;
c) $\frac{51}{3! \cdot 2!}$;
d) none of the above answers are correct.
|
144
| |
Let \( T \) be the set of all positive divisors of \( 60^{100} \). \( S \) is a subset of \( T \) such that no number in \( S \) is a multiple of another number in \( S \). Find the maximum value of \( |S| \).
|
10201
| |
In Zuminglish, words consist only of the letters $M, O,$ and $P$. $O$ is a vowel, while $M$ and $P$ are consonants. A valid Zuminglish word must have at least two consonants between any two $O's$. Determine the number of valid 7-letter words in Zuminglish and find the remainder when this number is divided by $1000$.
|
912
| |
If $\tan \theta = 4,$ then find $\tan 3 \theta.$
|
\frac{52}{47}
| |
If $M(\frac{p}{2}$,$p)$ is a point on the parabola $y^{2}=2px\left(p \gt 0\right)$, and the distance from $M$ to the point $\left(1,0\right)$ is 1 greater than the distance to the $y$-axis, find:<br/>
$(1)$ The equation of the parabola.<br/>
$(2)$ Let line $l$ intersect the parabola at points $A$ and $B$. The circle with diameter $AB$ passes through point $M$. Find the maximum distance from point $\left(1,0\right)$ to line $l$.
|
2\sqrt{5}
| |
Find the sum of the roots of the equation \[(2x^3 + x^2 - 8x + 20)(5x^3 - 25x^2 + 19) = 0.\]
|
\tfrac{9}{2}
| |
Jordan wants to divide his $\frac{48}{5}$ pounds of chocolate into $4$ piles of equal weight. If he gives one of these piles to his friend Shaina, how many pounds of chocolate will Shaina get?
|
\frac{12}{5}
| |
You are standing at the edge of a river which is $1$ km wide. You have to go to your camp on the opposite bank . The distance to the camp from the point on the opposite bank directly across you is $1$ km . You can swim at $2$ km/hr and walk at $3$ km-hr . What is the shortest time you will take to reach your camp?(Ignore the speed of the river and assume that the river banks are straight and parallel).
|
\frac{2 + \sqrt{5}}{6}
| |
For a $5 \times 5$ chessboard colored as shown below, place 5 different rooks on black squares such that no two rooks can attack each other (rooks attack if they are in the same row or column). How many different ways are there to do this?
|
1440
| |
How many distinct triangles can be drawn using three of the dots below as vertices, where the dots are arranged in a grid of 2 rows and 4 columns?
|
48
| |
How many positive integers less than 1998 are relatively prime to 1547 ? (Two integers are relatively prime if they have no common factors besides 1.)
|
1487
|
The factorization of 1547 is \(7 \cdot 13 \cdot 17\), so we wish to find the number of positive integers less than 1998 that are not divisible by 7, 13, or 17. By the Principle of Inclusion-Exclusion, we first subtract the numbers that are divisible by one of 7, 13, and 17, add back those that are divisible by two of 7, 13, and 17, then subtract those divisible by three of them. That is, \(1997-\left\lfloor\frac{1997}{7}\right\rfloor-\left\lfloor\frac{1997}{13}\right\rfloor-\left\lfloor\frac{1997}{17}\right\rfloor+\left\lfloor\frac{1997}{7 \cdot 13}\right\rfloor+\left\lfloor\frac{1997}{7 \cdot 17}\right\rfloor+\left\lfloor\frac{1997}{13 \cdot 17}\right\rfloor-\left\lfloor\frac{1997}{7 \cdot 13 \cdot 17}\right\rfloor\) or 1487.
|
Find the infinite sum of \(\frac{1^{3}}{3^{1}}+\frac{2^{3}}{3^{2}}+\frac{3^{3}}{3^{3}}+\frac{4^{3}}{3^{4}}+\cdots\).
求 \(\frac{1^{3}}{3^{1}}+\frac{2^{3}}{3^{2}}+\frac{3^{3}}{3^{3}}+\frac{4^{3}}{3^{4}}+\cdots\) 無限項之和。
|
\frac{33}{8}
| |
The complex numbers \( \alpha_{1}, \alpha_{2}, \alpha_{3}, \) and \( \alpha_{4} \) are the four distinct roots of the equation \( x^{4}+2 x^{3}+2=0 \). Determine the unordered set \( \left\{\alpha_{1} \alpha_{2}+\alpha_{3} \alpha_{4}, \alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{4}, \alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}\right\} \).
|
\{1 \pm \sqrt{5},-2\}
|
Employing the elementary symmetric polynomials \( \left(s_{1}=\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=\right. -2, s_{2}=\alpha_{1} \alpha_{2}+\alpha_{1} \alpha_{3}+\alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}+\alpha_{2} \alpha_{4}+\alpha_{3} \alpha_{4}=0, s_{3}=\alpha_{1} \alpha_{2} \alpha_{3}+\alpha_{2} \alpha_{3} \alpha_{4}+\alpha_{3} \alpha_{4} \alpha_{1}+\alpha_{4} \alpha_{1} \alpha_{2}=0 \) and \( s_{4}=\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}=2 \) we consider the polynomial \( P(x)=\left(x-\left(\alpha_{1} \alpha_{2}+\alpha_{3} \alpha_{4}\right)\right)\left(x-\left(\alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{4}\right)\right)\left(x-\left(\alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}\right)\right) \). Because \( P \) is symmetric with respect to \( \alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4} \), we can express the coefficients of its expanded form in terms of the elementary symmetric polynomials. We compute \( P(x) =x^{3}-s_{2} x^{2}+\left(s_{3} s_{1}-4 s_{4}\right) x+\left(-s_{3}^{2}-s_{4} s_{1}^{2}+s_{4} s_{2}\right) =x^{3}-8 x-8 =(x+2)\left(x^{2}-2 x-4\right) \). The roots of \( P(x) \) are -2 and \( 1 \pm \sqrt{5} \), so the answer is \{1 \pm \sqrt{5},-2\}.
|
1. Convert the parametric equations of the conic curve $C$:
$$
\begin{cases}
x=t^{2}+ \frac {1}{t^{2}}-2 \\
y=t- \frac {1}{t}
\end{cases}
$$
($t$ is the parameter) into a Cartesian coordinate equation.
2. If the polar equations of two curves are $\rho=1$ and $\rho=2\cos\left(\theta+ \frac {\pi}{3}\right)$ respectively, and they intersect at points $A$ and $B$, find the length of segment $AB$.
|
\sqrt {3}
| |
What is the sum of all positive integers $\nu$ for which $\mathop{\text{lcm}}[\nu, 45]=180$?
|
292
| |
Let $n$ be a fixed integer, $n \geqslant 2$.
1. Determine the smallest constant $c$ such that the inequality
$$
\sum_{1 \leqslant i<j \leqslant n} x_i x_j (x_i^2 + x_j^2) \leqslant c \left(\sum_{i=1}^n x_i \right)^4
$$
holds for all non-negative real numbers $x_1, x_2, \cdots, x_n$.
2. For this constant $c$, determine the necessary and sufficient conditions for which equality holds.
|
\frac{1}{8}
| |
Fido's leash is tied to a stake at the center of his yard, which is in the shape of a regular hexagon. His leash is exactly long enough to reach the midpoint of each side of his yard. If the fraction of the area of Fido's yard that he is able to reach while on his leash is expressed in simplest radical form as $\frac{\sqrt{a}}{b}\pi$, what is the value of the product $ab$?
|
18
| |
Calculate the value of the expression \[(5^{1003}+6^{1004})^2-(5^{1003}-6^{1004})^2\] and express it in the form of $k\cdot30^{1003}$ for some positive integer $k$.
|
24
| |
Ali Baba and the 40 thieves decided to divide a treasure of 1987 gold coins in the following manner: the first thief divides the entire treasure into two parts, then the second thief divides one of the parts into two parts, and so on. After the 40th division, the first thief picks the largest part, the second thief picks the largest of the remaining parts, and so on. The last, 41st part goes to Ali Baba. For each of the 40 thieves, determine the maximum number of coins he can secure for himself in such a division irrespective of the actions of other thieves.
|
49
| |
Consider integers \( \{1, 2, \ldots, 10\} \). A particle is initially at 1. It moves to an adjacent integer in the next step. What is the expected number of steps it will take to reach 10 for the first time?
|
90
| |
Given a sequence $\{a_{n}\}$ that satisfies ${a}_{n+1}=\frac{1}{3}{a}_{n}$, if $a_{4}+a_{5}=4$, calculate $a_{2}+a_{3}$.
|
36
| |
Given that $a > 0$, $b > 0$, and $\frac{1}{a} + \frac{2}{b} = 2$.
(1) Find the minimum value of $ab$;
(2) Find the minimum value of $a + 2b$, and find the corresponding values of $a$ and $b$.
|
\frac{9}{2}
| |
A line with a slope of 2 and a line with a slope of -4 each have a $y$-intercept of 6. What is the distance between the $x$-intercepts of these lines?
|
\frac{9}{2}
|
The line with a slope of 2 and $y$-intercept 6 has equation $y=2x+6$. To find its $x$-intercept, we set $y=0$ to obtain $0=2x+6$ or $2x=-6$, which gives $x=-3$. The line with a slope of -4 and $y$-intercept 6 has equation $y=-4x+6$. To find its $x$-intercept, we set $y=0$ to obtain $0=-4x+6$ or $4x=6$, which gives $x=\frac{6}{4}=\frac{3}{2}$. The distance between the points on the $x$-axis with coordinates $(-3,0)$ and $\left(\frac{3}{2}, 0\right)$ is $3+\frac{3}{2}$ which equals $\frac{6}{2}+\frac{3}{2}$ or $\frac{9}{2}$.
|
A line segment is divided into four parts by three randomly selected points. What is the probability that these four parts can form the four sides of a quadrilateral?
|
1/2
| |
What is the probability, expressed as a decimal, of drawing one marble which is either red or blue from a bag containing 3 red, 2 blue, and 5 yellow marbles?
|
0.5
| |
Compute the surface area of a cube inscribed in a sphere of surface area $\pi$.
|
2
|
The sphere's radius $r$ satisfies $4 \pi r^{2}=\pi \Rightarrow r=1 / 2$, so the cube has body diagonal 1 , hence side length $1 / \sqrt{3}$. So, its surface area is $6(1 / \sqrt{3})^{2}=2$.
|
Remove all perfect squares and perfect cubes from the set
$$
A=\left\{n \mid n \leqslant 10000, n \in \mathbf{Z}_{+}\right\}
$$
and arrange the remaining elements in ascending order. What is the 2014th element of this sequence?
|
2068
| |
Given $\sin (30^{\circ}+\alpha)= \frac {3}{5}$, and $60^{\circ} < \alpha < 150^{\circ}$, solve for the value of $\cos \alpha$.
|
\frac{3-4\sqrt{3}}{10}
| |
Marty and three other people took a math test. Everyone got a non-negative integer score. The average score was 20. Marty was told the average score and concluded that everyone else scored below average. What was the minimum possible score Marty could have gotten in order to definitively reach this conclusion?
|
61
|
Suppose for the sake of contradiction Marty obtained a score of 60 or lower. Since the mean is 20, the total score of the 4 test takers must be 80. Then there exists the possibility of 2 students getting 0, and the last student getting a score of 20 or higher. If so, Marty could not have concluded with certainty that everyone else scored below average. With a score of 61, any of the other three students must have scored points lower or equal to 19 points. Thus Marty is able to conclude that everyone else scored below average.
|
27 people went to a mall to buy water to drink. There was a promotion in the mall where three empty bottles could be exchanged for one bottle of water. The question is: For 27 people, the minimum number of bottles of water that need to be purchased so that each person can have one bottle of water to drink is $\boxed{18}$ bottles.
|
18
| |
Given that the sequence $\{a_n\}$ is a geometric sequence, and the sequence $\{b_n\}$ is an arithmetic sequence. If $a_1-a_6-a_{11}=-3\sqrt{3}$ and $b_1+b_6+b_{11}=7\pi$, then the value of $\tan \frac{b_3+b_9}{1-a_4-a_3}$ is ______.
|
-\sqrt{3}
| |
To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\ \hline 2 & 3 & &\ \hline & &&4\ \hline & &&\ \hline\end{tabular}\]
|
1
|
We are given a 4x4 grid where each digit from 1 through 4 must appear exactly once in each row and once in each column. The initial grid is:
\[
\begin{array}{|c|c|c|c|}
\hline
1 & & 2 & \\
\hline
2 & 3 & & \\
\hline
& & & 4 \\
\hline
& & & \\
\hline
\end{array}
\]
We need to determine the number in the lower right-hand square.
1. **Fill in the first row**: The first row already contains the numbers 1 and 2. The numbers missing are 3 and 4. Since there is a 3 in the second row, second column, the 3 cannot be placed in the second column of the first row. Therefore, the only place for 3 in the first row is the fourth column. The remaining number for the first row, second column is 4. The updated grid is:
\[
\begin{array}{|c|c|c|c|}
\hline
1 & 4 & 2 & 3 \\
\hline
2 & 3 & & \\
\hline
& & & 4 \\
\hline
& & & \\
\hline
\end{array}
\]
2. **Fill in the second row**: The second row already contains the numbers 2 and 3. The numbers missing are 1 and 4. Since there is a 4 in the third row, fourth column, the 4 cannot be placed in the fourth column of the second row. Therefore, the only place for 4 in the second row is the third column. The remaining number for the second row, fourth column is 1. The updated grid is:
\[
\begin{array}{|c|c|c|c|}
\hline
1 & 4 & 2 & 3 \\
\hline
2 & 3 & 4 & 1 \\
\hline
& & & 4 \\
\hline
& & & \\
\hline
\end{array}
\]
3. **Fill in the third row**: The third row already contains the number 4. The numbers missing are 1, 2, and 3. Since there is a 1 in the second row, fourth column, and a 2 in the first row, third column, the only place for 1 in the third row is the second column, and the only place for 2 is the third column. The remaining number for the third row, first column is 3. The updated grid is:
\[
\begin{array}{|c|c|c|c|}
\hline
1 & 4 & 2 & 3 \\
\hline
2 & 3 & 4 & 1 \\
\hline
3 & 1 & 2 & 4 \\
\hline
& & & \\
\hline
\end{array}
\]
4. **Fill in the fourth row**: The fourth row is missing all numbers. Since the numbers 1, 2, 3, and 4 must appear exactly once in each column, and considering the numbers already placed in each column, the only arrangement that satisfies all conditions is 4 in the first column, 2 in the second column, 3 in the third column, and 1 in the fourth column. The final grid is:
\[
\begin{array}{|c|c|c|c|}
\hline
1 & 4 & 2 & 3 \\
\hline
2 & 3 & 4 & 1 \\
\hline
3 & 1 & 2 & 4 \\
\hline
4 & 2 & 3 & 1 \\
\hline
\end{array}
\]
Thus, the number in the lower right-hand square is $\boxed{\textbf{(A)}\ 1}$.
|
In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
|
7
|
Let $c$ be the number of correct answers Olivia had, and $w$ be the number of incorrect answers. Since there are 10 problems in total, we have:
\[ c + w = 10 \]
For each correct answer, Olivia gains 5 points, and for each incorrect answer, she loses 2 points. Therefore, her total score can be expressed as:
\[ 5c - 2w = 29 \]
We can solve these equations simultaneously. First, express $w$ in terms of $c$ from the first equation:
\[ w = 10 - c \]
Substitute this expression for $w$ into the second equation:
\[ 5c - 2(10 - c) = 29 \]
\[ 5c - 20 + 2c = 29 \]
\[ 7c - 20 = 29 \]
\[ 7c = 49 \]
\[ c = 7 \]
Thus, Olivia had 7 correct answers. To verify, calculate the score with 7 correct answers:
\[ 5 \times 7 - 2 \times (10 - 7) = 35 - 6 = 29 \]
This confirms that the calculations are correct. Therefore, the number of correct answers Olivia had is $\boxed{\text{(C)}\ 7}$. $\blacksquare$
|
Given a tetrahedron \( P-ABC \) with its four vertices on the surface of sphere \( O \), where \( PA = PB = PC \) and \( \triangle ABC \) is an equilateral triangle with side length 2. \( E \) and \( F \) are the midpoints of \( AC \) and \( BC \) respectively, and \( \angle EPF = 60^\circ \). Determine the surface area of sphere \( O \).
|
6\pi
| |
Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?
|
252
|
Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: \[0.4p+20 = 0.42(p+20)\] Solving for p gives us $p=580$, so the solution is $0.4p+20 = \boxed{252}$
|
Insert a square into an isosceles triangle with a lateral side of 10 and a base of 12.
|
4.8
| |
Given that \( x \) is a four-digit number and the sum of its digits is \( y \). When the value of \( \frac{x}{y} \) is minimized, \( x = \) _______
|
1099
| |
Given a point M$(x_0, y_0)$ moves on the circle $x^2+y^2=4$, and N$(4, 0)$, the point P$(x, y)$ is the midpoint of the line segment MN.
(1) Find the trajectory equation of point P$(x, y)$.
(2) Find the maximum and minimum distances from point P$(x, y)$ to the line $3x+4y-86=0$.
|
15
| |
From Moscow to city \( N \), a passenger can travel by train, taking 20 hours. If the passenger waits for a flight (waiting will take more than 5 hours after the train departs), they will reach city \( N \) in 10 hours, including the waiting time. By how many times is the plane’s speed greater than the train’s speed, given that the plane will be above this train 8/9 hours after departure from the airport and will have traveled the same number of kilometers as the train by that time?
|
10
| |
What is the sum of the odd positive integers less than 50?
|
625
| |
Label one disk "$1$", two disks "$2$", three disks "$3$", ..., fifty disks "$50$". Put these $1+2+3+ \cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
|
415
|
1. **Understanding the Problem:**
We are given disks labeled from 1 to 50, where the label corresponds to the number of disks with that label. We need to determine the minimum number of disks that must be drawn to ensure that at least ten disks with the same label are drawn.
2. **Calculating Total Disks:**
The total number of disks is the sum of the first 50 natural numbers:
\[
1 + 2 + 3 + \cdots + 50 = \frac{50 \times (50 + 1)}{2} = 1275
\]
3. **Applying the Pigeonhole Principle:**
To guarantee that at least ten disks with the same label are drawn, consider the worst-case scenario where we try to avoid this as long as possible.
4. **Drawing Disks with Labels 1 to 9:**
The sum of disks from labels 1 to 9 is:
\[
1 + 2 + 3 + \cdots + 9 = \frac{9 \times (9 + 1)}{2} = 45
\]
Drawing all these disks, we still do not have ten disks of any label.
5. **Drawing Disks with Labels 10 to 50:**
For labels 10 to 50, each label has at least 10 disks. To avoid drawing ten disks of any one label, we can draw up to 9 disks from each of these labels. There are 41 labels from 10 to 50 (inclusive):
\[
50 - 10 + 1 = 41
\]
Drawing 9 disks from each of these 41 labels gives:
\[
41 \times 9 = 369
\]
6. **Total Disks Drawn Without Reaching Ten of Any Label:**
Adding the disks drawn from labels 1 to 9 and 10 to 50:
\[
45 + 369 = 414
\]
At this point, we have not yet drawn ten disks of any one label.
7. **Drawing One More Disk:**
Drawing one more disk will ensure that we have at least ten disks of some label, because we have maximized the number of disks drawn without reaching ten for any label.
8. **Conclusion:**
Therefore, the minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is:
\[
\boxed{415}
\]
|
Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly $m$ minutes. The probability that either one arrives while the other is in the cafeteria is $40 \%,$ and $m = a - b\sqrt {c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a + b + c.$
|
87
| |
Given that Chloe's telephone numbers have the form $555-ab-cdef$, where $a$, $b$, $c$, $d$, $e$, and $f$ are distinct digits, in descending order, and are chosen between $1$ and $8$, calculate the total number of possible telephone numbers that Chloe can have.
|
28
| |
There are 12 different-colored crayons in a box. How many ways can Karl select four crayons if the order in which he draws them out does not matter?
|
495
| |
Expand $(2z^2 + 5z - 6)(3z^3 - 2z + 1)$.
|
6z^5+15z^4-22z^3-8z^2+17z-6
| |
Let $m/n$, in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$. Find $m + n$.
|
634
|
$10^{99} = 2^{99}5^{99}$, so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$. Our probability is $\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}$, and $m + n = \boxed{634}$.
|
Two circles of radius \( r \) are externally tangent to each other and internally tangent to the ellipse defined by \( x^2 + 4y^2 = 5 \). The centers of the circles are located along the x-axis. Find the value of \( r \).
|
\frac{\sqrt{15}}{4}
| |
A box contains 5 white balls and 6 black balls. A ball is drawn out of the box at random. What is the probability that the ball is white?
|
\dfrac{5}{11}
| |
Given vectors $$\overrightarrow {m}=(\cos \frac {x}{3}, \sqrt {3}\cos \frac {x}{3})$$, $$\overrightarrow {n}=(\sin \frac {x}{3}, \cos \frac {x}{3})$$, and $$f(x)= \overrightarrow {m}\cdot \overrightarrow {n}$$.
(Ⅰ) Find the monotonic intervals of the function $f(x)$;
(Ⅱ) If the graph of $f(x)$ is first translated to the left by $\varphi$ ($\varphi>0$) units, and then, keeping the ordinate unchanged, the abscissa is scaled to $\frac {1}{3}$ of its original, resulting in the graph of the function $g(x)$. If $g(x)$ is an even function, find the minimum value of $\varphi$.
|
\frac {\pi}{4}
| |
Calculate the value of $\frac{17!}{7!10!}$.
|
408408
| |
The set \( A = \left\{ x \left\lvert\, x = \left\lfloor \frac{5k}{6} \right\rfloor \right., k \in \mathbb{Z}, 100 \leqslant k \leqslant 999 \right\} \), where \(\left\lfloor x \right\rfloor\) denotes the greatest integer less than or equal to \( x \). Determine the number of elements in set \( A \).
|
750
| |
The arithmetic mean, geometric mean, and harmonic mean of $a$, $b$, $c$ are $8$, $5$, $3$ respectively. What is the value of $a^2+b^2+c^2$?
|
326
| |
Calculate:<br/>$(1)4.7+\left(-2.5\right)-\left(-5.3\right)-7.5$;<br/>$(2)18+48\div \left(-2\right)^{2}-\left(-4\right)^{2}\times 5$;<br/>$(3)-1^{4}+\left(-2\right)^{2}\div 4\times [5-\left(-3\right)^{2}]$;<br/>$(4)(-19\frac{15}{16})×8$ (Solve using a simple method).
|
-159\frac{1}{2}
| |
Calculate: (1) $(\sqrt[3]{1.5} \times \sqrt[6]{12})^{2} + 8{1}^{0.75} - (-\frac{1}{4})^{-2} - 5 \times 0.12{5}^{0}$; (2) $\lg 25 + \lg 2 \cdot \lg 50 + (\lg 2)^{2} - e^{3\ln 2}$.
|
-6
| |
What is the sum of all the even integers between $200$ and $400$?
|
30100
| |
Evaluate $\sqrt{2 -\!\sqrt{2 - \!\sqrt{2 - \!\sqrt{2 - \cdots}}}}$.
|
1
| |
Alice has 4 sisters and 6 brothers. Given that Alice's sister Angela has S sisters and B brothers, calculate the product of S and B.
|
24
| |
Let \( n \) be a positive integer. When \( n > 100 \), what are the first two decimal places of the fractional part of \( \sqrt{n^2 + 3n + 1} \)?
|
49
| |
How many triangles with positive area are there whose vertices are points in the $xy$-plane whose coordinates are integers $(x, y)$ satisfying $1 \leq x \leq 5$ and $1 \leq y \leq 5$?
|
2164
| |
Given an isosceles triangle DEF with DE = DF = 5√3, a circle with radius 6 is tangent to DE at E and to DF at F. If the altitude from D to EF intersects the circle at its center, find the area of the circle that passes through vertices D, E, and F.
|
36\pi
| |
Petra has 49 blue beads and one red bead. How many beads must Petra remove so that 90% of her beads are blue?
|
40
| |
Let $\triangle PQR$ be a right triangle with angle $Q$ as the right angle. A circle with diameter $QR$ intersects side $PR$ at point $S$. If the area of $\triangle PQR$ is $192$ and $PR = 32$, what is the length of $QS$?
|
12
| |
Let $a < b < c < d < e$ be real numbers. We calculate all possible sums in pairs of these 5 numbers. Of these 10 sums, the three smaller ones are 32, 36, 37, while the two larger ones are 48 and 51. Determine all possible values that $e$ can take.
|
27.5
| |
Evaluate $|\omega^2 + 7\omega + 40|$ if $\omega = 4 + 3i$.
|
15\sqrt{34}
| |
How many integers \(n\) satisfy \((n+5)(n-9) \le 0\)?
|
15
| |
There are four more girls than boys in Ms. Raub's class of $28$ students. What is the ratio of number of girls to the number of boys in her class?
|
4 : 3
|
1. **Define Variables:**
Let $g$ be the number of girls in Ms. Raub's class. Since there are four more girls than boys, the number of boys, $b$, can be expressed as $b = g - 4$.
2. **Set Up the Equation:**
The total number of students in the class is given as $28$. Therefore, we can write the equation:
\[
g + b = 28
\]
Substituting $b = g - 4$ into the equation, we get:
\[
g + (g - 4) = 28
\]
3. **Solve for $g$:**
Simplify and solve the equation:
\[
2g - 4 = 28
\]
\[
2g = 28 + 4
\]
\[
2g = 32
\]
\[
g = \frac{32}{2} = 16
\]
4. **Find the Number of Boys:**
Substitute $g = 16$ back into the expression for $b$:
\[
b = g - 4 = 16 - 4 = 12
\]
5. **Calculate the Ratio:**
The ratio of the number of girls to the number of boys is:
\[
\frac{g}{b} = \frac{16}{12}
\]
Simplifying the ratio by dividing both the numerator and the denominator by their greatest common divisor, which is $4$, we get:
\[
\frac{16}{12} = \frac{16 \div 4}{12 \div 4} = \frac{4}{3}
\]
6. **Conclusion:**
The ratio of the number of girls to the number of boys in Ms. Raub's class is $\boxed{\textbf{(B)}~4:3}$.
|
In triangle \( ABC \), it is given that \( AC = 5\sqrt{2} \), \( BC = 5 \), and \( \angle BAC = 30^\circ \). What is the largest possible size in degrees of \( \angle ABC \)?
|
135
| |
Petrov writes down odd numbers: \(1, 3, 5, \ldots, 2013\), and Vasechkin writes down even numbers: \(2, 4, \ldots, 2012\). Each of them calculates the sum of all the digits of all their numbers and tells it to the star student Masha. Masha subtracts Vasechkin's result from Petrov's result. What is the outcome?
|
1007
| |
For all positive integers $x$, let \[f(x)=\begin{cases}1 & \text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}\] and define a sequence as follows: $x_1=x$ and $x_{n+1}=f(x_n)$ for all positive integers $n$. Let $d(x)$ be the smallest $n$ such that $x_n=1$. (For example, $d(100)=3$ and $d(87)=7$.) Let $m$ be the number of positive integers $x$ such that $d(x)=20$. Find the sum of the distinct prime factors of $m$.
|
511
|
We backcount the number of ways. Namely, we start at $x_{20} = 1$, which can only be reached if $x_{19} = 10$, and then we perform $18$ operations that either consist of $A: (-1)$ or $B: (\times 10)$. We represent these operations in a string format, starting with the operation that sends $f(x_{18}) = x_{19}$ and so forth downwards. There are $2^9$ ways to pick the first $9$ operations; however, not all $9$ of them may be $A: (-1)$ otherwise we return back to $x_{10} = 1$, contradicting our assumption that $20$ was the smallest value of $n$. Using complementary counting, we see that there are only $2^9 - 1$ ways.
Since we performed the operation $B: (\times 10)$ at least once in the first $9$ operations, it follows that $x_{10} \ge 20$, so that we no longer have to worry about reaching $1$ again.
However, we must also account for a sequence of $10$ or more $A: (-1)$s in a row, because that implies that at least one of those numbers was divisible by $10$, yet the $\frac{x}{10}$ was never used, contradiction. We must use complementary counting again by determining the number of strings of $A,B$s of length $18$ such that there are $10$ $A$s in a row. The first ten are not included since we already accounted for that scenario above, so our string of $10$ $A$s must be preceded by a $B$. There are no other restrictions on the remaining seven characters. Letting $\square$ to denote either of the functions, and $^{[k]}$ to indicate that the character appears $k$ times in a row, our bad strings can take the forms: \begin{align*} &\underbrace{BA^{[10]}}\square \square \square \square \square \square \square\\ &\square \underbrace{BA^{[10]}}\square \square \square \square \square \square \\ &\qquad \quad \cdots \quad \cdots \\ &\square \square \square \square \square \square \underbrace{BA^{[10]}} \square \\ &\square \square \square \square \square \square \square \underbrace{BA^{[10]}} \\ \end{align*}
There are $2^7$ ways to select the operations for the $7$ $\square$s, and $8$ places to place our $BA^{[10]}$ block. Thus, our answer is $2^9(2^9-1)-8\cdot 2^7 = 2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509$, and the answer is $\boxed{511}$.
Note: This solution is quick and most similar to the official solution; however, neither this nor the official solution prove that the final results of these inverted operations are all distinct. A more sophisticated argument, such as the one below, is needed to do so.
|
What is the sum of all positive integers $\nu$ for which $\mathop{\text{lcm}}[\nu,20]=60$?
|
126
| |
Let $1=d_1<d_2<d_3<\dots<d_k=n$ be the divisors of $n$ . Find all values of $n$ such that $n=d_2^2+d_3^3$ .
|
68
| |
How, without any measuring tools, can you measure 50 cm from a string that is $2/3$ meters long?
|
50
| |
( Reid Barton ) An animal with $n$ cells is a connected figure consisting of $n$ equal-sized square cells. ${}^1$ The figure below shows an 8-cell animal.
A dinosaur is an animal with at least 2007 cells. It is said to be primitive if its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.
Animals are also called polyominoes . They can be defined inductively . Two cells are adjacent if they share a complete edge . A single cell is an animal, and given an animal with cells, one with cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.
|
\[ 4 \cdot 2007 - 3 = 8025 \]
|
Solution 1
Let a $n$ -dino denote an animal with $n$ or more cells.
We show by induction that an $n$ -dino with $4n-2$ or more animal cells is not primitive. (Note: if it had more, we could just take off enough until it had $4n-2$ , which would have a partition, and then add the cells back on.)
Base Case: If $n=1$ , we have two cells, which are clearly not primitive.
Inductive Step: Assume any $4n-2$ cell animal can be partitioned into two or more $n$ -dinos.
For a given $(4n+2)$ -dino, take off any four cells (call them $w,\ x,\ y,\ z$ ) to get an animal with $4n-2$ cells.
This can be partitioned into two or more $n$ -dinos, let's call them $A$ and $B$ . This means that $A$ and $B$ are connected.
If both $A$ and $B$ are $(n+1)$ -dinos or if $w,\ x,\ y,\ z$ don't all attach to one of them, then we're done.
So assume $A$ has $n$ cells and thus $B$ has at least $3n-2$ cells, and that $w,\ x,\ y,\ z$ are added to $B$ . So $B$ has $3n+2$ cells total.
Let $C$ denote the cell of $B$ attached to $A$ . There are $3n+1$ cells on $B$ besides $C$ . Thus, of the three (or less) sides of $C$ not attached to $A$ , one of them must have $n+1$ cells by the pigeonhole principle . It then follows that we can add $A$ , $C$ , and the other two sides together to get an $(n+1)$ dino, and the side of $C$ that has $n+1$ cells is also an $n+1$ -dino, so we can partition the animal with $4n+2$ cells into two $(n+1)$ -dinos and we're done.
Thus, our answer is $4(2007) - 3 = 8025$ cells.
Example of a solution Attempting to partition solution into dinosaurs
Solution 2
For simplicity, let $k=2007$ and let $n$ be the number of squares . Let the centers of the squares be vertices , and connect any centers of adjacent squares with edges. Suppose we have some loops . Just remove an edge in the loop. We are still connected since you can go around the other way in the loop. Now we have no loops. Each vertex can have at most 4 edges coming out of it. For each point, assign it the quadruple : $(a,b,c,d)$ where $a$ , $b$ , $c$ , $d$ are the numbers of vertices on each branch, WLOG $a\ge b\ge c\ge d$ . Note $a+b+c+d=n-1$ .
Claim: If $n=4k-2$ , then we must be able to divide the animal into two dinosaurs.
Chose a vertex, $v$ , for which $a$ is minimal (i.e. out of all maximal elements in a quadruple, choose the one with the least maximal element). We have that $4a \ge a+b+c+d=4k-3$ , so $a\ge k$ . Hence we can just cut off that branch, that forms a dinosaur.
But suppose the remaining vertices do not make a dinosaur. Then we have $b+c+d+1\le k-1 \iff n-a\le k-1\iff a\ge 3k-1$ . Now move to the first point on the branch at $a$ . We have a new quadruple $p,\ q,\ r,\ b+c+d+1$ ) where $p+q+r=a-1\ge 3k-2$ .
Now consider the maximal element of that quadruple. We already have $b+c+d+1\le k-1$ . WLOG $p\ge q\ge r\ge 0$ , then $3p\ge p+q+r=a-1\ge 3k-2\implies p\ge k$ so $p>k-1=b+c+d+1$ , so $p$ is the maximal element of that quadruple.
Also $a-1=p+q+r\ge p+0+0$ , so $p<a$ . But that is a contradiction to the minimality of $a$ . Therefore, we must have that $b+c+d+1\ge k$ , so we have a partition of two dinosaurs.
Maximum: $n=4k-3$ .
Consider a cross with each branch having $k-1$ verticies. Clearly if we take partition $k$ vertices, we remove the center, and we are not connected.
So $k=2007$ : $4\cdot 2007-3=8025$ .
Solution 3 (Generalization)
Turn the dinosaur into a graph (cells are vertices , adjacent cells connected by an edge) and prove this result about graphs. A connected graph with $V$ vertices, where each vertex has degree less than or equal to $D$ , can be partitioned into connected components of sizes at least $\frac{V-1}{D}$ . So then in this special case, we have $D = 4$ , and so $V = 2006 \times 4+1$ (a possible configuration of this size that works consists of a center and 4 lines of cells each of size 2006 connected to the center). We next throw out all the geometry of this situation, so that we have a completely unconstrained graph. If we prove the above-mentioned result, we can put the geometry back in later by taking the connected components that our partition gives us, then filling back all edges that have to be there due to adjacent cells. This won't change any of the problem constraints, so we can legitimately do this.
Going, now, to the case of arbitrary graphs, we WOP on the number of edges. If we can remove any edge and still have a connected graph, then we have found a smaller graph that does not obey our theorem, a contradiction due to the minimality imposed by WOP. Therefore, the only case we have to worry about is when the graph is a tree. If it's a tree, we can root the tree and consider the size of subtrees. Pick the root such that the size of the largest subtree is minimized. This minimum must be at least $\frac{V-1}{D}$ , otherwise the sum of the size of the subtrees is smaller than the size of the graph, which is a contradiction. Also, it must be at most $\frac{V}{2}$ , or else pick the subtree of size greater than $\frac{V}{2}$ and you have decreased the size of the largest subtree if you root from that vertex instead, so you have some subtree with size between $\frac{V-1}{D}$ and $\frac V2$ . Cut the edge connecting the root to that subtree, and use that as your partition.
It is easy to see that these partitions satisfy the contention of our theorem, so we are done.
Solution 4
Let $s$ denote the minimum number of cells in a dinosaur; the number this year is $s = 2007$ .
Claim: The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1$ .
First, a primitive dinosaur can contain up to $4(s - 1) + 1$ cells. To see this, consider a dinosaur in the form of a cross consisting of a central cell and four arms with $s - 1$ cells apiece. No connected figure with at least $s$ cells can be removed without disconnecting the dinosaur.
The proof that no dinosaur with at least $4(s - 1) + 2$ cells is primitive relies on the following result.
Lemma. Let $D$ be a dinosaur having at least $4(s - 1) + 2$ cells, and let $R$ (red) and $B$ (black) be two complementary animals in $D$ , i.e. $R\cap B = \emptyset$ and $R\cup B = D$ . Suppose $|R|\leq s - 1$ . Then $R$ can be augmented to produce animals $\~{R}\subset R$ (Error compiling LaTeX. Unknown error_msg) and $\{B} = D\backslash\{R}$ (Error compiling LaTeX. Unknown error_msg) such that at least one of the following holds:
(i) $|\{R}|\geq s$ (Error compiling LaTeX. Unknown error_msg) and $|\~{B}|\geq s$ (Error compiling LaTeX. Unknown error_msg) ,
(ii) $|\{R}| = |R| + 1$ (Error compiling LaTeX. Unknown error_msg) ,
(iii) $|R| < |\{R}|\leq s - 1$ (Error compiling LaTeX. Unknown error_msg) .
Proof. If there is a black cell adjacent to $R$ that can be made red without disconnecting $B$ , then (ii) holds. Otherwise, there is a black cell $c$ adjacent to $R$ whose removal disconnects $B$ . Of the squares adjacent to $c$ , at least one is red, and at least one is black, otherwise $B$ would be disconnected. Then there are at most three resulting components $\mathcal{C}_1, \mathcal{C}_2, \mathcal{C}_3$ of $B$ after the removal of $c$ . Without loss of generality, $\mathcal{C}_3$ is the largest of the remaining components. (Note that $\mathcal{C}_1$ or $\mathcal{C}_2$ may be empty.) Now $\mathcal{C}_3$ has at least $\lceil (3s - 2)/3\rceil = s$ cells. Let $\{B} = \mathcal{C}_3$ (Error compiling LaTeX. Unknown error_msg) . Then $|\{R}| = |R| + |\mathcal{C}_1| + |\mathcal{C}_2| + 1$ (Error compiling LaTeX. Unknown error_msg) . If $|\{B}|\leq 3s - 2$ (Error compiling LaTeX. Unknown error_msg) , then $|\{R}|\geq s$ (Error compiling LaTeX. Unknown error_msg) and (i) holds. If $|\{B}|\geq 3s - 1$ (Error compiling LaTeX. Unknown error_msg) then either (ii) or (iii) holds, depending on whether $|\{R}|\geq s$ (Error compiling LaTeX. Unknown error_msg) or not. $\blacksquare$
Starting with $|R| = 1$ , repeatedly apply the Lemma. Because in alternatives (ii) and (iii) $|R|$ increases but remains less than $s$ , alternative (i) eventually must occur. This shows that no dinosaur with at least $4(s - 1) + 2$ cells is primitive.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
|
Two dice are tossed. What is the probability that the sum is greater than three?
|
\frac{11}{12}
| |
A sequence consists of $2010$ terms. Each term after the first is 1 larger than the previous term. The sum of the $2010$ terms is $5307$. When every second term is added up, starting with the first term and ending with the second last term, what is the sum?
|
2151
| |
A circle is inscribed in a right triangle. The point of tangency divides the hypotenuse into two segments of lengths $6 \mathrm{~cm}$ and $7 \mathrm{~cm}$. Calculate the area of the triangle.
|
42
| |
Find the projection of the vector $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$ onto the vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}.$
|
\begin{pmatrix} 4 \\ 0 \end{pmatrix}
| |
Two numbers $a$ and $b$ with $0 \leq a \leq 1$ and $0 \leq b \leq 1$ are chosen at random. The number $c$ is defined by $c=2a+2b$. The numbers $a, b$ and $c$ are each rounded to the nearest integer to give $A, B$ and $C$, respectively. What is the probability that $2A+2B=C$?
|
\frac{7}{16}
|
By definition, if $0 \leq a<\frac{1}{2}$, then $A=0$ and if $\frac{1}{2} \leq a \leq 1$, then $A=1$. Similarly, if $0 \leq b<\frac{1}{2}$, then $B=0$ and if $\frac{1}{2} \leq b \leq 1$, then $B=1$. We keep track of our information on a set of axes, labelled $a$ and $b$. The area of the region of possible pairs $(a, b)$ is 1, since the region is a square with side length 1. Next, we determine the sets of points where $C=2A+2B$ and calculate the combined area of these regions. We consider the four sub-regions case by case. In each case, we will encounter lines of the form $a+b=Z$ for some number $Z$. We can rewrite these equations as $b=-a+Z$ which shows that this equation is the equation of the line with slope -1 and $b$-intercept $Z$. Since the slope is -1, the $a$-intercept is also $Z$. Case 1: $A=0$ and $B=0$ For $C$ to equal $2A+2B$, we need $C=0$. Since $C$ is obtained by rounding $c$, then we need $0 \leq c<\frac{1}{2}$. Since $c=2a+2b$ by definition, this is true when $0 \leq 2a+2b<\frac{1}{2}$ or $0 \leq a+b<\frac{1}{4}$. This is the set of points in this subregion above the line $a+b=0$ and below the line $a+b=\frac{1}{4}$. Case 2: $A=0$ and $B=1$ For $C$ to equal $2A+2B$, we need $C=2$. Since $C$ is obtained by rounding $c$, then we need $\frac{3}{2} \leq c<\frac{5}{2}$. Since $c=2a+2b$ by definition, this is true when $\frac{3}{2} \leq 2a+2b<\frac{5}{2}$ or $\frac{3}{4} \leq a+b<\frac{5}{4}$. This is the set of points in this subregion above the line $a+b=\frac{3}{4}$ and below the line $a+b=\frac{5}{4}$. Case 3: $A=1$ and $B=0$ For $C$ to equal $2A+2B$, we need $C=2$. Since $C$ is obtained by rounding $c$, then we need $\frac{3}{2} \leq c<\frac{5}{2}$. Since $c=2a+2b$ by definition, this is true when $\frac{3}{2} \leq 2a+2b<\frac{5}{2}$ or $\frac{3}{4} \leq a+b<\frac{5}{4}$. This is the set of points in this subregion above the line $a+b=\frac{3}{4}$ and below the line $a+b=\frac{5}{4}$. Case 4: $A=1$ and $B=1$ For $C$ to equal $2A+2B$, we need $C=4$. Since $C$ is obtained by rounding $c$, then we need $\frac{7}{2} \leq c<\frac{9}{2}$. Since $c=2a+2b$ by definition, this is true when $\frac{7}{2} \leq 2a+2b<\frac{9}{2}$ or $\frac{7}{4} \leq a+b<\frac{9}{4}$. This is the set of points in this subregion above the line $a+b=\frac{7}{4}$ and below the line $a+b=\frac{9}{4}$. We shade the appropriate set of points in each of the subregions. The shaded regions are the regions of points $(a, b)$ where $2A+2B=C$. To determine the required probability, we calculate the combined area of these regions and divide by the total area of the set of all possible points $(a, b)$. This total area is 1, so the probability will actually be equal to the combined area of the shaded regions. The region from Case 1 is a triangle with height $\frac{1}{4}$ and base $\frac{1}{4}$, so has area $\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{32}$. The region from Case 4 is also a triangle with height $\frac{1}{4}$ and base $\frac{1}{4}$. This is because the line $a+b=\frac{7}{4}$ intersects the top side of the square (the line $b=1$) when $a=\frac{3}{4}$ and the right side of the square (the line $a=1$) when $b=\frac{3}{4}$. The regions from Case 2 and Case 3 have identical shapes and so have the same area. We calculate the area of the region from Case 2 by subtracting the unshaded area from the area of the entire subregion (which is $\frac{1}{4}$). Each unshaded portion of this subregion is a triangle with height $\frac{1}{4}$ and base $\frac{1}{4}$. We can confirm this by calculating points of intersection as in Case 4. Therefore, the area of the shaded region in Case 2 is $\frac{1}{4}-2 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$. Therefore, the combined area of the shaded regions is $\frac{1}{32}+\frac{1}{32}+\frac{3}{16}+\frac{3}{16}=\frac{14}{32}=\frac{7}{16}$. Thus, the required probability is $\frac{7}{16}$.
|
Given non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ satisfy $|\overrightarrow{a}|=2|\overrightarrow{b}|$, and $(\overrightarrow{a}-\overrightarrow{b})\bot \overrightarrow{b}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is ______.
|
\frac{\pi}{3}
| |
In a right triangle $DEF$ with $\angle D = 90^\circ$, we have $DE = 8$ and $DF = 15$. Find $\cos F$.
|
\frac{15}{17}
| |
From the numbers \\(1, 2, \ldots, 100\\) totaling \\(100\\) numbers, three numbers \\(x, y, z\\) are chosen in sequence. The probability that these three numbers satisfy \\(x+z=2y\\) is __________.
|
\dfrac{1}{198}
|
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