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3 years ago

What is the length of the hypotenuse of the triangle?

Mathematics

2 answers:

saw5 [17]3 years ago

8 0

Answer:

37.7

Step-by-step explanation:

Hypotenuse is the longest side

Eduardwww [97]3 years ago

3 0

Answer:

answer is b

Step-by-step explanation:

we need a picture of the triangle

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Solve. 4/5x-33/4x=21/2 A. -1 9/50 B. -50/59 C.34/75 D.19/50

vaieri [72.5K]

I think the answer is choice D, but don't trust me I didn't do the math.

6 0

3 years ago

Mary ran 2 miles in about 23 minutes. If she continued at the same pace, how long will it take her to run 10 miles?

ZanzabumX [31]

Answer:115 minutes

Step-by-step explanation:

Set you problem 2/23 = 10/x

X represents the number of minutes you are trying to figure out. Cross multiply to get 2x=230. Decide both sides by 2 to get x=115

8 0

3 years ago

The sum of two positive integers, x and y, is not more than 40. The difference of the two integers is at least 20. Chaneece choo

Butoxors [25]

Answer:

The value of x lies between 20 and 40 and value of y lies between -20 and 10

Step-by-step explanation:

We are given that

$y\leq 40-x$ ....(1)

$y\leq x-20$ ...(2)

First we convert inequality equation into equality equation to find the solution of the given system of inequality equation.

Therefore, we can write as

$y=-x+40$ ...(3)

$y=x-20$ ...(4)

Adding equation (3) and (4) we get

$2y=20$

$y=10$

Substitute y=10 in equation (3) we get

$10=40-x$

$x=40-10=30$

(30,10) is the intersect point of two equation.

Put x=0 in equation (3)

$y=40$

Substitute y=0 in equation (3)

$x=40$

Substitute x=0 in equation (4)

$y=-20$

Substitute y=0 in equation (4)

$x=20$

Substitute x=0 and y=40 and in equation (1)

$40\leq 40$

Hence, the equation is true.Therefore, the shaded region below the line.

Substitute x=0 and y=-20 in equation (2)

$-20\leq -20$

The equation is true. Hence, the shaded region is below the line.

Hence, the value of x lies between 20 and 40 and value of y lies between -20 and 10.

5 0

3 years ago

Read 2 more answers

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

Please help me thank u so much

alex41 [277]

Answer:

28/5= 5 3/5 15/7= 2 1/7 21/4=5 1/4

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers