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2 years ago

Can you help me solve if x=6214

Mathematics

1 answer:

Verdich [7]2 years ago

4 0

Pretty big number for x, lol.

Anyways, the hypotenuse is 2×6214= 12428.

The short leg is 6214.

The long leg is 6214×sqrt 3= 10763 rounded. Good luck!

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17. Jing has $800 to deposit into two different savings accounts.

Olin [163]

Answer:

$864.88

Step-by-step explanation:

(1. $600 + $49.50 = $649.50

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$649.50 + $215.38 = $864.88

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7 0

3 years ago

Help me with this pleaae

marissa [1.9K]

Answer:

(D) 30pi inches

Step-by-step explanation:

First, we use the given volume, the given height, and the formula of the volume of a cylinder to find the radius of the base. Then we use the radius of the base to find the circumference of the base.

$volume = \pi r^2 h$

$volume = 6750 \pi~in.^3$

We set the formula equal to the volume and replace h with 30 in.

$\pi r^2 h = 6750 \pi~in.^3$

$\pi r^2 \times 30 ~in. = 6750 \pi~in.^3$

Divide both sides by 30pi in.

$r^2 = 225 ~in.^2$

Take the square root of each side.

$\sqrt{r^2} = \sqrt{225 ~in.^2}$

$r = 15~in.$

The radius of the base is 15 in.

Now we use the radius of the base and the formula of the circumference of a circle to find the answer.

$circumference = 2 \pi r$

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6 0

3 years ago

What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$