Please help for Algebra II

lord [1]

Answer:-9765625

Step-by-step explanation: multiply the exponents first and then include the base. So, you would have 5x2=10, then se -5 to the 10th power (use a calculator).

7 0

3 years ago

Read 2 more answers

Use the diagram to solve for the three variables so that the quadrilateral will be a rhombus. Show all of your work for full cre

satela [25.4K]

Answer:

x = 4
y = 4
z = 7

Step-by-step explanation:

Each triangle is a right triangle, and all are congruent. So ...

90 -m∠CAB = (1/2)m∠CDA

90 -(4x +33) = (1/2)(12x +34)

57 -4x = 6x +17 . . . . . . . eliminate parentheses

40 = 10x . . . . . . . . . . . . .add 4x-17

4 = x

We also have ...

2m∠CAD = m∠BCD

2(3z +28) = 10z +28

6z +56 = 10z +28 . . . . . eliminate parentheses

28 = 4z . . . . . . . . . . . . . .subtract 6z+28

7 = z . . . . . . . . . . . . . . . . divide by 4

Since BC = CD ...

x = y = 4

The values of the three variables are ...

x = 4, y = 4, z = 7.

3 0

3 years ago

Please help with this

ArbitrLikvidat [17]

Answer:

$c = 10$

Step-by-step explanation:

This can be solved with Pythagreon Thereom.

${a}^{2} + {b}^{2} = {c}^{2}$

Follow the steps below. A and B are the heaight an base, C is they hypotonuese.

${6}^{2} + {8}^{2} = {c}^{2} \\ 36 + 64 = {c}^{2} \\ 100 = {c}^{2} \\ 10 = c$

7 0

3 years ago

A set of normally distributed data has a mean of 485 and a standard deviation of 11.6. Find the probability of randomly selectin

Zinaida [17]

Let $X_i$ denote a data point taken from the distribution, where $1\le i\le40$ , and let $Y$ denote the average.

You want to find

$\mathbb P\left(\displaystyle\frac1{40}\sum_{i=1}^{40}X_i$

First, let's recall a few things. The PDF of a normal distribution with mean $\mu$ and variance $\sigma^2$ is

$f(x;\mu,\sigma^2)=\displaystyle\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$

Each of the $X_i$ are presumably independently selected, so they are i.i.d. random variables.

The MGF of a normal distribution is

$M_X(t)=\mathbb E(e^{tX})$
$M_X(t)=\displaystyle\int_{-\infty}^\infty e^{tx}f_X(x)\,\mathrm dx$
$M_X(t)=\exp\left(\mu t+\dfrac12\sigma^2t^2\right)$

The MGF of a linear combination of i.i.d. random variables is

$M_{c_1X_1+\cdots+c_nX_n}=M_{X_1}(c_1t)\times\cdots\times M_{X_n}(c_nt)=\displaystyle\prod_{i=1}^nM_{X_i}(c_it)$

In this case, each $c_i=\dfrac1{40}$ . This product of MGFs reduces to an MGF of a normal distribution because the $X_i$ are i.i.d..

$M_Y(t)=\displaystyle\prod_{i=1}^{40}M_{X_i}(t)=\exp\left(\mu\left(\dfrac t{40}\right)+\frac12\sigma^2\left(\dfrac t{40}\right)^2\right)\times\cdots\times\exp\left(\mu\left(\dfrac t{40}\right)+\frac12\sigma^2\left(\dfrac t{40}\right)^2\right)$
$M_Y(t)=\exp\left(\mu t+\dfrac12\left(\dfrac\sigma{\sqrt{40}}\right)^2t^2\right)$

which is indeed the MGF of a normal distribution with mean $\displaystyle\mu\sum_{i=1}^{40}\frac1{40}=\mu$ and variance $\sigma^2\displaystyle\sum_{i=1}^{40}\left(\frac1{40}\right)^2=\frac{\sigma^2}{40}$

So, the PDF of $Y$ , given that $\mu=485$ and $\sigma=11.6$ , is

$f_Y(y)=\displaystyle\frac1{\frac{11.6}{\sqrt{40}}\sqrt{2\pi}}\exp\left(-\frac{(y-485)^2}{2\left(\frac{11.6}{\sqrt{40}}\right)^2}\right)$

Now,

$\mathbb P(Y$

or, using the CDF of $Y$ ,

$\mathbb P(Y$

5 0

3 years ago

Calculate 4,235 divided by 11

love history [14]

Answer:385

Step-by-step explanation:

5 0

2 years ago