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4 years ago

Kelly is using a compass and straightedge to perform the Geometric construction below.

Mathematics

2 answers:

coldgirl [10]4 years ago

7 0

Answer: A line parallel to RS through Q

Step-by-step explanation:

KATRIN_1 [288]4 years ago

3 0

Answer:

A.) a line perpendicular to RS through S

Explanation:

a compass has perpendicular lines, not parallel.

and the image shows perpendicular lines

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Find the smallest n for which the numbers 1^2, 2^2, 3^2, 4^2, ....n^2 can be split into two groups with the same sum.

yan [13]

Answer:

yjfh

Step-by-step explanation:

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3 years ago

What fraction is equivalent to 0.7 repeated

pishuonlain [190]

.7 repeating would make it 7/9

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4 years ago

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please help with this. I got the first choice when I picked wrong so i would like help on a second attempt. thanks

SpyIntel [72]

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Step-by-step explanation:

im pretty sure that's the right answer!! let me know if I'm right

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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 $Y$ ,

$\mathbb P(Y$

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3 years ago

Can some help me pls find the angle measure for each one

Marianna [84]

Answer:

1 = 140 2 = 70

Step-by-step explanation

1 is 140 because it is the same as 140 and 2 is 70 because it is the same angle as 70.

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3 years ago

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