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3 years ago

Find the perimeter.

Mathematics

2 answers:

serg [7]3 years ago

7 0

Answer:

b) 70 feet

Step-by-step explanation:

let me know if you want an explanation please :)

olga55 [171]3 years ago

3 0

Answer:

the answer will be a 375

Step-by-step explanation:

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How do you find the area of a cone?

tamaranim1 [39]

Surface area of cone = πr(r+√(h2+r2))

where r is the radius of the circular base

h is the height of coneOr

Surface area of cone =πr(r+L)

where L is the slant height of the cone

4 0

3 years ago

Read 2 more answers

Which choices are equivalent to the expression below? Check all that apply.<br>3√6

Ulleksa [173]

Answer:

$\sqrt 54$

$\sqrt 9.\sqrt 6$

$\sqrt {27}.\sqrt 2$

Step-by-step explanation:

$3 \sqrt{6} = \sqrt{ {3}^{2} \times 6 } = \sqrt{9 \times 6} = \sqrt{54} \\$

Also,

$\sqrt 9.\sqrt 6$

$\sqrt {27}.\sqrt 2$

6 0

3 years ago

Read 2 more answers

In a triangle ΔABC, the median AM is extended beyond point M to point N so that MN = AM. Find the distances NB and NC, if AB = c

Troyanec [42]

Answer:

it's just opposite image

nb=C, nc=b

8 0

3 years ago

20 PTS!!! NEED HELP NOW PLEASE!!!<br><br> If 6(2x-3)=72+8x

Helga [31]

6(2x-3) = 72+8x
12x-18=72+8x
12=72+8x
4x=90
4/4 90/4
x=22.5
x=22.5

3 0

4 years ago

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma

nata0808 [166]

Answer:

a) $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c) $P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d) $P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e) $P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

$X \sim N(110000,40000)$

Where $\mu=110000$ and $\sigma=40000$

If we select a sample size of n =35 the standard error is given by:

$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

$P(X > 112000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

$P(X > 100000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

$P(100000$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

8 0

4 years ago