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oksian1 [2.3K]
2 years ago
15

PLEASE HELP! Find the surface area of the right triangular prism.

Mathematics
1 answer:
slega [8]2 years ago
4 0

Answer:

  1200 in²

Step-by-step explanation:

The surface area of the prism is the sum of the lateral area and the areas of the bases.

__

<h3>lateral area</h3>

The lateral area is the total area of the three rectangular faces. It can be computed from ...

  LA = Ph

where P is the perimeter of the triangular base, and h is the height of the prism (the distance between triangular bases).

The perimeter is the sum of the edge lengths of the base:

  P = 30 in + 17 in + 17 in = 64 in

Then the lateral area is ...

  LA = (64 in)(15 in) = 960 in²

__

<h3>base area</h3>

The two bases are congruent triangles, so have a total area of ...

  A = 2(1/2bh) = bh

where b is the base of the triangle, and h is its height.

  A = (30 in)(8 in) = 240 in²

__

<h3>total surface area</h3>

The sum of lateral area and base area is ...

  total area = (960 in²) +(240 in²) = 1200 in²

The surface area of the triangular prism is 1200 square inches.

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Step-by-step explanation:

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Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy
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Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

df = n-1=75-1 =74

The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

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There are 30 children eating ice cream. 2/5 of the children have chocolate. 1/3 of them have vanilla. The rest have strawberry.
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Answer:

28/2

Step-by-step explanation:

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