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azamat
3 years ago
7

Find the volume of the prism.CmCm

Mathematics
1 answer:
spin [16.1K]3 years ago
7 0
V or rectangular prism=LWH
doesn't matter which, multiplication is commutative

3/2 times 5/4 times 7/3=(3*5*7)/(2*4*3)=105/24=4 and 9/25 cm^3
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Answer:

(a)\ \bar x_1 - \bar x_2 = 2.0

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Solving (b): 90% confidence interval

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c = 90\%

c = 0.90

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1 - \alpha = c

1 - \alpha = 0.90

\alpha = 0.10

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z_{\alpha/2} = z_{0.05} =1.645

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2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.645 * \sqrt{0.0735+0.2571}

2.0 \± 1.645 * \sqrt{0.3306}

2.0 \± 0.9458

Split

(2.0 - 0.9458) \to (2.0 + 0.9458)

(1.0542) \to (2.9458)

Hence, the 90% confidence interval is:

CI =(1.0542,2.9458)

Solving (c): 95% confidence interval

We have:

c = 95\%

c = 0.95

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.95

\alpha = 0.05

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.05/2}

z_{\alpha/2} = z_{0.025}

The z score is:

z_{\alpha/2} = z_{0.025} =1.96

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.96 * \sqrt{0.0735+0.2571}

2.0 \± 1.96* \sqrt{0.3306}

2.0 \± 1.1270

Split

(2.0 - 1.1270) \to (2.0 + 1.1270)

(0.8730) \to (2.1270)

Hence, the 95% confidence interval is:

CI = (0.8730,2.1270)

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