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3 years ago

Find the volume of this figure.

Mathematics

1 answer:

Bogdan [553]3 years ago

4 0

Answer:

The volume of the figure is $320\ units^{3}$

Step-by-step explanation:

we know that

The figure is a pyramid

The volume of a pyramid is equal to

$V=(1/3)BH$

where

B is the area of the rectangular base of pyramid

H is the height of pyramid

The area of the base B is equal to

$B=AB*AC$

we have that

$AB=8\ units$ ---> see the graph

$AC=10\ units$ ---> see the graph

substitute

$B=8*10=80\ units^{2}$

we have

$H=EF=12\ units$ ---> see the graph

<em>Find the volume</em>

$V=(1/3)(80*12)=320\ units^{3}$

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a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

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Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

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c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

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$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

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