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Leona [35]
3 years ago
7

How do you go about doing this surface area problem?

Mathematics
1 answer:
hammer [34]3 years ago
5 0
Break it down into 2-Dimensional shapes. Then add the areas together.

From the picture you can see;
front & back rectangles are 2*(4 x 8) = 64 m²
2 side rectangles are 2*(4 x 12) = 56 m²
2 triangular front & back pieces are (1/2)*8*3 = 12 m²
2 roof rectangles are 2*(5 x 12) = 120 m²

total Surface area = 64 m² + 56 m² + 12 m² + 120 m²
= 252 m²

For the volume; break it up into 3-dimenssional shapes and add the volumes together.

From the picture you can see;
Rectangular box volume is 4 x 8 x 12 = 384 m³
Triangular roof volume is area of front triangle multiplied through the length. (1/2)*8*3* 12 = 144 m³

Total volume = 384 m³ + 144 m³
= 528 m³
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Answer:

12.5 minutes of commercials

Step-by-step explanation:

9:30 - 7:00 = 2 and a half hours on air = 150 minutes.

150 / 30 = 5

5 x 2.5 = 12.5

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Step-by-step explanation:

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Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the
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Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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