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3 years ago

Please answer this correctly

Mathematics

2 answers:

Nastasia [14]3 years ago

5 0

Answer:

864 ft^2.

Step-by-step explanation:

Area of the base = 16^2 = 256.

Area of the triangular sides = 4 * 1/2 * 16 * 19

Total area = 256 + 608

= 864 ft^2.

ZanzabumX [31]3 years ago

5 0

Answer:

1728ft^2

Step-by-step explanation:

Surface area = 2(wl+hl+hw)

w = width

l = length

h = height

2((16 x 16) + (19+16) + (19 + 16))

= 1728ft^2

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1 and 2 are supplementary.If m1 = (3x-17) and m2= (5x+21) find the value of x

torisob [31]

Answer:

Step-by-step explanation:

m1 + m2 = 180

3x - 17 + 5x + 21 = 180

3x + 5x + 21 - 17 = 180

8x + 4 = 180

8x = 180 - 4

8x = 176

x = 176 / 8

x = 22

7 0

3 years ago

Given the function f(x)=x^2+2x+1, find a f(−7)

tatuchka [14]

Answer:

Step-by-step explanation:

f(x)=x^2+2x+1

Let x = -7

f(-7)=(-7)^2+2*-7+1

= 49 -14+1

= 36

3 0

3 years ago

Please help me

olganol [36]

Ok so I'm pretty sure you multiply 22.45×5 then multiply 25.45×7 then add your answer

3 0

4 years ago

A delivery truck traveled 133 miles in 3.5 hours. What was the average speed of the delivery truck in miles per hour

Korolek [52]

The average speed of the delivery truck would be 136.5

8 0

4 years ago

Read 2 more answers

Small Sample:

agasfer [191]

Answer:

The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).

A sample size of 8852 is needed.

Step-by-step explanation:

First question:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Only 2 of the 11 companies were planning to increase their workforce.

This means that $n = 11, \pi = \frac{2}{11} = 0.1818$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 - 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.033$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 + 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.331$

The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).

Second question:

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A poll taken in July 2010 estimates this proportion to be 0.36.

This means that $\pi = 0.36$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.01?

This is n for which M = 0.01. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.96\sqrt{\frac{0.36*0.64}{n}}$

$0.01\sqrt{n} = 1.96\sqrt{0.36*0.64}$

$\sqrt{n} = \frac{1.96\sqrt{0.36*0.64}}{0.01}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.36*0.64}}{0.01})^2$

$n = 8851.04$

Rounding up(as a sample of 8851 will have a margin of error slightly over 0.01):

A sample size of 8852 is needed.

4 0

3 years ago