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Gnoma [55]
2 years ago
15

An Native American tepee is conical tent. Find the number of skins needed to cover a teepee 10ft in diameter and 12ft high. each

skin covers 15 sq. feet (use pi=3.14)​
Mathematics
1 answer:
nirvana33 [79]2 years ago
6 0

Answer:

19 skins

Step-by-step explanation:

Total surface area of the cone = πr(r+l)

r is the radius = 10ft/2 = 5ft

l is the slant height

l^2 = 12^2 + 5^2

l^2 = 144+25

l^2 = 169

l = 13ft

Get the surface area

TSA = 3.14(5)(5+13)

TSA = 15.7*18

TSA = 282.6 ft^2

If each skin covers 15 sq. feet

Number of skin needed = 282.6 /15

Number of skin needed = 18.84

Number of skin needed is approximately 19 skins

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According to Okun's law, if the unemployment rate goes from 5% to 3%, what will be the effect on the GDP?
olga2289 [7]

Answer:

D. It will increase by 1%.

Step-by-step explanation:

Given

u_1 = 5\% --- initial rate

u_2 = 3\% --- final rate

Required

The effect on the GDP

To calculate this, we make use of:

\frac{\triangle Y}{Y} = u_1 - 2\triangle u

This gives:

\frac{\triangle Y}{Y} = 5\% - 2(5\% - 3\%)

\frac{\triangle Y}{Y} = 5\% - 2(2\%)

\frac{\triangle Y}{Y} = 5\% - 4\%

\frac{\triangle Y}{Y} = 1\%

<em>This implies that the GDP will increase by 1%</em>

5 0
3 years ago
1. The national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15. The dean of a college wants to kno
Nat2105 [25]

Answer:

We conclude that the mean IQ of her students is different from the national average.

Step-by-step explanation:

We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.

The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113

Let, Null Hypothesis, H_0 : \mu = 100 {means that the mean IQ of her students is same as of national average}

Alternate Hypothesis, H_1 : \mu\neq 100  {means that the mean IQ of her students is different from the national average}

(a) The test statistics that will be used here is One sample z-test statistics;

               T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ N(0,1)

where, Xbar = sample mean score = 113

              s = population standard deviation = 15

             n = sample of students = 144

So, test statistics = \frac{113-100}{\frac{15}{\sqrt{144} } }

                             = 10.4

Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean IQ of her students is different from the national average.

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