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2 years ago

Can someone help please?

Mathematics

1 answer:

Aliun [14]2 years ago

5 0

Answer:

L=7.295

Step-by-step explanation:

I can't type in a square root symbol so here's the sequence:

divide both sides by 2 Pi

square both sides

now you have:

.227973=L/32

multiply both sides by 32

L=7.295

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Solve the given equation by completing the square.

klasskru [66]

Answer:

Step-by-step explanation:

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3 years ago

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

adelina 88 [10]

Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

The mean age is $\= x = 20 \ years$

The standard deviation is $\sigma = 4 \ years$

Generally the degree of freedom for this data set is mathematically represented as

$df = n - 1$

substituting values

$df = 64 - 1$

$df = 63$

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$\alpha = 100 - 90$

$\alpha =$ 10 %

$\alpha = 0.10$

Now $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

$19.15< L < 20.85$

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Answer:

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