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12345 [234]
4 years ago
11

Write an equation. Twice the length of a segment x is 9ft

Mathematics
1 answer:
dimaraw [331]4 years ago
4 0
2x=9 
x meaning the segment
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A student researcher compares the ages of cars owned by students and cars owned by faculty at a local state college. A sample of
diamong [38]

Answer:

The point estimate for the true difference between the population means is 0.13.

The 90% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -0.35 years and 0.61 years.

Step-by-step explanation:

To solve this question, before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When we subtract two normal variables, the mean is the subtraction of the means while the standard deviation is the square root of the sum of the variances.

A sample of 263 cars owned by students had an average age of 7.25 years. The population standard deviation for cars owned by students is 3.77 years.

This means that:

\mu_s = 7.25, \sigma_s = 3.77, n = 263, s_s = \frac{3.77}{\sqrt{263}} = 0.2325

A sample of 291 cars owned by faculty had an average age of 7.12 years. The population standard deviation for cars owned by faculty is 2.99 years.

This means that:

\mu_f = 7.12, \sigma_f = 2.99, n = 291, s_f = \frac{2.99}{\sqrt{291}} = 0.1753

Difference between the true mean ages for cars owned by students and faculty.

Distribution s - f. So

\mu = \mu_s - \mu_f = 7.25 - 7.12 = 0.13

This is also the point estimate for the true difference between the population means.

s = \sqrt{s_s^2+s_f^2} = \sqrt{0.2325^2+0.1753^2} = 0.2912

90% confidence interval for the difference:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = zs = 1.645*0.2912 = 0.48

The lower end of the interval is the sample mean subtracted by M. So it is 0.13 - 0.48 = -0.35 years

The upper end of the interval is the sample mean added to M. So it is 0.13 + 0.48 = 0.61 years.

The 90% confidence interval for the difference between the true mean ages for cars owned by students and faculty is between -0.35 years and 0.61 years.

5 0
3 years ago
Free points everyone<br><br><img src="https://tex.z-dn.net/?f=10%20%2B%2062%20%5Cdiv%202" id="TexFormula1" title="10 + 62 \div 2
Brut [27]
41 is the right answer
8 0
3 years ago
Read 2 more answers
What is the estimated probability that at least two of the puppies will be
goldfiish [28.3K]
Bubble B. 8/10 80%
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8 0
3 years ago
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Jada has a stand in the marketplace where she sells ground cumin. Her weekly expenses are $300, and she sells her cumin at a fix
Irina18 [472]
Profit = revenue (income) - expenses

600 = 120x - 300
600 + 300 = 120x
900 = 120x
900/120 = x
7.50 = x ....this is how much 1 kg costs

so to cover her weekly expenses of 300, she would have to sell :
300/7.5 = 40 kg of cumin <===
8 0
3 years ago
Find all real solutions<br> x + 5 = 14 - <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20" id="TexFormula1" title="
rosijanka [135]
Multiply all terms by 2 to get rid of fractions
2x+10=28-x

Add x to both sides
3x+10=28

Subtract 10 from both sides
3x=18

Divide both sides by 3
x=6

Final answer: x=6
8 0
4 years ago
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