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Vlad1618 [11]
3 years ago
15

Do I have it correct

Mathematics
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

yes  sir i do belive so

Step-by-step explanation:

You might be interested in
I'm not sure how to find angle 5 and 6​
Arte-miy333 [17]

Answer:

angle 5 = 17 degrees

angle 6 = 129 degrees

Step-by-step explanation:

Start by finding angle 5 by doing 90+39+x+34=180

You get 17 degrees for angle 5

Then, find angle 6 by doing 17+34+x=180, which is 129 degrees.

6 0
4 years ago
GUYS I NEED YOUR HELP THIS IS SUPER HARD PLZ SOMEONE HELP ME I BEEN WORKING ON THIS QUESTION FOR A WEEK AND I STILL DONT KNOW TH
sasho [114]

Answer:

1

Step-by-step explanation:

8 0
3 years ago
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
Find a function that describes the sequence. Use y for the term in the sequence and n for its place in the list. Then use the fu
sammy [17]
B. y=n+20; 25,26,27 because if you plug in 1 into the equation it equals 21. 2+20=22, etc. so to find the next three terms of the equation plug in 5: y=5+20=25; plug in 6: y=6+20=26; and plug in 7: y=7+20=27
6 0
3 years ago
Can someone help I’m confused
exis [7]

Answer:

D and E

Step-by-step explanation:

Let's double check:

D. 9 ÷ 3 = 3

E. 4 × 9 = 36

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Have a great summer :)

3 0
3 years ago
Read 2 more answers
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