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Luba_88 [7]
3 years ago
9

Please help will mark brainliest and thank you

Mathematics
2 answers:
Brilliant_brown [7]3 years ago
7 0
The answer is 5.66 because you take the value of 8 and take a square root of 2 and that equals your a and b values.
Law Incorporation [45]3 years ago
3 0

Answer:

40

Step-by-step explanation:

You might be interested in
Assume the average height for American women is 64 inches with a standard deviation of 2 inches. A sorority on campus wonders if
satela [25.4K]

Answer:

a) s = 0.4

b) Z = 2.5

c) 99th percentile.

d) The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Assume the average height for American women is 64 inches with a standard deviation of 2 inches

This means that \mu = 64, \sigma = 2

A) Calculate the standard error for the distribution of means.

Sample of 25 means that n = 25, so s = \frac{2}{\sqrt{25}} = 0.4.

B) Calculate the z statistic for the sorority group.

Sample mean of 65 means that X = 65.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{0.4}

Z = \frac{65 - 64}{0.4}

Z = 2.5

C) What is the approximate percentile for this sample? Enter as a whole number.

Z = 2.5 has a p-value of 0.9938, so 0.99*100 = 99th percentile.

D) If the sorority actually had 36 members (still with an average of 65 inches), would you expect the percentile value to increase or decrease? why?

The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.

3 0
2 years ago
Can you help me with Math
Nonamiya [84]

Answer:

42+106=x

148=x

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
If 25% of an amount is 10% What is<br>75%​
Katena32 [7]

Answer:

30%

i guess

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Halp me halp me <br> halp me halp me
atroni [7]

Answer:

8.72

Step-by-step explanation:

sqrt of 76 is 8.717 which rounded to the nearest hundereth is 8.72

8 0
2 years ago
Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based
Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for \mathrm{sinc}\,x about x=0:

\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}

Then the Taylor series for

f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt

is obtained by integrating the series above:

f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}

We have f(0)=0, so C=0 and so

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}

which converges by the ratio test if the following limit is less than 1:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}

Like in the linked problem, the limit is 0 so the series for f(x) converges everywhere.

7 0
3 years ago
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