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3 years ago

WILL MARK BRAINLIEST AND THANK YOU!!!! EXTRA POINTS!!!

Mathematics

2 answers:

evablogger [386]3 years ago

8 0

Answer:

The first is a Postive slope. the Second is a Zero slope

shepuryov [24]3 years ago

7 0

Answer:

The first is a Postive slope. the Second is a Zero slope

Step-by-step explanation:

Becuase the first gradully increases. meaning positive

and the 2nd is a zero because there is no slope. it's a stright line. level ground. no up hill or down hill

Hope this helps!

You might be interested in

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

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.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0

3 years ago

4. [5 pts] Describe how and why the formula for permutations differs from the formula for combinations.

ivann1987 [24]

Answer:

They are different because of the order in the permutation matters. In combination, the order doesn't matter. In other words in a permutation 123 and 132 are different but in a combination are the same group (they have the same digits 1,2, and 3).

Step-by-step explanation:

The formula of the permutation is $P(n,r)=\frac{n!}{(n-r)!}$ , when you are performing a permutation you pick r objects from a total of n, for the first pick you can choose from n, but for the second you have n-1, and this continues to your pick number r in which you will choose from n-r+1, and the total of permutation is the multiplicación of this number of choices for each pick, like this:

$n(n-1)(n-2)...(n-r+1)$

If $n!=n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)$ and $(n-r)!=(n-r)(n-r-1)(n-r-2)...(1)$

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}$

The factor equals above and under cancel each other.

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}\\\frac{n!}{(n-r)!}=n(n-1)(n-2)...(n-r+1)$

In combination, the order of the element isn't important, so from the total of permutation you have to eliminate the ones with the same objects with different order and counting just once each group, when choosing r objects the total of permutation for a single group of r objects is: $r(r-1)(r-2)...(1)=r!$ . If you divide the total of permutations of n taking r by r! you get the combinations (where the order is not important). The formula of the combination is $C(n,r)=\frac{n!}{r!(n-r)!}$ .