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3 years ago

5,-5/2,5/4,-5/8 what is the 5th term

Mathematics

2 answers:

Bumek [7]3 years ago

8 0

Answer: your answer would be 5/16th :)

valentina_108 [34]3 years ago

3 0

Answer:

5/16

Step-by-step explanation:

The common ratio is -5/2 / 5 = -1/2

Xn = 5 × (-1/2)^(n-1)

X5 = 5 × (-1/2)^(5-1)

X5 = 5 × (-1/2)^(4) = 5/16

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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

2 years ago

Sara has 23 more colored pencils than Dan. If s represents the number of colored pencils Sara has and d represents the number of

kotykmax [81]

The answer is 31 your welcome!!!!!!!!!!!!

6 0

3 years ago

A migrating bird flies 390 miles in 15 hours. How many miles does it fly in 4 hours?

viva [34]

Answer:

104

Step-by-step explanation:

390/15=26

26 times 4 is 104

6 0

3 years ago

Choose an American adult at random. The probability that you choose a woman is 0.52 . 0.52. The probability that the person you

Nikitich [7]

Answer:

The probability that the person you choose is either a woman or has never been married (or both) is therefore about 0.66

Step-by-step explanation:

Given Data:

Probability of choosing a women = P(W) = 0.52

Probability that chosen person has never married = P(M) = 0.26

Probability of choosing a women that has never married = P(W and M) = 0.11

We need to find the probability that the person you choose is either a woman or has never been married (or both). The addition rule of probabilities of two events is given as:

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) is the probability of occurrence of either A or B or both. P(A and B) is the probability of occurrence of A and B at the same time.

Re-writing the addition formula for given case, we get:

P(W or M) = P(W) + P(M) - P(W and M)

Substituting the given values results in:

P(W or M) = 0.52 + 0.26 - 0.11

P(W or M) = 0.67

Therefore, the probability that the person you choose is either a woman or has never been married (or both) is 0.67.

Note: There is either typo in given options or the questions. The question I found on Google has 0.25 as the probability that the person you choose has never married ( i.e. P(M) = 0.25 ).

So, in this case the correct answer would be option (b) 0.66

4 0

3 years ago

Suppose that Marlie had decided to apply for a private loan rather than a federal loan. She has

34kurt

Using simple interest, and considering a loan of $5,000, it is found that:

a) Her monthly payment is of $74.17.

b) She will pay back $8,900.

c) The total interest amount is of $3,900.

The amount of money after t years in simple interest is modeled by:

$A(t) = A(0)(1 + rt)$