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andriy [413]
3 years ago
7

Suppose you begin a job with an annual salary of $32,900. Each year you are assured of a 5.5% raise. What its the total amount t

hat you can earn in 15 years? A) $34,815 B) $51,751 C) $737,245 D) $1,682,920
Mathematics
1 answer:
olga2289 [7]3 years ago
4 0

Answer: the total amount that you can earn in 15 years is $737245. Option C

Step-by-step explanation:

You receive an annual salary of $32,900 and each year, you are assured of a 5.5% raise. Assuming there was no raise, you get 100% of your previous salary each year. With a raise of 5.5%, you will get 100 + 5.5 = 105.5% of your previous salary for each year. This is a geometric progression and we want to determine the sum of 15 terms(15 years).

The formula for the sum of terms in a geometric progression is

Sn = [a(r^n - 1)]/ r - 1

Sn = sum of n terms

a = the first term

n = number of terms

r = common ratio

From the information given,

a = 32900

n = 15

r = 105.5/100 = 1.055

S15 = [32900(1.055^15 - 1)] / 1.055 - 1

S15 = [32900(2.23247649 - 1)] / 0.055

S15 = 32900 × 1.23247649) / 0.055

S15 = 737245.0277

S15 = $737245

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You have $32 to spend at the music store. Each cassette tape costs $5 and each cd costs $9. Write a linea inequality that repres
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Answer:

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Step-by-step explanation:

Let x be the number of cassette tapes you purchase.

Let y be the number of CDs you purchase.

The problem is you only have $32 to spend.

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3 years ago
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Answer:

10.08-1.64\frac{0.1}{\sqrt{6}}=10.013    

10.08+1.64\frac{0.1}{\sqrt{6}}=10.147    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

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\sigma =0.1 represent the population standard deviation

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Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=10.08

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

Now we have everything in order to replace into formula (1):

10.08-1.64\frac{0.1}{\sqrt{6}}=10.013    

10.08+1.64\frac{0.1}{\sqrt{6}}=10.147    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

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Answer:

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10/3 / 13/5

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