Josh has $500 in his savings account. He deposits $25 into his savings account each month. He purchased home theater equipment f
or $1,600. The store he bought the equipment from offered him a $1,600 interest-free loan for 24 months. He pays off $80 of the loan every month from his checking account (not his savings account). How many months will it take for Josh to have enough money in his savings account to pay off the entire outstanding amount for his purchase? Round your answer to the nearest whole number. A. 8 months B. 10 months C. 20 months D. 24 months
1 answer:
Answer:
B. 10 months
Step-by-step explanation:
The balance on the loan will be ...
b = 1600 - 80t . . . . . . where t is the number of months of payments
The balance in the savings account will be ...
s = 500 + 25t
The savings account balance will be at least as much as the loan balance when ...
s ≥ b
500 +25t ≥ 1600 -80t . . . substitute the account balance expressions
105t ≥ 1100 . . . . . . . . . . . . add 80t -500
t ≥ 1100/105 ≈ 10.48 ≈ 10
It will take Josh 10 months to have enough savings to pay the loan in full.
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<em>Comment on rounding</em>
IMO, it makes no sense to round down, as Josh will NOT have enough in 10 months. He will have enough after he makes one more payment of $80. At 10 months, the loan balance is $50 more than the savings balance. It will be 11 months before there is enough savings to pay off the loan.
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Answer:
a) The 99% confidence interval would be given by (346.708;453.292)
b) The 99% confidence interval would be given by (338.445;461.555)
Step-by-step explanation:
1) 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".
represent the sample mean for the sample
population mean (variable of interest)
represent the population standard deviation
n=15 represent the sample size
2) Part a
The confidence interval for the mean is given by the following formula:
(1)
Since the Confidence is 0.99 or 99%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 99% confidence interval would be given by (346.708;453.292)
3) Part b
For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.
The confidence interval for the mean is given by the following formula:
(1)
We need to find the degrees of freedom first df=n-1=15-1=14
Since the Confidence is 0.99 or 99%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 99% confidence interval would be given by (338.445;461.555)