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Sonbull [250]
2 years ago
11

Esteban bought a new stove for $986 on his credit card. He used the stove for eleven years before replacing it. The stove cost h

im an average of $0. 14 per day in electricity. Esteban had preventive maintenance done on the stove, costing $24. 25 each year for the eleven years. Esteban’s credit card has an APR of 9. 26%, compounded monthly. He paid off his balance by making identical monthly payments for five years. Sales tax in Esteban’s area is 8. 22%. Assuming that Esteban made no other purchases or payments with his credit card, what was the lifetime total cost of the stove? (Assume that two of the years Esteban had the stove were leap years, and round all dollar values to the nearest cent. ) a. $2,534. 57 b. $2,166. 53 c. $2,234. 23 d. $2,064. 53.
Mathematics
1 answer:
vivado [14]2 years ago
4 0

Answer:

b) $2,166

Step-by-step explanation:

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The weight of people in a small town in Missouri is known to be normally distributed with a mean of 186 pounds and a standard de
OleMash [197]

Answer:

the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

Step-by-step explanation:

The summary of the given statistical data set are:

Sample Mean = 186

Standard deviation = 29

Maximum capacity 3,417 pounds or 17 persons.

sample size = 17

population mean =3417

The objective is to determine the probability  that a random sample of 17 persons will exceed the weight limit of 3,417 pounds

In order to do that;

Let assume X to be the random variable that follows the normal distribution;

where;

Mean \mu = 186 × 17 = 3162

Standard deviation = 29* \sqrt{17}

Standard deviation = 119.57

P(X>3417) = P(\dfrac{X - \mu}{\sigma}>\dfrac{X - \mu}{\sigma})

P(X>3417) = P(\dfrac{3417 - \mu}{\sigma}>\dfrac{3417 - 3162}{119.57})

P(X>3417) = P(Z>\dfrac{255}{119.57})

P(X>3417) = P(Z>2.133)

P(X>3417) =1- 0.9834

P(X>3417) =0.0166

Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

5 0
3 years ago
if set A contains 4 distinct numbers and set B contains 7 distinct numbers, how many elements are in ( A u B)
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7 0
3 years ago
WILL MARK BRAINILEST!!!!!
kirill [66]
To solve this we are going to use the compound interest formula with periodic deposits: A=P(1+ \frac{r}{n} )^{nt}+P_{d}( \frac{(1+ \frac{r}{n})^{nt}-1 }{ \frac{r}{n} } )(1+ \frac{r}{n} )
where 
A is the final amount after t years 
P is the initial investment 
P_{d} is the periodic deposits
r is the interest rate in decimal form 
n is the number of times the interest is compounded per year 
t is the time in years

Since he is going to save from 27 years old  until 65 years old, t=65-27=38. We know that hes is opening his IRA with $0, so P=0; We also know that he is going to invest $200 at the beginning of each month, so P_{d}=200. To convert the interest rate to decimal form, we are going to divide it by 100: r= \frac{2.65}{100} =0.0265, and since the interest is compounded monthly, n=12. Lets replace all the values in our formula to find A:
A=P(1+ \frac{r}{n} )^{nt}+P_{d}( \frac{(1+ \frac{r}{n})^{nt}-1 }{ \frac{r}{n} } )(1+ \frac{r}{n} )
A=0(1+ \frac{0.0265}{12} )^{(12)(38)}+200( \frac{(1+ \frac{0.0265}{12})^{(12)(38)}-1 }{ \frac{0.0265}{12} } )(1+ \frac{0.0265}{12} )
A=157419.04

We can conclude that Rick will have $157,419.04 in his IRA account by the time when he retires.

4 0
3 years ago
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