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3 years ago

9

an electronic store had a 20% off sale. If a stareo originally cost $125, how much did the stareo cost during the sale?

1 answer:

spayn [35]3 years ago

6 0

Answer:

100

Step-by-step explanation:

You multiply 125 by 0.80. This is because 125 is the original amount nd to find out how much you have to pay, you would subtract 20% from 100% to get 80%. Now, you know that you have to pay for 80% of the stereo. In order to multiply the two numbers together, you have to convert 80% to a decimal which is 0.80. Lastly, you multiply to get 1oo. Therefore, the stereo costs $100.00 during the sale.

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How long will it take for $2500 to double if it is invested at 6% annual interest compounded 6 times a year? Enter exact calcula

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Answer:

Part 1) $t=11.610\ years$

Part 2) $t=11.552\ years$

Step-by-step explanation:

Part 1) we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$A=\$5,000\\P=\$2,500\\ r=6\%=6/100=0.06\\n=6$

substitute in the formula above

$5,000=2,500(1+\frac{0.06}{6})^{6t}$

$2=(1.01)^{6t}$

Apply log both sides

$log(2)=log[(1.01)^{6t}]$

$log(2)=(6t)log(1.01)$

solve for t

$t=log(2)/[6log(1.01)]$

$t=11.610\ years$

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$A=\$5,000\\P=\$2,500\\ r=6\%=6/100=0.06$

substitute in the formula above

$5,000=2,500(e)^{0.06t}$

$2=(e)^{0.06t}$

Apply ln both sides

$ln(2)=ln[(e)^{0.06t}]$

$ln(2)=(0.06t)ln(e)$

$ln(2)=(0.06t)$

$t=ln(2)/(0.06)$

$t=11.552\ years$

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