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Roman55 [17]
3 years ago
5

Diana invested $3000 in a savings account for 3 years. She earned $450 in interest over that time period. What interest rate did

she earn? Use the formula I=Prt to find your answer, where I is interest, P is principal, r is rate and t is time. Enter your solution in decimal form rounded to the nearest hundredth. For example, if your solution is 12%, you would enter 0.12.
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
3 0
\bf ~~~~~~ \textit{Simple Interest Earned}
\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\to &\$450\\
P=\textit{original amount deposited}\to& \$3000\\
r=rate\to r\%\to \frac{r}{100}\\
t=years\to &3
\end{cases}
\\\\\\
450=(3000)(r)(3)\implies \cfrac{450}{(3000)(3)}=r\implies \cfrac{1}{20}=r
\\\\\\
0.05=r\implies r\%=0.05\cdot 100\implies r=\stackrel{\%}{5}
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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
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The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

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