Mathematics, 16.01.2020 18:31 josierednour285

Find the time required for an investment of 5000 dollars to grow to 6800 dollars at an interest rate of 7.5 percent per year, compounded quarterly. round your answer to two decimal places.

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4.14 years required for an investment of 5000 dollars to grow to 6800 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

Step-by-step explanation:

We are asked to find time it will take for an investment of 5000 dollars to grow to 6800 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

We will use compound interest formula to solve our given problem.

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

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years

Upon substituting our given values in above formula, we will get:

$$6800=5000((1+\frac{0.075}{4} )^{4t}\\$

$\frac{6800}{5000} =\frac{5000}{5000} ((1+\frac{0.075}{4}) ^{4t} )$

$1.36=(1+0.01865)^{4t}$

$1.36=(1.01875)^{4t}$

Take natural log of both sides:

$ln(1.36)=ln (1.01875)^{4t}$

ln(1.36)=4t .(1.01875)

$0.3074846997479606=4t. 0.01857638557293\\t=\frac{0.3074846997479606}{4. 0.01857638557293}$

t=4.13811258574

t ≈ 4.14 (up to two decimal places)

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Answer from: jasminebailey829

4.14 years.

Step-by-step explanation:

We are asked to find time it will take for an investment of 5000 dollars to grow to 6800 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

We will use compound interest formula to solve our given problem.

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

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years

$7.5\%=\frac{7.5}{100}=0.075$

Upon substituting our given values in above formula, we will get:

$\$6800=\$5000(1+\frac{0.075}{4})^{4*t}$

$\frac{\$6800}{\$5000}=\frac{\$5000(1+\frac{0.075}{4})^{4t}}{\$5000}$

$1.36=(1+0.01875)^{4t}$

$1.36=(1.01875)^{4t}$

Take natural log of both sides:

$\text{ln}(1.36)=\text{ln}((1.01875)^{4t})$

$\text{ln}(1.36)=4t\cdot \text{ln}(1.01875)$

$0.3074846997479606=4t\cdot 0.0185763855729354$

$0.3074846997479606=t\cdot 0.0743055422917416$

$t=\frac{0.3074846997479606}{0.0743055422917416}$

t=4.13811258574

$t\approx 4.14$

Therefore, it will take approximately 4.14 years for an investment of 5000 dollars to grow to 6800 dollars.

Answer from: Quest

answer:

125 yards

explanation:

75+50=125. that is my answer

Answer from: Quest

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote

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Find the time required for an investment of 5000 dollars to grow to 6800 dollars at an interest rate...