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3 years ago

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Bernie deposited $30,000 in a savings account with simple interest. One year later, he had earned $3,300 in interest. What was t

he interest rate?

1 answer:

Paraphin [41]3 years ago

7 0

Answer:

The interest rate is 11%.

Step-by-step explanation:

Let's use PRT/100 = Interest to solve this problem

<u>We know that:</u>

Formula: Principle x rate x time/100 = Interest

<u>Work:</u>

30,000 x r x 1/100 = 3,300
=> 300 x r x 1 = 3,300
=> 300r = 3,300
=> r = 3,300/300
=> r = 11

Hence, the interest rate is 11%.

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For his birthday, Tyrone's parents gave him $7,790.00 which they put into a savings account that earns 15% interest compounded m

torisob [31]

Answer:

5 years and 5 months

Step-by-step explanation:

<u />

<u>Compound Interest Formula</u>

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

where:

A = final amount
P = principal amount
r = interest rate (in decimal form)
n = number of times interest applied per time period
t = number of time periods elapsed

Given:

A = $17,474.00
P = $7,790.00
r = 15% = 0.15
n = 12
t = number of years

Substitute the given values into the formula and solve for t:

$\implies \sf 17474=7790\left(1+\dfrac{0.15}{12}\right)^{12t}$

$\implies \sf \dfrac{17474}{7790}=\left(1.0125}\right)^{12t}$

$\implies \sf \ln\left(\dfrac{17474}{7790}\right)=\ln \left(1.0125}\right)^{12t}$

$\implies \sf \ln\left(\dfrac{17474}{7790}\right)=12t \ln \left(1.0125}\right)$

$\implies \sf t=\dfrac{\ln\left(\frac{17474}{7790}\right)}{12 \ln \left(1.0125}\right)}$

$\implies \sf t=5.419413037...\:years$

Therefore, the money was in the account for 5 years and 5 months (to the nearest month).

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2 years ago

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artcher [175]

Answer:

$91p^{1+3q^{2r}}q^{2r^3}$

Step-by-step explanation:

$(-7pq^2r^3) * (-13p^3q^2r)$

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= $91p^{1+3q^{2r}}q^{2r^3}$

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2 years ago

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3 years ago

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Answer:

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