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kobusy [5.1K]
4 years ago
12

An account earned interest of 3% per year. The beginning balance was $150. The equation t=log1.03 (E/150) represents the situati

on, where t is the time in years and E is the ending balance. If the account was open for 8 years, what was the ending balance?
Mathematics
1 answer:
barxatty [35]4 years ago
3 0

Answer:

the ending balance is <em>$190</em>.

Step-by-step explanation:

Step 1:

The given information is:

Interest rate = 3% per year

Beginning balance = 150 $

Equation:

t = log₁.₀₃ (E / 150)

Step 2:

Solve for E:

8 = log₁.₀₃ (E / 150)

E / 150 = (1.03)⁸

E = (1.03)⁸ (150)

<em>E = 190 $</em>

<em></em>

Therefore, the ending balance is <em>$190</em>.

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7(1 - y) = -3(y-2)<br> What is the answer
Fynjy0 [20]

Answer:

\huge y =  \frac{1}{4}

Step-by-step explanation:

7(1 - y) = -3(y-2)

<u>Expand the terms in the bracket</u>

That's

7 - 7y =  - 3y + 6

<u>Add 3y to both sides of the equation</u>

7 - 7y + 3y = 3y - 3y + 6 \\ 7 - 4y = 6

Subtract 7 from both sides by the equation

7 - 7  - 4y = 6 - 7 \\  - 4y =  - 1

<u>Divide both sides by - 4</u>

\frac{ - 4y}{ - 4}  =  \frac{ - 1}{ - 4}  \\

We have the final answer as

y =  \frac{1}{4}  \\

Hope this helps you

4 0
3 years ago
Which function forms an arithmetic sequence when x = 1, 2, 3, ...?
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The first one is arithmetic sequence  as each term decreases by same amount 
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5 0
3 years ago
Read 2 more answers
Yet another question.
inna [77]

Answer:

t= 3 years

Step-by-step explanation:

So far we have 2 useful relationshipsP_{t}= P_{o} e^{r*t} and P_{t}=2P_{0}

Now, clearing t

P_{t}= P_{o} e^{r*t} \\\frac{ P_{t}}{P_{o}} = e^{r*t}\\\frac{2 P_{0}}{P_{o}} = e^{r*t}\\2 = e^{r*t}

I apply logarithm

log(2)=r*t\\ t=\frac{log(2)}{r}\\t=\frac{0.3}{0.1} \\ t=3 years

Done

4 0
4 years ago
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Answer:

12

Step-by-step explanation:

4x3=12

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3 years ago
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Answer:
1

Explanation:
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