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Anika [276]
2 years ago
13

Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest

and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000?
Mathematics
1 answer:
hjlf2 years ago
8 0

Answer:

In the 8% option , $5,000 should be invested

In the 12% option, $15,000 should be invested

Step-by-step explanation:

Let’s start with the total return.

Since the total return expected is 11%, the amount expected to be returned will be;

11/100 * 20,000 = $2,200

Now, let the amount invested at 8% interest be $x while the amount invested at 12% interest be $y

Mathematically;

x + y = 20,000 ••••••(i)

Now let’s work with the interest part;

On the 8% part, amount of interest expected is 8/100 * x

on the 12% part, amount of interest expected is 12/100 * x

8% of $x + 12% of $y = 2,200

Writing this fully mathematically, we have;

(8/100 * x) + (12/100 * y) = 2,200

8x/100 + 12y/100 = 2,200

Multiply through by 100

8x + 12y = 220,000 ••••••(ii)

Now we have two equations to solve simultaneously;

x + y = 20,000

8x + 12y = 220,000

From i, x = 20,000-y

Substitute this into ii

8(20,000-y) + 12y = 220,000

160,000-8y + 12y = 220,000

4y = 220,000 - 160,000

4y = 60,000

y = 60,000/4

y = $15,000

x = 20,000 -y

x = 20,000 - 15,000

x = $5,000

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Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
2 years ago
Erica earned a total of $50,450 last year from her two jobs. The amount she earned from her job at the store was $1,250 more tha
Mariana [72]

Erica earned a total of $50,450 last year from her two jobs. The amount she earned from her job at the store was $1,250 more than four times the amount she earned from her job at the college.

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amount she earned at the store = 4 * amount earned at college + 1250

= 4x + 1250

Amount earned at college + amount earned at store = 50450

x + 4x + 1250 = 50450

5x + 1250 = 50450

Subtract 1250 from both sides

5x = 49200 (divide by 5)

x = 9840

she earn $9840 from her job at the college

5 0
2 years ago
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erma4kov [3.2K]
Negative dividing by negative gives positive,
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6 0
2 years ago
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Answer:

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Step-by-step explanation:

n - 2y = 3y- n/m,

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--------

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