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Anika [276]
4 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:
hjlf4 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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Step-by-step explanation:

Simplifying

2m + -6n + 5n + 3m = 0


Reorder the terms:

2m + 3m + -6n + 5n = 0


Combine like terms: 2m + 3m = 5m

5m + -6n + 5n = 0


Combine like terms: -6n + 5n = -1n

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5m + -1n = 0


Solving for variable 'm'.


Move all terms containing m to the left, all other terms to the right.


Add 'n' to each side of the equation.

5m + -1n + n = 0 + n


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The 1996 gss asked, "if the husband in a family wants children, but the wife decides that she does not want any children, is it
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The confidence interval would be
0.80\pm0.03.

We use the formula
p\pm z*\sigma_p, where

\sigma_p=\sqrt{\frac{p(1-p)}{N}}, with p being the sample proportion and N being the sample size.

First we find the z-score associated with this level of confidence:
Convert 95% to a decimal:  95/100 = 0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
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Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 1.96.

Since 578/720 said yes, this gives us p=0.80:
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This gives us
0.80\pm1.96(0.015)=0.80\pm 0.03
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