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Vinil7 [7]
3 years ago
14

Ken invests 6,000 into two accounts one account earns 8% interest and the other earns 12% interest after one year his total inte

rest earned from the two accounts totals $580 how much did ken invest in the 8%
Mathematics
1 answer:
Len [333]3 years ago
7 0

Answer: ken invested $3500 in the 8% account.

Step-by-step explanation:

Let x represent the amount invested in the account earning 8% interest.

Let y represent the amount invested in the account earning 12% interest.

Ken invests 6,000 into two accounts one account earns 8% interest and the other earns 12% interest. This means that

x + y = 6000

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

Considering the account earning 8%

I = (x × 8 × 1)/100 = 0.08x

Considering the account earning 12%

I = (y × 12 × 1)/100 = 0.12y

After one year, his total interest earned from the two accounts totaled $580 . This means that

0.08x + 0.12y = 580 - - - - - - - - - - 1

Substituting x = 6000 - y into equation 1, it becomes

0.08(6000 - y) + 0.12y = 580

480 - 0.08y + 0.12y = 580

- 0.08y + 0.12y = 580 - 480

0.04y = 100

y = 100/0.04

y = 2500

x = 6000 - y = 6000 - 2500

x = 3500

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Answer:

z=-0.994

And if we solve for a we got

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The answer rounded would be 2.6 years approximately  

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

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Solution to the problem

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P(X   (b)

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P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.994

And if we solve for a we got

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sp2606 [1]
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