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Anika [276]
3 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:
hjlf3 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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A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates
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Answer:

The probability is 0.971032

Step-by-step explanation:

The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.

The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

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So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:

P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)

We can also calculated that as:

P(x ≥ 5) = 1 - P(x ≤ 4)

Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)

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Finally, P(x ≥ 5) is:

P(x ≥ 5) = 1 - 0.028968

P(x ≥ 5) = 0.971032

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