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Mrac [35]
1 year ago
10

Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4%

Mathematics
1 answer:
bogdanovich [222]1 year ago
4 0

Mr. Jackson invested $800 at 6% per year and $ 2400  at 4 % per year

<h3><u>Solution:</u></h3>

Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4% per year.

Let the sum invested be ‘a’ and ‘3a’ at 6% per year and 4 % per year respectively

Also, his annual return totaled $144

We can form following equation on the basis of question:-

\begin{array}{l}{\text { Then, } \frac{a \times 6 \times 1}{100}+\frac{3 a \times 4 \times 1}{100}=\$ 144} \\\\ {\frac{6 a}{100}+\frac{12 a}{100}=144} \\\\ {\frac{6 a+12 a}{100}=144} \\\\ {\frac{18 a}{100}=144} \\\\ {18 a=14400} \\\\ {a=14400 \div 18}\end{array}

a = $800

The amount of money invested at 6% = a = 800

The amount of money invested at 4 % = 3a = 3(800) = 2400

So, the amount of money invested at 6% is $800 and the amount of money invested at 4% is $ 2400

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Step-by-step explanation:

Since f(0) = f(5) = f(8) = 0, we have f(x) = Ax(x - 5)(x - 8), where A is a real constant.

We know that f(10) = 17.

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Answer:

The probability that the next failure will not occur before 30 months have elapsed is 0.0454

Step-by-step explanation:

Using Poisson distribution  where

t= number of units of time

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λ= average number of occurrences per unit of time

P(x;λt) = e raise to power (-λt)  multiplied by λtˣ divided by x!

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Answer:

The number of different combinations of three students that are possible is 35.

Step-by-step explanation:

Given that three out of seven students in the cafeteria line are chosen to answer a survey question.

The number of different combinations of three students that are possible is given as:

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Therefore, the number of different combinations of three students that are possible is 35.

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