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Kay [80]
3 years ago
7

The sum of the ages of the father and son is 44 years .After 8 years the age of the father will be twice the age of the son . Fi

nd their present ages.
​
Mathematics
1 answer:
Vadim26 [7]3 years ago
8 0

Step-by-step explanation:

Let the present age of father = x years

and the present age of son = y years

according to the first condition

x + y =44 ........(i)

according to the second condition

x+8 = 2(y+8)

or x+ 8 = 2y + 16

or x- 2y = 8 ............(ii)

solve the equation (i) & (ii) we get

x + y =44

x- 2y = 8

••••••••••••••••••••••••

3y = 36

or y = 12

put the value of y in equation (i), we get

x + 12 =44

or x = 32

so the present age of father = x =32 years

and the present age of son = y = 12 years.

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Scores on a test are normally distributed with a mean of 81.2 and a standard deviation of 3.6. What is the probability of a rand
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<u>Answer:</u>

The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.

<u>Solution:</u>

Given, Scores on a test are normally distributed with a mean of 81.2  

And a standard deviation of 3.6.  

We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?

For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6  

Now probability of getting more than 88.4 = 1 - area of z – score of 88.4

\mathrm{Now}, \mathrm{z}-\mathrm{score}=\frac{88.4-\mathrm{mean}}{\text {standard deviation}}=\frac{88.4-81.2}{3.6}=\frac{7.2}{3.6}=2

So, probability of getting more than 88.4 = 1 – area of z- score(2)

= 1 – 0.9772 [using z table values]

= 0.0228.

Now probability of getting more than 77.6 = 1 - area of z – score of 77.6

\mathrm{Now}, \mathrm{z}-\text { score }=\frac{77.6-\text { mean }}{\text { standard deviation }}=\frac{77.6-81.2}{3.6}=\frac{-3.6}{3.6}=-1

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