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sattari [20]
3 years ago
9

3 years ago a father was 3 times as old as his son, in five years time he will be twice as old as his son, what will be the sum

of their years in four years time​
Mathematics
2 answers:
ivolga24 [154]3 years ago
5 0

Answer: 46 years

Step-by-step explanation:

Let the father's age be x and the son's age be y, then 3 years ago:

Father = x - 3

son     = y - 3

Then , from the first statement :

x - 3 = 3 ( y - 3 )

x - 3 = 3y - 9

x     = 3y - 9 + 3

x     = 3y - 6 .......................................... equation 1

In five years time

father = x + 5

son = y + 5

Then , from the second statement

x + 5 = 2 ( y + 5 )

x + 5 = 2y + 10

x       = 2y + 10 - 5

x       = 2y + 5 ........................ equation 2

Equating equation 1 and 2 , we have

3y -6 = 2y + 5

add 6 to both sides

3y = 2y + 5 + 6

subtract 2y from both sides

3y - 2y = 11

y = 11

substitute y = 11 into equation 1 to find the value of x

x = 3y - 6

x = 3(11) - 6

x = 33 - 6

x = 27

This means that the father is presently 27 years and the son is presently 11 years.

In four years time

father = 27 + 4 = 31

son = 11 + 4      = 15

sum of their ages in four years time will be

31 + 15 = 46 years

docker41 [41]3 years ago
4 0

Answer:

46 years

Step-by-step explanation:

Answer:

Step-by-step explanation:

In solving this, we'll make some keen assumptions to solve the question.

We'll represent the father's age with "a"

And represent the sons age as "b"

Lets interpret the question.

3 years ago, (means in the past), we'll subtract.

Therefore:

Father's age = a - 3

Sons age = b - 3

The sentence said, " 3 years ago, father was 3 times the sons age*. Interpreting that becomes:

a - 3 = 3 times (b - 3)

Simplifying that gives :

a - 3 = 3(b - 3)

a - 3 = 3b - 9

a = 3b - 9 + 3

a = 3b - 6 ( FIRST EXPRESSION )

The next line says: In five years time (that's in the future, hence we add) he'll be twice his sons age.

Therefore I become:

Father : a + 5

Son : b + 5

a + 5 = 2(b + 5)

a + 5 = 2b + 10

a = 2b + 5 ( SECOND EXPRESSION)

Equating the first and second equation to solve for "a" - the father's age

3b - 6 = 2b + 5

b = 5 + 6

b = 11 years (Sons age)

Substitute b = 1 in the first expression.

a = 3b - 6

a = 3(11) - 6

a = 33 - 6

a = 27 years (Father's age)

Let resolve, the last sentence:

What will be the sum of their ages in 4 years time. (SINCE IT'S IJ THE FUTURE, WE ADD)

Father's age in four years time: 27 + 4 = 31

Son's age in four years time: 11 + 4 = 15.

The sum of their ages in same four years time becomes:

31 + 15 (years)

46 years

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The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g
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Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

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This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 \\ n \geq 30. Mathematically

\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean \\ \mu = 0 and a population standard deviation \\ \sigma = 1.

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

From the question, we know that

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  • The population standard deviation is \\ \sigma = 14 WPM

We also know the size of the sample for this case: \\ n = 137 sixth graders.

We need to estimate the probability that a sample mean being greater than \\ \overline{X} = 89.87 WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}

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\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}

\\ Z = 1.5634 \approx 1.56

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\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

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