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3 years ago

Karen is 6 years older than her sister Michelle, and Michelle is 2 years younger than their brother David. If the sum of their a

ges is 62 , how old is Michelle how do i get the answer

Mathematics

1 answer:

ale4655 [162]3 years ago

6 0

$K\ \rightarrow\ the\ Karen's\ age\\M\ \rightarrow\ the\ Michelles\ age \\D\ \rightarrow\ the\ David's\ age\\\\K=M+6\\M=D-2\ \ \ \Rightarrow\ \ \ D=M+2\\K+M+D=62\\\\(M+6)+M+(M+2)=62\\3M+8=62\\3M=54\ /:3\\M=18\ \ \ \Rightarrow\ \ \ K=18+6=24\ \ \ and\ \ \ D=18+2=20\\\\Ans.\ Michelle\ is\ 18,\ David\ is\ 20,\ Karen\ is\ 24.$

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You could simplify this work by factoring "3" out of all four terms, as follows:

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3 years ago

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Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 96% confidence interval for the population mean, $\mu$ is ;

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree

of freedom are -2.114 & 2.114 with P = 2%}

P(-2.114 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.114) = 0.96

P( $-2.114 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $55-2.114 \times {\frac{3.4}{\sqrt{101} } }$ , $55+2.114 \times {\frac{3.4}{\sqrt{101} } }$ ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

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