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4 years ago

Michael is 4 times as old as Brandon and is also 27 years older than Brandon.

2 answers:

4 0

Let's represent Brandon's age by x and Michael is 4 times as old as Brandon so 4x andx+ 27. The equation is
4x=x+27 Therefore Brandon is 9 years old.
4x-x=x-x+27 4x=4*9=36
3x=27 x+27=9+27=36
3x/3=27/3
x=9

gtnhenbr [62]4 years ago

3 0

M=4b
m=27+b

subsitute
4b=27+b
minus b
3b=27
divide 3
b=9
sub back
4b=m
4(9)=m
36=m

michel=36 and brandon is 9

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(a) $\hat p\sim N(0.90,\ 0.0212^{2}})$

(b) $\hat q\sim N(0.10,\ 0.0212^{2}})$

Step-by-step explanation:

The information provided is:

The age at first startup for 90% of entrepreneurs was 29 years of age or less.
The age at first startup for 10% of entrepreneurs was 30 years of age or more.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

(a)

Let p represent the proportion of entrepreneurs whose first startup was at 29 years of age or less.

A sample of n = 200 entrepreneurs is selected.

As n = 200 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be approximated by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat p}=p=0.90\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.90(1-0.90)}{200}}=0.0212$

So, $\hat p\sim N(0.90,\ 0.0212^{2}})$ .

(b)

Let q represent the proportion of entrepreneurs whose first startup was at 30 years of age or more.

A sample of n = 200 entrepreneurs is selected.

As n = 200 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be approximated by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat q}=q=0.10\\\\\sigma_{\hat q}=\sqrt{\frac{q(1-q)}{n}}=\sqrt{\frac{0.10(1-0.10)}{200}}=0.0212$

So, $\hat q\sim N(0.10,\ 0.0212^{2}})$ .

(c)

The standard deviation of sample proportions is also known as the standard error.

The standard deviation of p is, 0.0212.

The standard deviation of q is, 0.0212.

Thus, the standard errors of the sampling distributions in parts (a) and (b) are same.

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3 years ago