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3 years ago

15 POINTS AND BRANLIEST!!! X = ?

Mathematics

1 answer:

PIT_PIT [208]3 years ago

7 0

Answer:

x = 7

Step-by-step explanation:

The 2 angles are Alternate exterior angles and are congruent, so

2x - 3 = x + 4 ( subtract x from both sides )

x - 3 = 4 ( add 3 to both sides )

x = 7

For the lines to be parallel, then x = 7

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The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was 29 years of age or less and the age

Hitman42 [59]

Answer:

(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:

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