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2 years ago

Please help I will give brainliest

Mathematics

1 answer:

pentagon [3]2 years ago

7 0

The option that represents the answers for 6, 8, 10, and 14 is the second option- 31, 41, 51, 71

<h3>Sequence and Series </h3>

From the question, we are to determine the option that represents the answer for pictures 6, 8, 10, and 14

In the given diagram,

The first picture requires 6 matches

The 2nd picture requires additional 5 matches

the 3rd picture requires additional 5 matches

and so on

That is

1 → 6

2 → 6 + 5 = 11

3 → 11 + 5 = 16

4 → 16 + 5 = 21

5 → 21 + 5 = 26

6 → 26 + 5 = 31

7 → 31 + 5 = 36

8 → 36 + 5 = 41

9 → 41 + 5 = 46

10 → 46 + 5 = 51

11 → 51 + 5 = 56

12 → 56 + 5 = 61

13 → 61 + 5 = 66

14 → 66 + 5 = 71

Hence, the option that represents the answers for 6, 8, 10, and 14 is the second option- 31, 41, 51, 71

Learn more on Sequence and Series here: brainly.com/question/10321751

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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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