Answer these 3 questions and ill give u brainliest and 5 stars and 15 points and a thank you

Answers

Answer 1

Answer: A=8
B=1/8
C=ht,hh,tt,th

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Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gender of any other children find the probability of at most three boys in ten births

Answers

The probability of at most three boys in ten births is approximately 0.17139, or about 17.14%.

What is Probability?

It is a branch of mathematics that deals with the occurrence of a random event.

This is a binomial probability problem with n = 10 (number of births) and p = 0.5 (probability of a boy or a girl).

We want to find the probability of at most three boys in ten births, which is equivalent to finding the probability of 0, 1, 2, or 3 boys.

To calculate this probability, we can use the binomial probability formula:

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 0.00098 + 0.00977 + 0.04395 + 0.11719

= 0.17139

Therefore, the probability of at most three boys in ten births is approximately 0.17139, or about 17.14%.

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Answer:

Step-by-step explanation:

This is a binomial distribution

The probability of at most 3 boys=

P(exactly 0 boys)+P(exactly 1 boy)+P(exactly 2 boys)+P(exactly 3 boys)

.171875

What is the answer for this please help

Answers

Answer:

graph attached

Step-by-step explanation:

How many seats are in theatre

Answers

When we solve the equation, we find that there are 26 seats in the theatre. Option C

How do we form and solve an equation from the given scenario?

We will consider the number of seats in the theatre as x.

The total cost of renting the theatre and the seats is $520 for the rental plus $30 × the number of seats should be expressed as 30x.

If every seat is sold, the total income from selling the tickets would be $50 × the number of seats is expressed as (50x)

The equation becomes

520 + 30x = 50x

520 = 50x - 30x

520 = 20x

x = 520/20

x = 26

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Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 99% confidence interval for p given that p-hat = 0.34 and n= 500. Point estimate ___________ (2 decimal places) Margin of error __________ (3 decimal places) The 99% confidence interval is ________ to _______ (3 decimal places)

Answers

Answer:

(a) The point estimate for the population proportion p is 0.34.

(b) The margin of error for the 99% confidence interval of population proportion p is 0.055.

Step-by-step explanation:

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean $\bar x$ is a point estimate of the population mean μ.

Similarly, the the point estimate of the population proportion of a characteristic, p is the sample proportion $\hat p$ .

The (1 - α)% confidence interval for the population proportion p is:

$CI=\hat p\pm z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}$

The margin of error for this interval is:

$MOE= z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}$

The information provided is:

$\hat p=0.34\nn=500\n(1-\alpha)\%=99\%$

(a)

Compute the point estimate for the population proportion p as follows:

Point estimate of p = $\hat p$ = 0.34

Thus, the point estimate for the population proportion p is 0.34.

(b)

The critical value of z for 99% confidence level is:

$z={\alpha/2}=z_(0.01/2)=z_(0.005)=2.58$

*Use a z-table for the value.

Compute the margin of error for the 99% confidence interval of population proportion p as follows:

$MOE= z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}$

$=2.58\sqrt{(0.34(1-0.34))/(500)}$

$=2.58* 0.0212\n=0.055$

Thus, the margin of error for the 99% confidence interval of population proportion p is 0.055.

(c)

Compute the 99% confidence interval of population proportion p as follows:

$CI=\hat p\pm z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}$

$CI=\hat p\pm MOE$

$=0.34\pm 0.055\n=(0.285, 0.395)$

Thus, the 99% confidence interval of population proportion p is (0.285, 0.395).

Final answer:

The point estimate for p is 0.34. The margin of error, calculated using a z-score of 2.576, is 0.034. The 99% confidence interval is from 0.306 to 0.374.

Explanation:

This question is about calculating a confidence interval for a proportion using the normal distribution. The best point estimate for p is the sample proportion, p-hat, which is 0.34.

For a 99% confidence interval, we use a z-score of 2.576, which corresponds to the 99% confidence level in a standard normal distribution. The formula for the margin of error (E) is: E = Z * sqrt[(p-hat(1 - p-hat))/n]. Substituting into the formula, E = 2.576 * sqrt[(0.34(1 - 0.34))/500] = 0.034.

The 99% confidence interval for p is calculated by subtracting and adding the margin of error from the point estimate: (p-hat - E, p-hat + E). The 99% confidence interval is (0.34 - 0.034, 0.34 + 0.034) = (0.306, 0.374).