PLEASE HELP QUICKLY! IF 2 ANSWERS I WILL VOTE ONE BRAINLIEST

Does anyone get this I accidentally pressed that by accident

sergiy2304 [10]

Answer:

this has happened to me before. unfortunately I had to refresh the page.

Step-by-step explanation:

write down the answers you already have done.

4 0

2 years ago

Given h(x) = x^ - 2, what is the value of h(-3) ? ^ = is exponent 2

ella [17]

Answer:

1/9

Step-by-step explanation:

Given

h(x) = x^ - 2
h(-3) = ?

Substitute x with -3

h(-3) = (-3)^-2 = 3^-2 = 9^-1 = 1/9

3 0

2 years ago

Read 2 more answers

g "They hired an analyst who collected a random sample of 40,361 Game of Thrones fans, and found that 8,337 of those fans said t

Viktor [21]

Answer:

The lower bound for a 90% confidence interval is 0.2033.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 40361, \pi = \frac{8337}{40361} = 0.2066$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2066 - 1.645\sqrt{\frac{0.2066*0.7934}{40361}} = 0.2033$

The lower bound for a 90% confidence interval is 0.2033.

6 0

2 years ago

Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the

Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 9.6, \sigma = 0.5$ .