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2 years ago

I need help plzzz it is due in 5 minutes

Mathematics

2 answers:

gladu [14]2 years ago

5 0

Answer:

Just tried this on a calculator

)) i think its 7/2!

Step-by-step explanation:

jeka942 years ago

4 0

Answer:

7/2

Step-by-step explanation:

Multiply one factor at a time.

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An air transport association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maxim

Flura [38]

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the population mean, when the population standard deviation is not provided is:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

The sample selected is of size, n = 50.

The critical value of t for 95% confidence level and (n - 1) = 49 degrees of freedom is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, 49}\approx t_{0.025, 60}=2.000$

*Use a t-table.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552$

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

$=6.76\pm 2.000\times \frac{2.552}{\sqrt{50}}\\\\=6.76\pm 0.722\\\\=(6.038, 7.482)\\\\\approx (6.0, 7.5)$

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

7 0

3 years ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central limit theorem

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

3 years ago

A seamstress uses 12 yards of fabric to make 3 costumes for students in the chorus. How many yards will she need to make costume

Step2247 [10]

Answer:

100 yards

Step-by-step explanation:

It takes 4 yards of fabric to make a costume for one student. She needs to make costumes for every student in chorus, so you just multiply this number by 25 and get 100 yards.

7 0

3 years ago

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Which ordered pair is also on the line relating the area of a face of a cube and the surface area of the cube?

Alexxx [7]

What cube how big is the cube

7 0

3 years ago

There are 93 calories in a small candy bar how many calories are ther in a half dozen small candy bars?

Tju [1.3M]

Half dozen = 6

6 × 93 = 558

Your answer is 558

6 0

3 years ago

Read 2 more answers