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UkoKoshka [18]
2 years ago
15

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 - <em>α</em>)% 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, <em>n</em> = 50.

The critical value of <em>t</em> for 95% confidence level and (<em>n</em> - 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 <em>t</em>-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
Read 2 more answers
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
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