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3 years ago

Free 50 points hurry up and i will mark brainliest

Mathematics

2 answers:

svetlana [45]3 years ago

8 0

Yes please merry Christmas

natita [175]3 years ago

6 0

YYAYAYYAYYAYYAGDGVEGHVHEWEVHEW

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Match the real-world descriptions with the features they represent within the context of Melissa's garden. Not all tiles will

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Can’t see photo so I can answer

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3 years ago

I WILL GIVE BRAINLIEST!

Nesterboy [21]

Answer:

Step-by-step explanation:

So we get 5 hrs.

1____2____3____4____5

$2 + $1 + $1 + $1 + $1 = $6

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4 years ago

What is the remainder when 1,095 is divided by 7?<br> A) 3 <br> B) 4 <br> C) 5 <br> D) 6

Rina8888 [55]

The answer is a or 3

5 0

3 years ago

Read 2 more answers

A survey of 35 people was conducted to compare their self-reported height to their actual height. The difference between reporte

monitta

Answer:

The test statistic is t = 3.36.

Step-by-step explanation:

You're testing the claim that the mean difference is greater than 0.7.

At the null hypothesis, we test if it is 0.7 or less, that is:

$H_0: \mu \leq 0.7$

At the alternate hypothesis, we test if it is greater than 0.7, that is:

$H_1: \mu > 0.7$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

0.7 is tested at the null hypothesis:

This means that $\mu = 0.7$

Survey of 35 people. From the sample, the mean difference was 0.95, with a standard deviation of 0.44.

This means that $n = 35, X = 0.95, s = 0.44$

Calculate the test statistic

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{0.95 - 0.7}{\frac{0.44}{\sqrt{35}}}$

$t = 3.36$

The test statistic is t = 3.36.

7 0

3 years ago

A random sample of 35 business students required an average of 50.7 minutes to complete a statistics exam. Assume that the popul

love history [14]

Answer:

(47.3, 54.1)

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{10.4}{\sqrt{35}} = 3.4$

The lower end of the interval is the sample mean subtracted by M. So it is 50.7 - 3.4 = 47.3

The upper end of the interval is the sample mean added to M. So it is 50.7 + 3.4 = 54.1

The answer is (47.3, 54.1).

6 0

3 years ago