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maw [93]
3 years ago
14

Which relation is a function of x?

Mathematics
2 answers:
Orlov [11]3 years ago
8 0
The function of the relation is: x3

Hope this helps you!! :)
stiv31 [10]3 years ago
7 0

Answer:

x3

Step-by-step explanation:

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Give two ways to write each algebraic expression in words. 4+×
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3 years ago
Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa
4vir4ik [10]

Using the z-distribution, it is found that since the <u>test statistic is greater than the critical value</u>, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is <u>tested if the mean length of jail time is still of 2.5 years</u>, that is:

H_0: \mu = 2.5

At the alternative hypothesis, it is <u>tested if it has increased</u>, that is:

H_1: \mu > 2.5

We have the <u>standard deviation for the population</u>, thus, the z-distribution is used. The test statistic is given by:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • \sigma is the standard deviation of the sample.
  • n is the sample size.

For this problem, the values of the <u>parameters</u> are: \overline{x} = 3, \mu = 2.5, \sigma = 1.5, n = 26

Hence, the value of the <u>test statistic</u> is:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}

z = 1.7

The critical value for a <u>right-tailed test</u>, as we are testing if the mean is greater than a value, with a <u>significance level of 0.05</u>, is of z^{\ast} = 1.645

Since the <u>test statistic is greater than the critical value</u>, it can be concluded that the mean length of jail time has increased.

A similar problem is given at brainly.com/question/24166849

5 0
2 years ago
Many people use scanners to read documents and store them into pdf file. To help determine which brand of scanner to buy, a stud
konstantin123 [22]

The data is missing in the question. The data is provided below :

Document : 1     2      3      4      5     6      7      8

Brand A       17  29    18    14    21   25    22    29

Brand B       21  38    15    19    22   30    31   37

Solution :

State of the hypothesis of the null hypothesis and alternate hypothesis.

Null hypothesis : $h_A = h_B$

Alternate hypothesis : $h_A > h_B$

These hypothesis is a one tailed test. The null hypothesis will get rejected when the mean difference between the sample means is very small.

Significance level = 0.05

Therefore the standard error is :  $SE = \sqrt{(\frac{s^2_1}{n_1})+(\frac{s^2_2}{n_2})}$

                                                         = 3.602

And the degree of freedom, DF = 14

$t=\frac{(x_1-x_2)-d}{SE}$

 = -1.319

Here, $s_1$ = standard deviation of the sample 1

        $s_2$ = standard deviation of the sample 2

         $n_1$ = size of the sample 1

        $n_2$ = size of the sample 2

         $x_1$ = mean of the sample 1

        $x_2$ = mean of the sample 2      

          d = the hypothesis difference between the population mean

The observed difference in a sample means t static of -1.32. From t distribution calculator to determine P($t \leq -1.32$) = 0.1042  

Since the P value of 0.1042 is greater than significance level o 0.05, we therefore cannot reject the null hypothesis.

But from the test, we have no sufficient evidence that supports that Brand A is better than Brand B.      

8 0
3 years ago
Answerrr pleaseee fastttt
steposvetlana [31]
1: 3, 2: 5, 4:14 ther u go
4 0
3 years ago
Read 2 more answers
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