Answer:
2 (x - 5)
I hope this helps u.
Step-by-step explanation:
Part A:
For a table to be considered a function, every x-value must have one y-value.
Each x-value in this table is unique, and has only one y-value, so this table
does represent a function.Part B:
Plug in 7 for every x in the relation:

The table's output when x = 7 is 11. Compare the two outputs:
11 < 27
The relation, 2x + 13, has a greater value when x = 7.
Part C:
Set the relation to equal 75:

Subtract 13 from both sides:

Divide both sides by 2 to get x by itself:

The x value that produces an output of 75 will be
31.
Complete question :
A data set includes data from student evaluations of courses. The summary statistics are nequals92, x overbarequals4.09, sequals0.55. Use a 0.10 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
Answer:
H0 : μ = 4.25
H1 : μ < 4.25
T = - 2.79
Pvalue =0.0026354
we conclude that there is enough evidence to conclude that population mean is different from 4.25 at 10%
Step-by-step explanation:
Given :
n = 92, xbar = 4.09, s = 0.55 ; μ = 4.25
H0 : μ = 4.25
H1 : μ < 4.25
The test statistic :
T = (xbar - μ) ÷ s / √n
T = (4.09 - 4.25) ÷ 0.55/√92
T = - 0.16 / 0.0573414
T = - 2.79
The Pvalue can be obtained from the test statistic, using the Pvalue calculator
Pvalue : (Z < - 2.79) = 0.0026354
Pvalue < α ; Hence, we reject the Null
Thus, we conclude that there is enough evidence to conclude that population mean is different from 4.25 at 10%