How many lines of reflectional symmetry does the trapizoid have
1 answer:
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Answer:
Null hypothesis is not rejecting at 0.05 level.
Step-by-step explanation:
We are given that a publisher reports that 63% of their readers own a personal computer. A random sample of 110 found that 56% of the readers owned a personal computer.
And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;
Null Hypothesis, : p = 0.63 {means that the percentage of readers who own a personal computer is same as reported 63%}
Alternate Hypothesis, : p
0.63 {means that the percentage of readers who own a personal computer is different from the reported 63%}
The test statistics we will use here is;
T.S. = ~ N(0,1)
where, p = actual % of readers who own a personal computer = 0.63
= percentage of readers who own a personal computer in a
sample of 110 = 0.56
n = sample size = 110
So, Test statistics =
= -1.48
Now, at 0.05 significance level, the z table gives critical value of -1.96 to 1.96. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.
Therefore, null hypothesis is not rejecting at 0.05 level.
Replace the f(x) with y and proceed to make x the subject:
y=1/3x+5
y-5=1/3x
We need to ÷1/3 on both sides, so ×3 on both sides
3(y-5)=x
3y-15=x
Now swap y with x and replace the isolated x with f(x) so the inverse function is:
y=1/3x+5
y-5=1/3x
We need to ÷1/3 on both sides, so ×3 on both sides
3(y-5)=x
3y-15=x
Now swap y with x and replace the isolated x with f(x) so the inverse function is: