The valid conclusions for the manager based on the considered test is given by: Option
<h3>When do we perform one sample z-test?</h3>
One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.
For this case, we're specified that:
- Population mean =
= $150 - Population standard deviation =
= $30.20 - Sample mean =
= $160 - Sample size = n = 40 > 30
- Level of significance =
= 2.5% = 0.025 - We want to determine if the average customer spends more in his store than the national average.
Forming hypotheses:
- Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get:

- Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically

where
is the hypothesized population mean of the money his customer spends in his store.
The z-test statistic we get is:

The test is single tailed, (right tailed).
The critical value of z at level of significance 0.025 is 1.96
Since we've got 2.904 > 1.96, so we reject the null hypothesis.
(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).
Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.
Learn more about one-sample z-test here:
brainly.com/question/21477856
Answer:
A = 5.5t + 9
Step-by-step explanation:
We assume that the relationship between t and A is linear.
We have two points on a line: (2, 20) and (4, 31).
We now find the equation of the line.
A = mt + b
m = (31 - 20)/(4 - 2) = 11/2 = 5.5
20 = 5.5(2) + b
20 = 11 + b
b = 9
A = 5.5t + 9
20 is the answer.
15% of 50= 7.5
45% of 50= 22.5
22.5+7.5=30
50-30=20.
This question is based on prime factorization method.
As given there are 12 peaches and 18 nectarines
Factors of 12 are : 2,3,4,6 and 12
Factors of 18 are : 2,3,6,9 and 18
As mentioned, fruits are divided into the same number of equal groups, so the groups can be formed as-
1. 6 groups of peaches with 2 in each and 6 groups of nectarines with 3 in each,
2. 3 groups of peaches with 4 in each and 3 groups of nectarines with 6 in each.
3. 2 groups of peaches with 6 in each and 2 groups of nectarines with 9 in each.
I’m not sure what your teacher wants you to do but I would do 110-50=60