Answer:
the second one dose I believe
Answer:
6
Step-by-step explanation:
you have to distributw the 4 to the 1/4 so essentially youre multiplying the exponents. think of it as 4*1/4 and youll have 4/4 which equals to 1. 6^1 is just 6 so thats the answer
Answer:
0.0764
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What proportion of boxes is underweight (i.e., weigh less than 32 oz)?
This is the pvalue of Z when X = 32. So



has a pvalue of 0.0764, which is the correct answr
Answer: 4 cents or .4
Step-by-step explanation:
Whenever in a problem like this where there is a fee, tax, and or bonus you move the decimal in front of the number and round to the nearest penny but if there is only one number in front of the decimal you don’t gotta do anything with it unless its .5 or higher but .4 and below you keep the same.
Answer:
By using hypothesis test at α = 0.01, we cannot conclude that the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single
Step-by-step explanation:
let p1 be the proportion of elementary teachers who were single
let p2 be the proportion of high school teachers who were single
Then, the null and alternative hypotheses are:
: p2=p1
: p2>p1
We need to calculate the test statistic of the sample proportion for elementary teachers who were single.
It can be calculated as follows:
where
- p(s) is the sample proportion of high school teachers who were single (
) - p is the proportion of elementary teachers who were single (
)
- N is the sample size (180)
Using the numbers, we get
≈ 1.88
Using z-table, corresponding P-Value is ≈0.03
Since 0.03>0.01 we fail to reject the null hypothesis. (The result is not significant at α = 0.01)