Ice thickness decreases at a rate of 0.2 meters per week. After 7 weeks,the ice is only 2.4 meters thick. What is the equation f
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<u>Given:</u>
The width of the snack box is
The length of the snack box is
The height of the snack box is 4 inches.
We need to determine the volume of the hamsta snack box.
<u>Volume of the box:</u>
The volume of the box can be determined using the formula,
where L is the length, W is the width and H is the height of the box.
Substituting the values, we get;
Simplifying, we get,
Thus, the volume of the Hamsta snack box is 15 cubic inches.
Answer:
(a) The approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution = 0.00738
The approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625
(b) pM - pF = 0.06
While pF - pM = -0.06
The difference in the standard deviation ≈ 0.01025
(c) The probability that a higher proportion of women than men worked last year is 0
Step-by-step explanation:
(a) The given information are;
The percentage of college men that were employed last summer = 88%
The percentage of college women that were employed last summer = 82%
The number of college men interviewed in the survey = 400
The number of college women interviewed in the survey = 400
Therefore, given that the proportion of women that worked = 0.82, we have for the binomial distribution;
p = 0.82
q = 1 - 0.82 = 0.18
n = 400
Therefore;
p × n = 0.82 × 400 = 328 > 10
q × n = 0.18 × 400 = 72 > 10
Therefore, the binomial distribution is approximately normal
We have;
Therefore, the approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution ≈ 0.01921
Similarly, given that the proportion of male that worked = 0.88, we have for the binomial distribution;
p = 0.88
q = 1 - 0.88 = 0.12
n = 400
Therefore;
p × n = 0.88 × 400 = 352 > 10
q × n = 0.12 × 400 = 42 > 10
Therefore, the binomial distribution is approximately normal
We have;
Therefore, the approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625
(b) Given two normal random variables, we have
The distribution of the difference the two normal random variable = A normal random variable
The mean of the difference = The difference of the two means = pM - pF = 0.88 - 0.82 = 0.06
While pF - pM = -0.06
The difference in the standard deviation, giving only the real values, is given as follows;
(c) When there is no difference between the the proportion of men and women that worked last summer, the probability that there is a difference = 0
Therefore, taking 0 as the standard score, we have;
Given that the maximum values for a cumulative distribution table is approximately 4, we have that the probability that a higher proportion of women than men worked last year is 0.