If it costs susan $25 to rent a truck for 1 hour and $100 to rent a truck for 4 hours, how much will it cost for her to rent a t
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The region(s) represent the intersection of Set A and Set B (A∩B) is region II
<h3>How to determine which region(s) represent the intersection of Set A and Set B (A∩B)?</h3>The complete question is added as an attachment
The universal set is given as:
Set U
While the subsets are:
- Set A
- Set B
The intersection of set A and set B is the region that is common in set A and set B
From the attached figure, we have the region that is common in set A and set B to be region II
This means that
The intersection of set A and set B is the region II
Hence, the region(s) represent the intersection of Set A and Set B (A∩B) is region II
Read more about sets at:
#SPJ1
The first part of the question is missing and it says;
Use these parameters: Men's heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.
Answer:
A) Percentage of women meeting the height requirement = 72.24%
B) Percentage of men meeting the height requirement = 0.875%
C) Corresponding women's height =67.42 inches while corresponding men's height = 72.19 inches
Step-by-step explanation:
From the question,
For men;
Mean μ = 68.6 in
Standard deviation σ = 2.8 in
For women;
Mean μ = 63.7 in
Standard deviation σ = 2.9 in
Now let's calculate the standardized scores;
The formula is z = (x - μ)/σ
A) For women;
Z = (62 - 63.7)/2.9 = - 0.59
Z = (78 - 63.7)/2.9 = 4.93
The original question cam be framed as;
P(62 < X < 78).
So thus, the probability of only women will take the form of;
P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)
From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996
And P(Z > - 0.59) = 0.277595
Thus;
P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.7224
So, percentage of women meeting the height requirement is 72.24%.
B) For men;
Z = (62 - 68.6)/2.8 = -2.36
Z = (78 - 68.6)/2.8 = 3.36
Thus, the probability of only men will take the form of;
P(-2.36 < Z < 3.36) = P(Z<3.36) - P(Z > - 2.36)
From the normal probability table attached, when we interpolate, we'll arrive at P(Z<3.36) = 0.99961
And P(Z > -2.36) = 0.99086
Thus;
P(Z<3.36) - P(Z > -2.36) 0.99961 - 0.99086 = 0.00875
So, percentage of women meeting the height requirement is 72.24%.
B)For women;
Z = (62 - 63.7)/2.9 = - 0.59
Z = (78 - 63.7)/2.9 = 4.93
The original question cam be framed as;
P(62 < X < 78).
So thus, the probability of only women will take the form of;
P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)
From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996
And P(Z > - 0.59) = 0.277595
Thus;
P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.00875
So, percentage of women meeting the height requirement is 0.875%
C) Since the height requirements are changed to exclude the tallest 10% of men and the shortest10% of women.
For women;
Let's find the z-value with a right-tail of 10%. From the second table i attached ;
invNorm(0.90) = 1.2816
Thus, the corresponding women's height:: x = (1.2816 x 2.9) + 63.7= 67.42 inches
For men;
We have seen that,
invNorm(0.90) = 1.2816
Thus ;
Thus, the corresponding men's height:: x = (1.2816 x 2.8) + 68.6 = 72.19 inches
