Determine the unit rate for the following problems : he pumped 54 gallons in 12 minutes
2 answers:
To find the unit rate, we are essentially going to reduce this ratio down to it's lowest possible terms. So in this case, we are going to find how many gallons are pumped every 1 minute.
To do this, we are going to divide 54 by 12:
So now that we have this in such terms, we can call it a unit rate. Therefore, the unit rate is 4.5 gallons/min.
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To calculate it use pythagoras: a^2+b^2=c^2
7^2+10^2=49+100=149
calculate the square root:
so the diagonal is about 12.2 inches long
7^2+10^2=49+100=149
calculate the square root:
so the diagonal is about 12.2 inches long
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
