
They are the same fraction, so they are proportional.
Answer: The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.
Step-by-step explanation:
Let x and y area the random variable that represents the heights of women and men.
Given : The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches.
i.e.

Since , 
Then, z-score corresponds to a woman 6 feet tall (i.e. x=72 inches).
[∵ 1 foot = 12 inches , 6 feet = 6(12)=72 inches]

Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.
i.e.

Then, z-score corresponds to a man 5'10" tall (i.e. y =70 inches).
[∵ 1 foot = 12 inches , 5 feet 10 inches= 5(12)+10=70 inches]

∴ The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.
5x squared+7x+2. That is the answer
Answer:
The results don't make sense
Step-by-step explanation:
We can solve by means of a 2x2 system of equations, we have to:
"x" is the number of children's tickets
"y" is the number of adult tickets
Thus:
8 * x + 8.75 * y = 259
x + y = 35 => x = 35 - y
replacing we have:
8 * (35 - y) + 8.75 * y = 259
280 - 8 * y + 8.75 * y = 259
- 8 * y + 8.75 * y = 259 - 280
0.75 * y = -21
y = -21 / 0.75
y = -28
Thus:
x = 35 - (-28) = 63
With these results we notice that the problem has inconsistency, since the value of the tickets cannot be given a negative number, I recommend reviewing the problem, since the approach is correct.
Answer:
0
Step-by-step explanation: