a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
Hello :
<span> -3 = -15 + a
add : 15
-3 +15 = -15 +15 +a
12 = a
a = 12</span>
Answer: Side AB= -7/4
BC= 1/7 CD=-5/3 AD= 1/2 and neither
Step-by-step explanation:
none of those sides are parallel to each other because they dont have the same slope
The constant term is 7. this is because x and y are unknowns and change depending on the other. As 6x^3y will change depending on what x or y is so will: 8x^2 and 5x therefore 7 is the constant term.
Answer:
The answer is C. Physical attractiveness
Step-by-step explanation:
Physical attractiveness is a term describing the way people judge the physical appearances of others,
