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.
Answer:
1st number = 17
2nd number = 18
Step-by-step explanation:
Unknown number = x
Since they are consecutive numbers, we add 1 to each other.
=> x + x + 1 = 35
=> 2x + 1 = 35
=> 2x + 1 - 1 = 35 - 1
=> 2x = 34
=> 2x/2 = 34/2
=> x = 17
So, the first number = x = 17
=> Second number = x + 1 = 17 + 1 = 18
1st number = 17
2nd number = 18
Answer:
y=2x+11
Step-by-step explanation:
You first have to distribute the 2 to (x+7) to get y+3=2x+14. after that all you have to do is subtract both sides of the equation by 3 to get y=2x+11
