The answer is C. The explanation in the previous comment section is correct
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:
0
Step-by-step explanation:
The slope is 1/0, and that equals 0.
F(x) = x² + 4
y = x² + 4
First, I reverse the equation to be
x² + 4 = y
Second, we need only x to remain on the left side and we need to move the others to the right side
x² + 4 = y
x² = y - 4
x = √(y - 4)
Last, change into invers function
f⁻¹(x) = √(x - 4)
Answer:
300%.
Step-by-step explanation:
Suppose the population is 100, then it rises to 300 after one hour.
So the requires percent is (300/100) * 100
= 300%.
