Answer:
5.714 (rounded to the nearest thousandth)
Step-by-step explanation:
Answer:
f(-3) = -1
g(-3) = 14
Step-by-step explanation:
plug -3 for x to get the answer
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:
Explanation:
Factor Out 
From 
= 
Factor Out Common Term 5x + 1
= 
Step-by-step explanation:
f(0) = 0 , f(2) = -4 , f(4) = -2
0 = f(2x) → f(0) = f(2x) → x = 0 (0,0)
-4 = f(2x) → f(2) = f(2x) → x = 1 (1,-4)
-2 = f(2x) → f(4) = f(2x) → x = 2 (2,-2)
shape of the graph remain without change in vertical axis , but it is compressed in the horizontal direction.
