If the mean of the weights is 37 ounces and the standard deviation is 3 ounces then 34.13% of the percentage lies between 37 ounces and 40 ounces.
Given that the mean of the weights is 37 ounces and the standard deviation is 3 ounces.
We are required to find out the percentage of weights that lie between 37 and 40 ounces.
Because the distribution is bell shaped so it is a normal distribution and because we donot know the sample size so we will use z statistics.
Z=(X-μ)/σ
Z=(40-37)/3
Z=3/3
Z=1
p value of Z=1 is 0.3413 so the percentage will be 34.13%.
Hence if the mean of the weights is 37 ounces and the standard deviation is 3 ounces then 34.13% of the percentage lies between 37 ounces and 40 ounces.
Learn more about z score at brainly.com/question/25638875
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Question is incomplete. It should includes the following line.
Calculate the percentage of weights that lie between 37 ounces and 40 ounces.
8+4x-12+3x<=24
-4+7x<=24
7x<=28
X<=4
Answer:
D. y-5 = 3(x+1)
Step-by-step explanation:
Slope of the perpendicular line is 3.
Slope intercept form is,
y-5 = 3 (x-(-1))
y-5 = 3(x+1)
Answer:
divided by -5
Step-by-step explanation:
Answer:8.4
Step-by-step explanation:
Ten percent of 14 is 1.4 so add 1.4 4 times or multiple 1.4 by 4 and you get 5.6, then subtract 14 by 5.6 and you get 8.4