Hey! :)
10y-3(y+5)
Distribute the -3.
-3*y= -3y -3*5= -15
10y-3y-15
7y-15
"A" is the answer.
~'Manda
The probability that the mean clock life would differ from the population mean by greater than 12.5 years is 98.30%.
Given mean of 14 years, variance of 25 and sample size is 50.
We have to calculate the probability that the mean clock life would differ from the population mean by greater than 1.5 years.
μ=14,
σ==5
n=50
s orσ =5/=0.7071.
This is 1 subtracted by the p value of z when X=12.5.
So,
z=X-μ/σ
=12.5-14/0.7071
=-2.12
P value=0.0170
1-0.0170=0.9830
=98.30%
Hence the probability that the mean clock life would differ from the population mean by greater than 1.5 years is 98.30%.
Learn more about probability at brainly.com/question/24756209
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There is a mistake in question and correct question is as under:
What is the probability that the mean clock life would differ from the population mean by greater than 12.5 years?
Answer:
-22
Step-by-step explanation:
18 + 4 = 22
then just add the negative to 22
-22
<span>Probability = 0.063
Fourth try = 0.0973
Let X be the number of failed attempts at passing the test before the student passes. This
is a negative binomial or geometric variable with x â {0, 1, 2, 3, . . .}, p = P(success) = 0.7
and the number of successes to to observe r = 1. Thus the pmf is nb(x; 1, p) = (1 â’ p)
xp.
The probability P that the student passes on the third try means that there were x = 2
failed attempts or P = nb(2, ; 1, .7) = (.3)2
(.7) = 0.063 . The probability that the student
passes before the third try is that there were two or fewer failed attmpts, so P = P(X ≤
2) = nb(0, ; 1, .7) + nb(1, ; 1, .7) + nb(2, ; 1, .7) = (.3)0
(.7) + (.3)1
(.7) + (.3)2
(.7) = 0.973 .</span>
Answer: 30%
Step-by-step explanation:
First, we will put completed over the total.
Now, we will divide.
0.3
Lastly, we will multiply by 100 to turn it into a percent.
0.3 * 100 = 30%
You have ridden 30% of the course.