Probability that no samples are mutated is 0.83, probability that at most one sample is mutated is 0.9812 and probability that more than half the samples are mutated is 0.
Given percentage of rejuvenated mitochondria defective is 1%, and sample size is 18.
Binomial distribution is the probability of exactly x successes on n repeated trials and X can have two outcomes.
P(X=x)=
percentage of defective rejuvendated mitochondria=1%
p=0.01
Sample size=18
n=18
a) No samples are mutated
This means P(X=0)=
=0.83
b) At most one sample is mutated.
P(X<=1)=P(X=0)+P(X=1)
so,
P(X=0)=
=0.83
P(X=1)=
=
=0.1512
P(X<=1)=0.83+0.1512
=0.9812
c) More than half the samples are mutated.
P(X>9)=P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)
Using two decimals digits precision all will be 0.
Hence Probability that no samples are mutated is 0.83, probability that at most one sample is mutated is 0.9812 and probability that more than half the samples are mutated is 0.
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If a = 4 and d = 3,



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Answer: The general equation for the nth term is 3n + 1.---------------------------------------------------------------------------------------
Since we have that the slope is m = 7/9 and the y-intercept is b = 12, we can write the equation of the line in slope-intercept form:

to find three coordinate points, we can use arbitrary values on x to get the y-coordinate. To make things easier, let's use x = 9, 18 and 27:

therefore, the line with slope m = 7/9 and y-intercept 12 passes through the three points (9,19), (18,26) and (27,33)