SOLUTION
The mean is 4min
standard deviation is 1min
the z score is
where
then we have
The probability the call lasted less than 3 min will be
Therefore, the probability that (z < -1 ) is
[tex]\begin{gathered} Pr(z<-1)=Pr(0
Hence, the percentage of the calls that lasted less than 3 min is 16%
First you must take into account the variable that is being defined for this case:
c = represents the number of puppies whose eyes are closed.
We can then write the following equation:
15 = 11 + c
Rewriting the equation:
11 + c = 15
Clearing c:
c = 15-11
c = 4
4 puppies have their eyes closed
Answer:
11 + c = 15
4 puppies have their eyes closed
P(picking one defective) = 3/10
P(picking a 2nd defective) = 2/9
P(1 and 2 defective) = 3/10 x 2/9 = 6/90 = 0.066
Second method using combination:
³C₂ / ¹⁰C₂ = 1/15 = 0.066
Answer:
Percent change from 15 to 18.
You need to divide the numbers, but which one goes on top?
You know the number increases from 15 to 18, so the percentage must be greater than 100%.
The only way to get a number greater than 100% is to put the larger number on top.
Therefore, the answer is 18/15 = 1.2
Multiply by 100 to get a %
1.2 * 100 = 120%
2. Calculate percentage change
from V1 = 5 to V2 = 8
ans= 60%
overall answer gain of 15 yards and 18 yards
Answer:
I think the first mistake was made in step 1
