Answer:
40 months
Step-by-step explanation:
Since it takes him 2 months to drive 15,000 miles
He drives 7,500 miles/month (15,000 miles / 2months)
Dividing the total distance by his speed yields the time it takes to drive that distance
300,000 miles / 7,500 miles/month = 40 months
Just write a number in between such as 0.118976453... or 0.128675,etc
Answer:
P = 0,0012 or P = 0.12 %
Step-by-step explanation:
We know for normal distribution that:
μ ± σ in that range we find 68.3 % of all values
μ ± 2σ ⇒ 95.5 % and
μ ± 3σ ⇒ 99.7 %
Fom problem statement
We have to find (approximately) % of cars that reamain in service between 71 and 83 months
65 + 6 = 71 ( μ + σ ) therefore 95.5 % of values are from 59 and up to 71 then by symmetry 95.5/2 = 49.75 of values will be above mean
Probability between 65 and 71 is 49.75 %
On the other hand 74 is a value for mean plus 1, 5 σ and
74 is the value limit for mean plus 1,5 σ and correspond to 49,85 (from z=0 or mean 65).
Then the pobabilty for 83 have to be bigger than 49.85 and smaller than 0,5 assume is 49.87
Finally the probability approximately for cars that remain in service between 71 and 83 months is : 0,4987 - 0.49.75
P = 0,0012 or P = 0.12 %
Answer: 3c - 5
Step-by-step explanation:
Since c represent the number of cookies that Mark made. We are then informed that Peter made five less than triple as many cookies as Mark. This will be:
= (3 × c) - 5
= 3c - 5
The expression is 3c - 5
Answer:
im pretty sure the answer would be 1/25.........???
Step-by-step explanation:
