Answer:
are not choice b 75%
75/100 *20 = 1500/100=15
Step-by-step explanation:
Answer: √20
Step-by-step explanation:
4.47213595 (the 5 repeats)
3) 2,000.376 inches of water
4) 30.375 for 1 box, 91.125 inches for all three
5) 4,536 feet
6) 24 feet
75 decreased by 40% is 45
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 %
