Answer:
ans=13.59%
Step-by-step explanation:
The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean 
and standard deviation 
, we have these following probabilities
In our problem, we have that:
The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months
So 
So:
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What is the approximate percentage of cars that remain in service between 64 and 75 months?
Between 64 and 75 minutes is between one and two standard deviations above the mean.
We have 
subtracted by is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.
To find just the percentage above the mean, we divide this value by 2
So:
The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.
We know that
L=8 ft
W=2 ft
H=3 ft
[<span>surface area of the doll house]=2*[W*H]+2*[L*H]+2*[W*L]
</span>[surface area of the doll house]=2*[2*3]+2*[8*3]+2*[2*8]
[surface area of the doll house]=2*[6]+2*[24]+2*[16]
[surface area of the doll house]=12+48+32--------> 92 ft²
the answer is 92 ft²
Mason:
time = t
V = 6 ft/s
Distance = v * t = 6t
Nora:
time = t - 60 [ I passed 1 minute to seconds]
V = 8 ft/s
Dsitance = v*t = 8 (t - 60)
Both distance are equal
6t = 8(t-60)
6t = 8t - 480
8t -6t = 480
2t = 480
t = 480/2
t = 240 s
Mason had been swimming for 240 seconds when both had swan the same distance.
Answer:
x=140
Step-by-step explanation:
solve for x
divide 10 on both sides:
x=140
Hope this helps!!!
