Answer:
i know i already answered this but, last option is correct :P
Step-by-step explanation:
Answer:
y = mx + b
Step-by-step explanation:
Answer:
2.35%
Step-by-step explanation:
Mean number of months (M) = 39 months
Standard deviation (S) = 10 months
According to the 68-95-99.7 rule, 95% of the data is comprised within two standard deviations of the mean (39-20 to 39+20 months), while 99.7% of the data is comprised within two standard deviations of the mean (39-30 to 39+30 months).
Therefore, the percentage of cars still in service from 59 to 69 months is:

The approximate percentage of cars that remain in service between 59 and 69 months is 2.35%.
What are you suppose to find i feel like it doesn't really say
Answer: in order to qualify, the time must be at least 15.93 minutes
Step-by-step explanation:
Let x be a random variable representing the average response times of fire departments to fire calls. Since it follows a normal distribution, we will determine the z score by applying the formula,
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = population standard deviation
From the information given,
µ = 12.8 minutes
σ = 3.7 minutes
in order to qualify for the special safety award, the probability value would be to the right of the z score corresponding to 80%
The z score corresponding to 80% on the normal distribution table is 0.845
Therefore,
0.845 = (x - 12.8)/3.7
0.845 × 3.7 = x - 12.8
3.13 = x - 12.8
x = 12.8 + 3.13 = 15.93