Answer: the probability that a truck drives between 166 and 177 miles in a day is 0.0187
Step-by-step explanation:
Since mileage of trucks per day is distributed normally, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = mileage of truck
µ = mean mileage
σ = standard deviation
From the information given,
µ = 100 miles per day
σ = 37 miles miles per day
The probability that a truck drives between 166 and 177 miles in a day is expressed as
P(166 ≤ x ≤ 177)
For x = 166
z = (166 - 100)/37 = 1.78
Looking at the normal distribution table, the probability corresponding to the z score is 0.9625
For x = 177
z = (177 - 100)/37 = 2.08
Looking at the normal distribution table, the probability corresponding to the z score is 0.9812
Therefore,
P(166 ≤ x ≤ 177) = 0.9812 - 0.9625 = 0.0187
If the problem is 6r•s•4rt•10rst, then the answer is 240•r^3•s^2•t^2.
Most of the question is cropped out
Angles LMO and NMO are congruent, of equal measures,
x=15
Answer:
-60-14x
Step-by-step explanation:
Distribute the 3: -3(20+5x)+x => 20(-3)+3(-5x)+x => -60 -15x+x
Simplify: -60-15x+x => -60-14x
