Answer: Yes it is
Step-by-step explanation:
It also is b=0.5 if you need decimal form.
308.3 divided by 15 could be estimated by rounding 308.3 down to 300 and dividing that by 15:
300
----- = 20
15
Answer: the probability of a student being overdrawn by more than $18.75 is 0.674
Step-by-step explanation:
Since the bank overdrafts of ASU student accounts are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = bank overdraft of Asu students.
µ = mean
σ = standard deviation
From the information given,
µ = $21.22
σ = $5.49
We want to find the probability of a student being overdrawn by more than $18.75. It is expressed as
P(x > 18.75) = 1 - P(x ≤ 18.75)
For x = 18.75,
z = (18.75 - 21.22)/5.49 = - 0.45
Looking at the normal distribution table, the probability corresponding to the z score is 0.326
Therefore,
P(x > 18.75) = 1 - 0.326 = 0.674
Answer:54
Step-by-step explanation:
72 classes in one year
75% spent outside
72x0.75=54
6(3m + -5n + 7)
Distribute the 6.
6*3m + 6*-5n + 6*7
Simplify each individual term.
18m -30n + 42
