Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation
Calculate z-scores for the following random variable and determine their probabilities from standard tables.
x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088
x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912
x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587
x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413
Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36
Answer: 27
Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36
Answer: 27
Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78
Answer: 204
Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150
Answer: 150
Answer:
x≥5
Step-by-step explanation:
3(x+15)≤5(x+2)+15
3x+45≤5x+10+15
3x+35≤5x+25
3x-5x≤25-35
-2x≤-10
x≥5
{x:xEx:6,7,8,9,10....}
Hope this helps ;) ❤❤❤
Answer:
probability of an electric power outage shipboard is 0.075 or 7.5 %
Step-by-step explanation:
Given the data in the question;
Probability of engine malfunction = 25% = 0.25
Probability of generator malfunction = 10% = 0.1
Probability of generator being repaired = 50% = 0.5
probability of an electric power outage shipboard = ?
To get the probability of an electric power outage shipboard, we use the expression;
probability of an electric power outage = 2
× p( Probability of generator malfunction ) × p( Probability of generator being repaired ) × ( 1 - p( Probability of engine malfunction ) )
so we substitute in our given values;
probability of an electric power outage = 2 × 0.1 × 0.5 × ( 1 - 0.25 )
probability of an electric power outage = 2 × 0.1 × 0.5 × 0.75
probability of an electric power outage = 0.075 or 7.5 %
Answer:
it will be c
Step-by-step explanation:
Inverse variation means y=k/x (direct variation is y=kx) so we can use the info given to find the constant k
y=k/x
6=k/900
k=6(900)
k=5400 so the equation is:
y=5400/x or more accurately for this problem
t(v)=5400/v then:
t(800)=5400/800
t(800)=6.75 hours