Answer:
a) 33.33%)
b) 135 minutes
c) 8.66 min
d) 50%
Step-by-step explanation:
a) the probability for a uniform distribution is
P(b<X<a) = (a-b)/(c-d) , where c and d are the maximum and minimum values
therefore the probability that the flight is more than 140 minutes ( and less than 150 since it is the maximum value)
P(140<X<150) = (a-b)/(c-d) = (150-140)/(150-120) = 10/30 = 1/3 (33.33%)
b) the mean (expected value) for a uniform probability distribution is
E(X) = (c+d)/2 = (120+150)/2 = 135 minutes
c) the standard deviation for a uniform probability distribution is
σ²(X)= (c-d)²/12 = (150-120)²/12 = 75 min²
σ = √75 min² = 8.66 min
b) following the same procedure as in a)
P(120<X<135) = (a-b)/(c-d) = (135-120)/(150-120) = 15/30 = 1/2 (50%)
Answer:
288 miles
If we have the distance, we divide by 36 instead to find how many gallons of gas he used.
Step-by-step explanation:
180/5=36 miles per gallon
8 gallonsx36 miles per gallon=288 miles
pls mark brainliest
Step-by-step explanation:
Below is an attachment containing the solution.
Hey there,
15 games = 7 hits
1 game = 7 / 15
= <span>0.46666666666
45 games = </span><span>0.46666666666 x 45
= 21
Hope this helps :))
<em>~Top♥</em>
</span>
Answer:
(a) 
The expected number in the sample that treats hazardous waste on-site is 0.383.
(b) 
There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.
Step-by-step explanation:
Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.
N = 209
Only eight of these facilities treated hazardous waste on-site.
r = 8
a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?
n = 10
The expected number in the sample that treats hazardous waste on-site is given by




Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.
b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site
The probability is given by
For the given case, we have
N = 209
n = 10
r = 8
x = 4




Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.