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%)
for 6
in a
given
slope(m)=1
point=(2,-4)=(x1,y1)
we know
equation of straight line
y-y1=m(x-x1)
y+4=1(x-2)
y+4=x-2
therefore x-y-6=0 is the required equation
substituting with ax+by+c=0 we get
a=1
b= -1
c= -6
The answer is 20. you'll have to multiply 10 times 3 and get 30. Then you'll subtract the 10 from the 30 and get 20
