Answer:
I really hope this helps
Step-by-step explanation:
brainliest please? :)
Answer:
Step-by-step explanation:
Let x be a random variable representing the flight arrival time from Boston to New York.
For a uniform probability distribution, the notation is
X U(a, b) where a is the lowest value of x and b is the lowest value of x
The probability density function, f(x) = 1/(b - a)
Mean, µ = (a + b)/2
Standard deviation, σ = √(b - a)²/12
From the information given, the time difference in minutes is 9:57 - 9:07 = 50 minutes. Therefore,
a = 0
b = 50
µ = (0 + 50)/2 = 25
σ = √(50 - 0)²/12 = 14.43
b) converting to minutes, it is 9:30 - 9:07 = 23 minutes
the probability that a flight arrives late(later than 9:30 am) is expressed as P(x > 23)
f(x) = 1/(50) = 0.02
P(x > 23) = (50 - 23)0.02 = 0.54
It is the distance between -14 and 0 on the number line
Answer:
22
Step-by-step explanation:
Pretend the 10 values in the first sentence are a,b,c,d,e,f,g,h,i,j
Pretend the addition 5 values is k,l,m,n,o
So the mean of all the 15 data is (a+b+c+d+e+f+g+h+i+j+k+l+m+n+o)/15=20
So the sum of all 15 data is a+b+c+d+e+f+g+h+i+j+k+l+m+n+o=300 since 15(20)=300
Now let's look at the first 10: We have their mean so we can write:
(a+b+c+d+e+f+g+h+i+j)/10=19
so a+b+c+d+e+f+g+h+i+j=190 since 10(19)=190
So that means using our first sum equation and our equation sum equation we have
190+k+l+m+n+o=300
k+l+m+n+o=300-190
k+l+m+n+o= 110
So the average of those 5 numbers mentioned in your problem is 110/5=22
Answer:
4pq(p³+q³)
Step-by-step explanation:
Exactly 5 game would be when 4 wins and 1 loss of a particular person
loss has to be one of the first 4 games
A wins: qp⁴ + pqp³ + p²qp² + p³qp
= qp⁴ + qp⁴ + qp⁴ + qp⁴ = 4qp⁴
B wins: pq⁴ + qpq³ + q²pq² + q³pq
= pq⁴ + pq⁴ + pq⁴ + pq⁴ = 4pq⁴
A wins or B wins:
4pq⁴ + 4qp⁴ = 4pq(q³+p³)
