The time (in minutes) until the next bus departs a major bus depot follows a uniform distribution from 25 to 46 minutes. Let X denote the time until the next bus departs. The distribution is and is . The mean of the distribution is μ= . The standard deviation of the distribution is σ= . The probability that the time until the next bus departs is between 30 and 40 minutes is P(30

Step-by-step explanation:

The distribution is Uniform and is continuous.

We are given a = 25 and b = 446

The mean of the distribution is

Mean = µ = (a + b) / 2

= (25 + 44) / 2

= 34.5

The standard deviation of the distribution is  

SD = σ = (b – a) / sqrt(12)

= (46 – 25) / sqrt(12)

= 21/3.464102

= 6.062minutes

SD = σ = 6.062minutes

Now, we have to find P(30<X<40)

P(30<X<40)

= (40 – 30) / (46 – 25)

= 10/19

= 0.4762

Required probability = 0.4762

Now, we have to find the value of a for which P(X≤a) = 0.9

We have, P(X≤a)

= (a – 25) / (46 – 25)

= (a – 25) / 21

(a – 25) / 21 = 0.9

(a – 25) = 0.9*21

So, a = 18.9 + 25 = 43.

a = 43.9

= 43.9minutes

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

answer: x=3 and y=4

step-by-step explanation: