The time spent waiting at a stoplight, assuming a positive amount of time is spent waiting and that the stoplight is on a timer and does not use a car sensor, can be well modeled by a uniform distribution. Suppose the time (in minutes) you spend waiting at a particular stoplight can be modeled by a Unif([0,2]) distribution. Compute and compare the average (expected value) and median amount of time you spend at the stoplight, and sketch the pdf of the distribution.

the mean and median = 1

Step-by-step explanation:

From the question we have a uniform distribution

We define

a = 0

b = 2

In a uniform distribution, we have the mean and the median to be one and the same, that is equal.

Mean = median = (a+b)/2

= 2/2 = 1

So the expected value (mean) and median amount of time spent at the particular spotlight is 1 and the same.

Then the probability density function, pdf, of the distribution

= 1/(b-a)

= 1/2

0 ⩽ x ⩽ 2

Please check attachment for the sketch. Thank you

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this