Answer:
Option "A" is the correct answer to the following question.
Step-by-step explanation:
Given:
Total number of experiment = 50
Find:
Probability to get less then 4:
Computation:
⇒ Total number can be get less then 4 = 4+2+3
⇒ Total number can be get less then 4 = 9
⇒ Probability of an outcome = Total number of outcomes / Total number of experiment
⇒ Probability to get less then 4 = Total number can be get less then 4 / Total number of experiment
⇒ Probability to get less then 4 = 9 / 50
Answer:
Terrance would take 4.2666 hours, Jesse would take 6.25 hours
Step-by-step explanation:
1h/12m per hour x 50m/1h is also 50m/12m per hour = 4.266 hours (round number)
1h/8m per hour is also 50m/8m per hour = 6.25 hours (round number)
Answer:
No. Dividing by 0 is not rational, no number works.
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.
Let's consider a simple example of estimating the value of a general definite integral,

Split up the interval
![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
into

equal subintervals,
![[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]](https://tex.z-dn.net/?f=%5Bx_0%2Cx_1%5D%5Ccup%5Bx_1%2Cx_2%5D%5Ccup%5Ccdots%5Ccup%5Bx_%7Bn-2%7D%2Cx_%7Bn-1%7D%5D%5Ccup%5Bx_%7Bn-1%7D%2Cx_n%5D)
where

and

. Each subinterval has measure (width)

.
Now denote the left- and right-endpoint approximations by

and

, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are

. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints,

.
So, you have


Now let

denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

Factoring out

and regrouping the terms, you have

which is equivalent to

and is the average of

and

.
So the trapezoidal approximation for your problem should be