Answer:
BE=15
Step-by-step explanation:
Since E is the center, BE=2x+1
And AC is 6x-12 with E being a midpoint.
Since AC is 2 segments and BE is only 1 then you multiply 2x+1 by 2 so that
you can find x. You multiply 2x+12 by 2 because E is the midpoint so that means ED is the same value as BE
6x-12=4x+2
BD=AC
6x-4x-12=2
2x=2+12
2x=14
x=7
BE= 2x+1
2(7)=14+1=15
Answer:
23.44%
Step-by-step explanation:
The probability of getting a 4 on the first 2 throws and different numbers on the last 5 throws = 1/6 * 1/6 * (5/6)^5
= 0.01116
There are 7C2 ways of the 2 4's being in different positions
= 7*6 / 2 = 21 ways.
So the required probability = 0.01116 * 21
= 0.2344 or 23.44%.
Can you tell me what the expressions, values and all that are
I ca't help with out the details.
Answer:
The probability that a student plays either basketball or soccer is 19% or 0.19.
Step-by-step explanation:
Let A be the event that student play basketball and B be the event that student play soccer.
![P(A)=8\%](https://tex.z-dn.net/?f=P%28A%29%3D8%5C%25)
![P(B)=14\%](https://tex.z-dn.net/?f=P%28B%29%3D14%5C%25)
It is given that 6 students play on both teams.
![P(A\cap B)=\frac{6}{200}\times 100=3\%](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3D%5Cfrac%7B6%7D%7B200%7D%5Ctimes%20100%3D3%5C%25)
We have to find the probability that a student plays either basketball or soccer.
![P(A\cup B)=P(A)+P(B)-(A\cap B)](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3DP%28A%29%2BP%28B%29-%28A%5Ccap%20B%29)
![P(A\cup B)=8\%+14\%-3\%](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3D8%5C%25%2B14%5C%25-3%5C%25)
![P(A\cup B)=19\%](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3D19%5C%25)
Therefore the probability that a student plays either basketball or soccer is 19% or 0.19.
Question: .9 + 4.15
Work: 4.15 + .90
You are always able to add a zero after the nonzero digit in the decimal places
4.15
+ .90
---------
5.05
Solution: <em>5.05</em>