I cant directly give u the answer but here is some help:
for part a add up and solve the equazions on the whiteboard.
for part b subtract 90 from what you got from part a.
Hope this helps. Sorry i couldnt help more.
Answer:
as a ratio is would be 1:5 so I think it would be I/5 but I'm not sure I'm sorry If you get it wrong
Answer:
y=4
Step-by-step explanation:
first you have y=mx+b
then b=4
and m=0
so it is
y=0x+4
or y=4
Answer:
Option D -
.
Step-by-step explanation:
Given : From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event.
To find : What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
Solution :
Choosing 4 people out of 8 volunteers is ![^8C_4](https://tex.z-dn.net/?f=%5E8C_4)
![^8C_4=\frac{8!}{4!(8-4)!}](https://tex.z-dn.net/?f=%5E8C_4%3D%5Cfrac%7B8%21%7D%7B4%21%288-4%29%21%7D)
![^8C_4=\frac{8\times 7\times 6\times 5\times 4!}{4!\times 4\times 3\times 2}](https://tex.z-dn.net/?f=%5E8C_4%3D%5Cfrac%7B8%5Ctimes%207%5Ctimes%206%5Ctimes%205%5Ctimes%204%21%7D%7B4%21%5Ctimes%204%5Ctimes%203%5Ctimes%202%7D)
![^8C_4=70](https://tex.z-dn.net/?f=%5E8C_4%3D70)
Choosing a group of 4 with Andrew and no karein is given by,
One position is fixed by Andrew and Karein the number of volunteer left is 6.
Rest 3 volunteers is chosen from 6.
Choosing 3 people out of 6 volunteers is ![^6C_3](https://tex.z-dn.net/?f=%5E6C_3)
![^6C_3=\frac{6!}{3!(6-3)!}](https://tex.z-dn.net/?f=%5E6C_3%3D%5Cfrac%7B6%21%7D%7B3%21%286-3%29%21%7D)
![^6C_3=\frac{6\times 5\times 4\times 3!}{3!\times 3\times 2}](https://tex.z-dn.net/?f=%5E6C_3%3D%5Cfrac%7B6%5Ctimes%205%5Ctimes%204%5Ctimes%203%21%7D%7B3%21%5Ctimes%203%5Ctimes%202%7D)
![^6C_3=20](https://tex.z-dn.net/?f=%5E6C_3%3D20)
The probability that Andrew will be among the 4 volunteers selected and Karen will not is given by,
![P=\frac{^6C_3}{^8C_4}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B%5E6C_3%7D%7B%5E8C_4%7D)
![P=\frac{20}{70}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B20%7D%7B70%7D)
![P=\frac{2}{7}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B2%7D%7B7%7D)
The probability that Andrew will be among the 4 volunteers selected and Karen will not is
.
Therefore, option D is correct.
Answer:
0.04845
Step-by-step explanation:
Probability that less than 7 but more than 4 bulbs from the sample are defective :
P(4 < X < 7) = P(x = 5) + P(x = 6)
Using the binomial probability formula :
P(x =x) = nCx * p^x * (1 - p)^(n - x)
From the question :
p = 0.2 ; 1 - p = 0.8 ; n = 11
P(x =5) = 11C5 * 0.2^5 * 0.8^6 = 0.03876
P(x = 6) = 11C6 * 0.2^6 * 0.8^5 = 0.00969
P(4 < X < 7) = P(x = 5) + P(x = 6)
P(4 < X < 7) = 0.03876 + 0.00969
P(4 < X < 7) = 0.04845