Answer: P(B|G) = 3/5 = 0.6
the probability that the guest is the friend of bride, P(bride | groom) is 0.6
Complete Question:
The usher at a wedding asked each of the 80 guests whether they werea friend of the bride or of the groom. The results are: 59 for Bride, 50 for Groom, 30 for both. Given that the randomly chosen guest is the friend of groom, what is the probability that the guest is the friend of bride, P (bride | groom)
Step-by-step explanation:
The conditional probability P(B|G), which is the probability that a guest selected at random who is a friend of the groom is a friend of the bride can be written as;
P(B|G) = P(B∩G)/P(G)
P(G) the probability that a guest selected at random is a friend of the groom.
P(G) = number of groom's friends/total number of guests sample
P(G) = 50/80
P(B∩G) = the probability that a guest selected at random is a friend is a friend of both the bride and the groom.
P(B∩G) = number of guests that are friends of both/total number of sample guest
P(B∩G) = 30/80
Therefore,
P(B|G) = (30/80)/(50/80) = 30/50
P(B|G) = 3/5 = 0.6
The formula is V= nR^2H
V=(3.14)(6.5)^2(16)
V=(3.14)(42.25)(16)
V=(3.14)(676)
V=2122.64cm ^2
330 would be the 6th term of the sequence
Best to factor 5 out of the first 2 terms:
5x^2-7x+2=0 => <span>5(x^2 - [7/5]x) +2=0
Take half of [-7/5]: That'd be -7/10.
Square 7/10, add it to </span>5(x^2 - [7/5]x) +2=0 and then subtract it:
5(x^2 - [7/5]x)+ 49/100 - 49/100 ) +2=0
Then we have 5(x- [7/10] )^2 - 49/100 ) + 200/100 = 0
5(x-7/5)^2 - 245/100 + 200/100 = 0
5(x-7/5)^2 - 45/100 = 0
Dividing all terms by 5, x-7/5 = plus or minus sqrt( 9/100 )
x-7/5 = plus or minus 3/10
Then one root is x = 7/5 + 3/10, or x= 17/10 (answer #1)
The other is x = 7/5 - 3/10, or x = 14/10 - 3/10, or x = 7/10 (answer #2)
Answer:histogram 4
Is this for stats? each number is going by 5 and looking at the numbers you can easily eliminate 2 and 3 as there is nothing in the area of 50-60 that only leaves 1 and 4 look at the numbers and how they add up in the frequency from 70-75 there are two numbers that meet the criteria. Histogram 1 only shows one number for that section so i assume that it should be number 4.
