Since you don't know the value of b about the best that you could do is:
(2b)^4
16b^4
16b*b*b*b
This question is incomplete because it was not written properly
Complete Question
A teacher gave his class two quizzes. 80% of the class passed the first quiz, but only 60% of the class passed both quizzes. What percent of those who passed the first one passed the second quiz? (2 points)
a) 20%
b) 40%
c) 60%
d) 75%
Answer:
d) 75%
Step-by-step explanation:
We would be solving this question using conditional probability.
Let us represent the percentage of those who passed the first quiz as A = 80%
and
Those who passed the first quiz as B = unknown
Those who passed the first and second quiz as A and B = 60%
The formula for conditional probability is given as
P(B|A) = P(A and B) / P(A)
Where,
P(B|A) = the percent of those who passed the first one passed the second
Hence,
P(B|A) = 60/80
= 0.75
In percent form, 0.75 × 100 = 75%
Therefore, from the calculations above, 75% of those who passed the first quiz to also passed the second quiz.
9.82 x 1/10 = 0.982
9.82 x 1/10 x 1/10 = 0.0982
9.82 x 1/10 x 1/10 x 1/10 = 0.00982
9.82 x 1/10 x 1/10 x 1/10 x 1/10 = 0.000982
Therefore:
9.82 x 10^(-4) = 0.000982
4*2=8, 4=number of sections on the spinner, 2=number of sides on a coin
Answer is 1/8 probability
Hope this helps you
I think The answer is 5100
