Answer: the probability that no sections receive a hard test is 0.1022
Step-by-step explanation:
Given that;
total number of exams = 20
hard exams = 8
reasonable = 7
easy exam = 5
so total exams that are bot hard = 7 + 5 = 12
we will make use of the probability mass function of hyper geometric distribution;
so the probability that no section of the received a hard test can be determined as follows;
P( x=0; N=20, n=4, k=8 ) = [ ⁸C₀ × ²⁰⁻⁸C₄₋₀ ] / ²⁰C₄
= [1 × ¹²C₄ ] / ²⁰C₄
= [1 × (12!/(4!×(12-4)!) ] / (20!/(4!×(20-4)!)
= [1 × 495 ] / 4845
= 495 / 4845
= 0.1022
Therefore the probability that no sections receive a hard test is 0.1022
Part A:
The area of the rectangle will be the length times the width, so 4•(x+3). Using distributive property, you'll multiply every term inside the parentheses by 4.
The area of the rectangle can be expressed by 4x+12.
Part B:
Multiply 4•2 and (x+3)•2, then multiply them together.
8(2x+6) Now use distributive property to find your answer simplified.
16x+48 is your answer
To find r, you can multiply both sides by πr²
You would then get:
513πr² = 79,000
Then to isolate r, you can divide 513π on both sides.
(513πr²)/(513π) = 79,000/(513π)
r² = 79,000/(513π)
To finally get r, you can take the square root of both sides.
C is the answer to your question
Ummm hmm this is a hard one..... I believe it’s 88?