I believe the answer would be C; but I'm not sure if this is the right category for this question?
Answer: a) z=1.43
b) 0.1528
Step-by-step explanation:
The given set of hypothesis :


Since the alternative hypothesis
is two-tailed , so we perform two-tailed test.
Also, it is given that : A random sample of n=76 Americans found 28 with brown eyes.
Sample proportion: 
a) The z-statistic would be :-


b) P-value for two-tailed test = 2P(Z>|z|)= 2P(z>|-1.43|)
=2P(z>1.43)
=2(1-P(z≤1.43)
=2-2P(z≤1.43)
= 2-2(0.9236)[Using standard z-table]
= 2-1.8472=0.1528
Hence, the P-value of the test= 0.1528
<span>Prime numbers are the numbers that are bigger than one and cannot be divided evenly by any other number except 1 and itself. If a number can be divided evenly by any other number not counting itself and 1, it is not prime and is referred to as a composite number. Prime numbers are whole numbers that must be greater than 1. Zero and one are not considered prime numbers. Learn how to determine which numbers are prime.
</span>This was not copied from a website or someone else. This was from my last year report.
step-by-step explanation:
every time we see "a number" or "the same number," we can add a variable (we'll use x here). rewriting the problem, we get: 8 less than 2 times x is gr8er than the sum of x and 9.
next, we can work backwards to write an inequality. we know from the word problem above that the number on the left side of the equation is 8 less than 2x, so we can rewrite that part like this: 2x-8.
for the right side of the equation, all we have to do is simplify "the sum of x and 9" to x + 9. so now the inequality should look like this: 2x - 8 > x + 9.
now we'll solve the inequality just like an equation: subtract x from both sides of the inequality, so now we have: x - 8 > 9.
then, we can add 8 to both sides to get the x by itself, and we get the answer: x > 15
hope this helps! <em>:)</em>
I believe the answer is $2.12 (rounded to the nearest cent) explanation: divide the price by the pound for each candy then subtract.