I think the answer is B. Or A.
Answer:
5 < x
Step-by-step explanation:
24<5x-1
Add 1 to each side
24+1<5x-1+1
25 < 5x
Divide by 5
25/5 < 5x/5
5 < x
The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.
95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.
Answer:
Step-by-step explanation:
I'm guessing at this a bit.
The final score will depend on what is par for each hole and/or for the entire course. The scores for the first four rounds were all negative. That means Ko required fewer strokes to hit the hole than the number considered "par" for the course.
Let P stand for the Par of the course. Then the score for each round would be:
P-3
P-4
P-7, and
P-2
The sum: 4P - 16 is her score for the first four rounds. I have no clue if there are more than 4 rounds, but if there are, the equation would be:
For n rounds:
nP - ∑
where ∑
is the sum of over/under pars for the n rounds.