Answer:
Step-by-step explanation:
6/14 = -9/21 = -12/28 = -15/35 = -18/42 = -21/49 = -24/56 = -27/63 = -30/70 = -33/77 = -36/84 = -39/91 = -42/98 = -45/105 = -48/112
Answer:
Yes, he earned a Varsity letter because he played in 61% of the games.
Step-by-step explanation:
To know the percent of games he played out of the total, we can do it by dividing the games he played by the total of games and multiply this by 100% to get the percent of games he played:

We solve:
0.611111*100% = 61.11%
So we know that he earned a letter because he played in 61.11% of the games, more than he needed.
Answer is <span>sqrt 17
I am 101% sure.
Thank me later</span>
Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:

In which:
is the sample proportion.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
The sample size and the estimate are given by:

Hence:


The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).
More can be learned about the z-distribution at brainly.com/question/25890103
When i calculated it , it says 993.813