Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).
<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.
We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

Hence the bounds of the interval are found as follows:


The 95% confidence interval is (0.2316, 0.3112).
More can be learned about the z-distribution at brainly.com/question/25890103
First divided 140 by 100 to find out how much 1 percent is
140 ÷ 100 = 1.4
multiply 1.4 by 45 to find 45 %; 1.4 x 45 = 63
then add 63, which is 45%, to 140, which is 100%, and you get 203, which is 140 + 45%, or 145%
please mark brainliest if this helped you. I hope it did!
Seven has the greatest value as a digit, but 1, 000 has the greatest standard form value.
Answer:
y=2x+2
Step-by-step explanation:
Well,5/6 is closer to one than 5/8 is,just add 1/6+5/6 and then you get 1,adding 1/8 to 5/8 is only 6/8,so that's how you can tell how it is greater! Hope this helps your daughter!