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
We can use ratios and the cross-multiply-divide to find this.
The ratio 21:5 is avaliable via the question. We then need to compare that to the ratio x:100, the 100 being the percent and the x being the number of airplane parts.
21/5 = x/100
We can then solve for x to find the number of airplane parts. First we multiply by 100 on both sides to get 2100/5 = x. Therefore x = 420 and there are 420 airplane parts.
Answer:
x = -7 and x = 3
Step-by-step explanation:
x² + 4x - 21 = 0 factors as follows: (x + 7)(x - 3) = 0.
Then x = -7 and x = 3.