Answer:

Step-by-step explanation:
Given (64 y Superscript 100 Baseline) Superscript one-half.
Let us write it into an equation.

Apply radical rule:
and 
![\begin{aligned}\left(64 y^{100}\right)^{\frac{1}{2}} &=\sqrt[2]{64 y^{100}} \\&=\sqrt[2]{8^{2} y^{50} y^{50}} \\&=\sqrt[2]{8^{2}\left(y^{50}\right)^{2}} \\&=8 y^{50}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cleft%2864%20y%5E%7B100%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%26%3D%5Csqrt%5B2%5D%7B64%20y%5E%7B100%7D%7D%20%5C%5C%26%3D%5Csqrt%5B2%5D%7B8%5E%7B2%7D%20y%5E%7B50%7D%20y%5E%7B50%7D%7D%20%5C%5C%26%3D%5Csqrt%5B2%5D%7B8%5E%7B2%7D%5Cleft%28y%5E%7B50%7D%5Cright%29%5E%7B2%7D%7D%20%5C%5C%26%3D8%20y%5E%7B50%7D%5Cend%7Baligned%7D)
Hence,
is equivalent to (64 y Superscript 100 Baseline) Superscript one-half.
<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
- The lower limit is 0.5.
- The upper limit is 0.7.
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of
.
60 out of 100 scores are passing scores, hence 
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
- The lower limit is 0.5.
- The upper limit is 0.7.
A similar problem is given at brainly.com/question/16807970