Answer:
The required confidence interval is (3.068,4.732)
Step-by-step explanation:
Consider the provided information.
He plans to use a 90% confidence interval. He surveys a random sample of 50 students. The sample mean is 3.90 alcoholic drinks per week. The sample standard deviation is 3.51 drinks and wants to construct 90% confidence interval.
Thus, n=50,
=3.90 σ=3.51
Now find degree of freedom.

The confidence level is 90% and df=49
Therefore,


Now by using t distribution table look at 49 df and alpha level on 0.05.

Calculate SE as shown:


Now multiply 1.67653 with 0.4964
Therefore, the marginal error is: 1.67653 × 0.4964≈ 0.832
Now add and subtract this value in given mean to find the confidence interval.

Hence, the required confidence interval is (3.068,4.732)
The value of y will be 18 or 144.
Given information:
The given expression is
.
It is required to find the values of y which are whole numbers.
Now, factorize 144 as,

So, for the value of given expression to be a whole number, the value of y should be,
or 144.
For the above values of y, the given expression will be,
![\sqrt[3]{\frac{144}{y} }=\sqrt[3]{\frac{144}{18} }\\=\sqrt[3]{8} =2\\\sqrt[3]{\frac{144}{y} }=\sqrt[3]{\frac{144}{144} }\\=\sqrt[3]{1} \\=1](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7By%7D%20%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7B18%7D%20%7D%5C%5C%3D%5Csqrt%5B3%5D%7B8%7D%20%3D2%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7By%7D%20%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B144%7D%7B144%7D%20%7D%5C%5C%3D%5Csqrt%5B3%5D%7B1%7D%20%5C%5C%3D1)
Therefore, the value of y will be 18 or 144.
For more details, refer to the link:
brainly.com/question/17429689
we have

we know that

so

two solutions
<u>first solution </u>


the first solution is the interval---------> (-1,∞)
<u>second solution</u>




the second solution is the interval---------> (-∞,-1)
therefore
the solution is all real numbers except the number 
the answer in the attached figure
Answer:
85.9
Step-by-step explanation: