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)
Answer:
10.84
Step-by-step explanation:
First we multiply
1.4+(3.2×2.95)
1.4+9.44=
=10.84
The y intercept is 1.5x+22.7 + the linear equation which is 22
(x)1100+(y)113 x depends on the amount of computers is sold so x will change. y will be the amount of hard drives sold and then you add the total amount together.
1. Perfect square trinomials, are 2nd degree polynomials, of the form

so that

, which can be written as perfect squares.
2. For example

3. Thus

are perfect square trinomials.
4.

5. In the first case -b=20, so b=-20. In the second case, -b=-20, so b=20.
6. b∈{-20, 20}