Answer:
The 95% confidence interval for the true proportion of university students who use laptop in class to take notes is (0.2839, 0.4161).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population proportion <em>P</em> is:

The information provided is:
<em>x</em> = number of students who responded as"yes" = 70
<em>n</em> = sample size = 200
Confidence level = 95%
The formula to compute the sample proportion is:

The R codes for the construction of the 95% confidence interval is:
> x=70
> n=200
> p=x/n
> p
[1] 0.35
> s=sqrt((p*(1-p))/n)
> s
[1] 0.03372684
> E=qnorm(0.975)*s
> lower=p-E
> upper=p+E
> lower
[1] 0.2838966
> upper
[1] 0.4161034
Thus, the 95% confidence interval for the true proportion of university students who use laptop in class to take notes is (0.2839, 0.4161).
This question does not make much sense. Brigid has already picked 112 bushels, and she picks at a rate of 58 bushels per hour, the scenario question does not make any sense. 58h + 112 = 5...? What does this mean? But whatever the left side of the equation is, first subtract 112 from both sides and divide by 58, then you will get the value of "h."
Since the equations for area is A= L*W
A= 9*4= 54
I have no clue what the answer is