Answer:5053.71
Step-by-step explanation:
Answer:
With a .95 probability, the margin of error is = 0.392
Step-by-step explanation:
Given -
In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.
Sample size ( n ) =81
Standard deviation 
= 1.8 hours
= 1 -.95 =.05
= 
= 1.96 (Using Z table)
With a .95 probability, the margin of error is
Margin of error = 
= 
= 
Margin of error = 0.392
Answer:
184 in²
Step-by-step explanation:
Given :
Width, w = 6 inches
Length, l = 10 inches
Height, h = 2 inches
To obtain how much wrapping paper is needed ; we take the surface area of the box
Surface area = 2(lw + lh + wh)
Surface area = 2((6*10) + (6*2) + (10*2))
Surface area = 2(60 + 12 + 20)
Surface area = 2(92)
Surface area = 184 in²
The amount of wrapping paper needed = 184 in²
Write 2 equations:
a +b = 39
and b = 1/2a + 6
Now replace b in the first equation with the second one:
a + 1/2a + 6 = 39
Combine like terms:
1 1/2a + 6 = 39
Subtract 6 from each side:
1 1/2a = 33
Divide both sides by 1 1/2:
a = 33 / 1 1/2 = 33 / 1.5
a = 22
Now replace a with 22 in the first equation and solve for b:
22 + b = 39
b = 39 -22
b = 17
The two numbers are 17 and 22.
The equivalent expressions are: A and C
