Answer:
Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .
Step-by-step explanation:
We are given that let X be the number that do some or all of their banking on the Internet.
Also; Mean, 
= 310/1000 or 0.31 and Standard deviation, 
= 14.63/1000 = 0.01463 .
We know that Z = 
~ N(0,1)
Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);
P(X <=0.13) = P( <= 
) = P(Z <= -12.303) = P(Z > 12.303)
Since this value is not represented in the z table as the value is very high and z table is limited to x = 4.4172.
So, after seeing the table we can say that this probability is approximately less than 0.0005% .
Answer:
15 1/2
Step-by-step explanation:
Answer:
37.5
Step-by-step explanation:
50+160=210.
210 is a 31.25% increase from 160
Multiply 120 by 1.3125 to get 157.5
subtract 120 from 157.5 to get a 37.5km increase or a 31.25% increase :)
Answer: Choice C) 124 square cm
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Explanation:
Let's calculate the area of the trapezoid shown
b1 and b2 are the parallel bases; h is the height of the 2D trapezoid
b1 = 2
b2 = 5
h = 1.5
A = h*(b1+b2)/2
A = 1.5*(2+5)/2
A = 1.5*7/2
A = 10.5/2
A = 5.25
The area of one 2D trapezoid is 5.25 sq cm
There are two of these trapezoids that form the base faces of the trapezoidal prism. So the total base area is 2*5.25 = 10.5 sq cm
Keep this value (10.5) in mind. We'll use it later.
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Now onto the lateral surface area (LSA)
It turns out that the formula for the LSA is
LSA = p*d
where
p = perimeter of the trapezoid shown
d = depth or height of the 3D trapezoid (I'm not using h as it was used earlier)
This formula works for any polygonal base. It doesn't have to be a trapezoid.
In this case the perimeter is,
p = 1.7+2+2.65+5
p = 11.35
So
LSA = p*d
LSA = 11.35*10
LSA = 113.5
Add this LSA to the base area found earlier
10.5+113.5 = 124
The total surface area is 124 square cm
You need 710 ft of fencing.
