HELP !! Calculate the cost of manufacturing a standard cereal box if cardboard costs $0.05 per square inch.
2 answers:
Answer:
The manufacturing cost is $14.50
Step-by-step explanation:
To calculate the cost for manufacturing a standard cereal box, we must take in account its surface area. The volume will be used to estimate the amount of cereal, the box can contain. The box is a cuboid as standard cereal boxes are cuboid.
So, we will consider the surface area. Given are:
Length- 8 inches
Width- 3 inches
Height- 11 inches
We will find the surface area as :
=
= square inches
Given is - the cost of manufacturing the box is $0.05 per square inch.
So, cost of manufacturing the box with 290 square inches area =
dollars.
The answer is $14.50
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Without replacement from a small (8) number of balls means the binomial distribution is NOT applicable since the probability of each pick is not constant.
We can either use the hypergeometric distribution, or solve from first principles.
For simplicity for the given problem, we will solve it using the hypergeometric distribution.
HypergeometricUse hypergeometric distribution where:a=number of target items selectedA=total number of target items in the "pool" (3 white balls)b=number of non-target items selected (B=total number of non-target items in the "pool" (5 blue balls)Then
where 
= combination of r items selected from n,A+B=total number of items = 3+5 = 8a+b=number of items selected=4
Case 1 : 2 white balls, => a=2, b=2, A=3, B=5
P(2W)=C(A,a)*C(B,b)/C(A+B, a+b)
=C(3,2)*C(5,2)/C(8,4)
=3*10/70
=3/7
Case 2: 3 white balls => a=3, b=1, A=3, B=5
P(3W)=C(3,3)*C(5,1)/C(8,4)
=1*5/70
=1/14
Therefore , by the rule of addition, the probability that two or three of the balls are white
P(2W ∪ 3W) = 3/7 + 1/14 = (6+1)/14 = 7/14 = 1/2
We can either use the hypergeometric distribution, or solve from first principles.
For simplicity for the given problem, we will solve it using the hypergeometric distribution.
HypergeometricUse hypergeometric distribution where:a=number of target items selectedA=total number of target items in the "pool" (3 white balls)b=number of non-target items selected (B=total number of non-target items in the "pool" (5 blue balls)Then
Case 1 : 2 white balls, => a=2, b=2, A=3, B=5
P(2W)=C(A,a)*C(B,b)/C(A+B, a+b)
=C(3,2)*C(5,2)/C(8,4)
=3*10/70
=3/7
Case 2: 3 white balls => a=3, b=1, A=3, B=5
P(3W)=C(3,3)*C(5,1)/C(8,4)
=1*5/70
=1/14
Therefore , by the rule of addition, the probability that two or three of the balls are white
P(2W ∪ 3W) = 3/7 + 1/14 = (6+1)/14 = 7/14 = 1/2