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Move over golfing fans. One of the 3rd floor teachers has taken a liking to the energetic and lightning quick paced style of bowling. She is so enthused about this sport, developed clearly for superior athletes, that she has purchased her own 6.5 - inch radius bowling ball, The Annihilator.

The diameter of the bowling ball is 13in. The dimensions of the cardboard box that the The Annihilator comes with are 13in for the length, 13in for the width, and 13 in for the height.

The Annihilator is sold in fancy cardboard box with packaging inside that total 14 oz. The box w/ sphere is a measly 20 Ibs. per cubic foot. Using only the dimensions of the cardboard box determine the weight of this pin splitter.

The Annihilator is a whopping $8972.27 per cubic meter (m^3). Using only the dimensions of the cardboard box determine the cost of The Annihilator.

2007-05-10 16:24:01 · 1 answers · asked by Right here Right now 1 in Education & Reference Homework Help

1 answers

We can find the volume of the box to start with, since we have length, width, and height.

V(box) = 13 * 13 * 13
V(box) = 2197 cu. in.
V(box) = 1.271 cu.ft. (just divide by 1728 to convert from cu. in. to cu. ft.)

Since the density of the entire package is 20 lb/cu. ft, we can see that the entire package weighs

m(all) = (1.271 cu. ft)(20 lbs/cu. ft)
m(all) = 25.428 lbs

And since the packaging weighs 14 oz (or 0.875 lbs), then the ball weighs

m(ball) = 25.428 lbs - 0.875 lbs
m(ball) = 24.553 lbs

...which is a little larger than regulation bowling balls!

As for the cost, you have $8927.27 per m^3 - is that for the entire box or for the ball itself? However you calculate it, you just need to convert your dimensions to meters:

V(box) = (2197 cu. in.)(1 m/39.37 in.)^3
V(box) = 0.036 m^3

Then you can figure your cost:

cost = ($8927.27/m^3)(0.036 m^3)
cost = $321.40

If you just need the volume of the ball to calculate the cost, then you'd use

V(ball) = (4/3)pi*r^3
V(ball) = (4/3)pi(6.5 in)^3
V(ball) = 1150.347 in^3
V(ball) = 0.019 m^3

...and redo the calculation above. Even so, that's an expensive ball!

2007-05-13 03:19:52 · answer #1 · answered by igorotboy 7 · 0 0

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