# A junior basketball has a diameter of approximately 7 in., and a regulation basketball has a diameter of approximately 9.5 in. about how many times as great is the volume of the regulation basketball as the volume of the junior basketball?

Answer: Scale factor  = 9.5 / 7
the ratio of the volumes will be 9.5^3 / 7^3

=  2.5  to nearest thousandth

So the volume the regulation ball is  2.5 times the volume of the junior one.

## Related Questions

what must be subtracted from 4x4 - 2x3 - 6x2 + x - 5 so that the answer is exactly divisible by x + x - 2

To solve this problem, what we have to do is to divide the whole equation 4 x^4 – 2 x^3 – 6 x^2 + x – 5 with the equation 2 x^2 + x – 1. Whatever remainder we get must be the value that we have to subtract from the main equation 4 x^4 – 2 x^3 – 6 x^2 + x – 5 for it to be exactly divisible by 2 x^2 + x – 1.

By using any method, I used long division we get a remainder of -6.

Therefore we have to subtract -6 from the main equation which results in:

4 x^4 – 2 x^3 – 6 x^2 + x – 5 – (-6) = 4 x^4 – 2 x^3 – 6 x^2 + x + 1

What is y-2=-2.5(x-1) in STANDARD FORM? (It has to be in Ax+By=C form)

Y-2=-2.5(x-1)
Y-2=-2.5x+2.5
2.5x+y=4.5

Make a conjecture about the sum of the first 15 positive even number

I conclude that the sum will be even because any even number can be represented by 2n where n is a whole number
and even numbers are 2 apart, so
the sum of the first 15 are
2n+2(n+1)+2(n+2) etc until we get to 2(n+14)
we can undistribute the 2 from all of them and get
2(n+n+1+n+2...n+14)
and we are sure that whatever is in the parenthasees is a whole number because whole+whole=whole
therefor, the sum is even

if you did want to find the sum then
an=2n
the 15th even number is 30
the first is 2
S15=15(2+30)/2=15(32)/2=15(16)=240
which is even

Two arcades have different pricing structures. Arcade A costs \$5 to enter and \$0.60 per game. Arcade B costs \$7 to enter and \$0.40 per game. How many games do you have to play in order for the two arcades to be the same price?

you would have to play 10 games

10 games

Step-by-step explanation:

Let C be cost and x be number of games

Arcade A: C = 5 + 0.6x

Arcade B: C = 7 + 0.4x

for both arcades to be the same price, Both C's must be equal, hence we equate the right side of both equations:

5 + 0.6x = 7 + 0.4x   (subtract 5 fromboth sides)

0.6x = 7 + 0.4x - 5

0.6x = 2 + 0.4x  (subtract 0.4x from both sides)

0.6x - 0.4x = 2

0.2x = 2 (divide both sides by 0.2)

x = 2 / (0.2)

x = 10