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Bunuel
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Pragya777
for the parabola
y= ax^2-bx, (a, b) lies above parabola, then
b>y(a)
= b> a^3-ab
=b>a^3/a+1

For a=4 b>9 (since a, b are both positive single digit integer),
a=1, 2, 3
when a=1, b>1/2, So, b can take any integer from 1 to 9
when a=2, b>8/3, So, b can take any integer from 3 to 9
when a=3, b>27/4, So, b can take any integer from 7 to 9

total possible values= 9*9=81
required probability= 19/81

I don't understand this topic very well; I understand how you get the inequality but I don't understand how you reached b>a^3/(a+1)? Can you please show your working for this solution?
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angysinghdhillon
Pragya777
for the parabola
y= ax^2-bx, (a, b) lies above parabola, then
b>y(a)
= b> a^3-ab
=b>a^3/a+1

For a=4 b>9 (since a, b are both positive single digit integer),
a=1, 2, 3
when a=1, b>1/2, So, b can take any integer from 1 to 9
when a=2, b>8/3, So, b can take any integer from 3 to 9
when a=3, b>27/4, So, b can take any integer from 7 to 9

total possible values= 9*9=81
required probability= 19/81

I don't understand this topic very well; I understand how you get the inequality but I don't understand how you reached b>a^3/(a+1)? Can you please show your working for this solution?

The given parabola is:

Y=aX^2-bX

To be above the parabola means that at a given X coordinate of the parabola the b value will be greater than the Y value of the parabola, meaning that:

b-[aX^2-bX]>0

But the X value of the parabola must also be equal to A in order for the B value to be directly above, so the above can be rewritten as:

B-A^3+AB>0, or rearranging:

B(A+1)>A^3, or

B>A^3/(A+1)


Then the number of cases where this is met is as described in the previous response, totalling 19

Posted from my mobile device
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Regor60
angysinghdhillon
Pragya777
for the parabola
y= ax^2-bx, (a, b) lies above parabola, then
b>y(a)
= b> a^3-ab
=b>a^3/a+1

For a=4 b>9 (since a, b are both positive single digit integer),
a=1, 2, 3
when a=1, b>1/2, So, b can take any integer from 1 to 9
when a=2, b>8/3, So, b can take any integer from 3 to 9
when a=3, b>27/4, So, b can take any integer from 7 to 9

total possible values= 9*9=81
required probability= 19/81

I don't understand this topic very well; I understand how you get the inequality but I don't understand how you reached b>a^3/(a+1)? Can you please show your working for this solution?

The given parabola is:

Y=aX^2-bX

To be above the parabola means that at a given X coordinate of the parabola the b value will be greater than the Y value of the parabola, meaning that:

b-[aX^2-bX]>0

But the X value of the parabola must also be equal to A in order for the B value to be directly above, so the above can be rewritten as:

B-A^3+AB>0, or rearranging:

B(A+1)>A^3, or

B>A^3/(A+1)


Then the number of cases where this is met is as described in the previous response, totalling 19

Posted from my mobile device

Thank you! This really helped!!
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