Alan

The work is force x distance = mgh = 0.35*9.81*45 = 154.5 joules

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39 = 3 x 13

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As a slight tweak to the problem, suppose the question was "what is the probability that at least one black ball is drawn in three b***s?"

Prob that 1st ball is non-black = 6/9

Prob that next ball is non-black = 5/8

Prob that final ball is non-black = 4/7

So probability of at least 1 black ball = 1 - 6*5*4/(9*8*7) = 16/21

Using DavidQD and Chris's values above:  Probability of at least 1 black ball = 64/84 = 16/21

Using my answer to the original problem:  Probability of at least 1 black ball = 384/504 = 16/21

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I agree, my answer assumed each ball was separately identifiable, whatever the colour.

What was the reasoning behind 69?  

8² - 11*8 + 24/8 - 8 = -29

-8*8 + 69 = 5

These aren't the same!

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They must start with either the 5 or the 6.

Consider those that start with a 5.  

There are 6 digits left that could go in the "tens" position.

 For each of these there are 5 digits left that could go in the "units" position.

 Hence there are 6*5 = 30 possible 3-digit numbers beginning with 5.  

Similar reasoning shows that there are also 30 possible 3-digit numbers beginning with 6.

So there are 60 possibilities in all.

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For the function to be continuous then both parts of it must have the same value at x = 8.

So you must have 8² - 11*8 + 24/8 -8 = -8*8 + k or -29 = -64 + k or k = 35

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Notice the gap when k isn't 35:

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Raka226 is correct, but here's a little more detail of the process:

2(2 - j) = 4 (j + 10)

Distribute the 2 over the 2-j and the 4 over the j + 10:

4 - 2j = 4j + 40

Add 2j to both sides

4 = 6j + 40

Subtract 40 from both sides:

-36 = 6j

Divide both sides by 6:

-6 = j

or just j = -6

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There are 3 black b***s and 6 non-black b***s.

First think of how many ways there are to draw three b***s (without replacement, presumably) such that you get no black b***s.

There are 6 ways to draw the first. For each of these there are 5 ways to draw the second. For each of these pairs there are 4 ways to draw the third.  Thus there are 6*5*4 = 120 ways to draw three b***s such that none are black.

In total there are 9*8*7 =  504 ways to draw three b***s from the bag, irrespective of colour, so there must be 504 - 120 = 384 ways to draw three b***s from the bag such that at least one is black.

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I suspect this is meant to be:

y = 25x + 75

where x is number of months and y is total amount.

If so, then because there are 48 months in 4 years the amount saved will be y = 25*48 + 75 = $1275

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