Math Is Fun Forum

It would be so straightforward if you just used a tree diagram (which is what I keep saying).

You actually missed three (because you listed S–D–C–F and S–G–B–A–E–F twice). Two of them are S–A–B–G–C–D–E–F and S–A–B–G–S–D–E–F. The other is a path beginning with S–G – there should be 7 distinct S–G paths but you listed only 6 distinct ones. Check your S–G paths again.

Just use a tree diagram. From S, it can go to A, G or D. If A, it can go from it to B or E; from B it can go to G or F, etc. Exploring all the possible branches from S–A, you find there are 7 paths. S–G is exactly the same as S–A, so there are also 7 paths for S–G. And S–D is simple: it’s either S–D–E–F or S–D–C–F, just 2 ways.

3.1%?

(i) One way to prove the statement wrong is with a counterexample.

Suppose you have 100 cows each producing 10 litres of milk. So you get 1000 litres of milk in total. If each cow produces 20% more milk, you would get 12 litres of milk per cow, giving 1200 litres in total.

But if you reduce your herd by 20%, you would have 80 cows; if each cow continues to produce 12 litres of milk, that would only yield 960 litres – less than the 1000 litres previously.

(ii) To put the error right, simply change the statement to the following one:

"If you produce 20 percent more milk per cow, you cannot decrease your herd by 20% to produce the same amount of milk."

(If you want to decrease your herd and produce at least the same amount of milk as before, you should decrease your herd by no more than (200⁄3)%. This I leave for you to work out as an exercise. )

No. The error is that it is 100 times more or 9900% more rather than 100% more.

I would probably say nothing to the shop assistant, but just pay the marked price less 40% and leave the shop quickly before the shop assistant discovers his error!

Let me explain why.

Okay, suppose each shirt costs 10 euros. A 20% reduction is a reduction of 20 ∕ 100 × 10 = 2 euros, so each shirt now costs 10 − 2 = 8 euros. I buy two shirts, which should cost me 16 euros. Now, if there were no reduction, I would have to pay 20 euros instead. The percentage difference is therefore (20 − 16) ∕ 20 × 100% = 20%, same as for just one shirt.

In fact, no matter how many shirts I buy, the percentage difference will always be 20%.

But, if the shop assistant wants to knock off 40% for two shirts, this would be a reduction of 40 ∕ 100 × 20 = 8 euros – which means I am only charged 20 − 8 = 12 euros! This would be less than I should have paid! So why should I complain …

0[sup]0[/sup] is defined as 1. Try it on a calculator.

And

so the graph is not right-continuous at x = 0.

Well, I plotted the graphs of y = 1.4[sup]x[/sup] and y = −x[sup]2[/sup]+4x+1 and found two solutions. One of them is x = 0; the other solution lies between x=3 and x=4.

Believe it or not, Mozart’s original first name was not Wolfgang, but Johannes.

He was christened Johannes Chrysostomus Wolfgangus Theophilus Mozart. Not surprisingly, he found it too long; so he shortened his first names to Wolfgang Amadeus (also replacing the Greek Theophilus (meaning “God lover”) with the Latin equivalent).

Why did Mozart’s parents gave him that awkward name on his birth? Well, maybe they were psychic and foresaw that their son was going to make a big name for himself.

There. That’s just some off-topic trivia for you.

Either n is divisible by 3 or it’s not.

If it is, fine. If not, then n = 3k+1 or n = 3k−1 for some integer k. It then follows that (n+2) or (n+1) respectively would be divisible by 3.

So one of n, (n+1), (n+2) is always divisible by 3. Hence the product n(n+1)(n+2) is always divisible by 3.

Well, in trying to compute P (the probability that the game never ends), I notice this pattern:
(i) If the sequence HT appears, the next throw must be T (otherwise Bill would win and the game would end)
(ii) If the sequence HH appears, the next two throws must be TT (in order for the game to continue)
(iii) If we have the sequence TT. the process can start all over again as if nothing had happened. This is indicated by an asterisk in the diagram below.

Still working on it …