Given that the first spinner has three equal sectors labelled 1, 2 and 3; and the second spinner has equal sectors labelled 3, 4, 5 and 6.

The number of possible outcomes that do not show a 1 on the first spinner is 2 (i.e. the first spinner shows 2 or 3).

The number of possible outcomes that the second spinner show the number 4 is 1 (i.e. the second spinner shows 4)

In probability, the word 'and' goes with multiplication.

Therefore, the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is given by 2 x 1 = 2 possible outcomes.

i.e. the first spinner shows the number 2 and the second spinner shows the number 4 or the first spinner shows the number 3 and the second spinner shows the number 4.

5 0

3 years ago

Prove the statement holds for all positive integers: 2 + 4 + 6 + ... + 2n = n² + n

crimeas [40]

Proving a relation for all natural numbers involves proving it for n = 1 and showing that it holds for n + 1 if it is assumed that it is true for any n.

The relation 2+4+6+...+2n = n^2+n has to be proved.

If n = 1, the right hand side is equal to 2*1 = 2 and the left hand side is equal to 1^1 + 1 = 1 + 1 = 2

Assume that the relation holds for any value of n.

2 + 4 + 6 + ... + 2n + 2(n+1) = n^2 + n + 2(n + 1)

= n^2 + n + 2n + 2

= n^2 + 2n + 1 + n + 1

= (n + 1)^2 + (n + 1)

This shows that the given relation is true for n = 1 and if it is assumed to be true for n it is also true for n + 1.

By mathematical induction the relation is true for any value of n.

6 0

2 years ago