Two of the desserts are sugar/free. Three of the drinks are sugar-free. A meal deal combination is selected at random. Work out the probability that both the desserts and the drink are sugar free. Give your answer as a fraction

8/15

Step-by-step explanation:

You need part a of the question to answer this question correctly. In part a of the question it is said there is 6 sandwiches, 3 desserts and 5 drinks.

Two of the desserts are sugar free (2/3).

Four of the drinks are sugar free (4/5)

Times these fractions together, 2/3x4/5= 8/15.

Hope this helps :)

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)