There are "n" delicious sweets in the bag. You know that 6 of them are orange and the rest are yellow. Hannah takes a random swe
et from the bag and eats it. Then she takes a second sweet from the bag and eats that one. You know that the probability that Hannah eats two of the orange sweets is 1/3. How many sweets were in the bag before Hannah ate any
At the point when Hannah takes her first sweet from the sack, there is a 6/n chance it is orange. This is because that there are 6 orange desserts and n desserts altogether. When Hannah takes out her second sweet, there is a 5/(n-1) chance that it is orange. This is because there are just 5 orange desserts let alone for an aggregate of n-1 desserts. The possibility of getting two orange desserts in succession is the main likelihood increased by the second one: 6/n x 5/n–1 The question lets us know that the shot of Hannah getting two orange desserts is 1/3. So: 6/n x 5/n–1 = 1/3 Now, rearrange this problem. (6x5)/n(n-1) = 1/3 This gets to be: 30/(n² – n) = 1/3 Times by 3 on both sides: 90/(n² – n) = 1 What's more, doing likewise with (n² – n): So (n² – n) = 90 Our answer is: n² – n – 90 = 0
it affects it because your going from smaller tires witch make it easier to make turns go faster things like that when you put bigger tires on the car it is gonna raise the car and have a whole differnt outlook
