Answer:
F(4) = 9
Step-by-step explanation:
Notice that for f(4), we need to use the function definition for the partitioned Domain that includes x = 4, and that is the expression :
Therefore:
Answer:
The probability that the last card dealt is an ace is 
.
Step-by-step explanation:
Given : A deck of ordinary cards is shuffled and 13 cards are dealt.
To find : What is the probability that the last card dealt is an ace?
Solution :
There are total 52 cards.
The total arrangement of cards is 52!.
There is 4 ace cards in total.
Arrangement for containing ace as the 13th card is 
.
The probability that the last card dealt is an ace is
Therefore, the probability that the last card dealt is an ace is .
Answer:
I think it would be 15
Step-by-step explanation:
Because if you split the number line into 6 different sections (with the 15 on the left side and the 16 on the right side) and shade/fill in 2 of those, you would be closer to 15.
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:
A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?
You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked
2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.
b - 1,000,000/2500 = 400
400 packages are defective each year
