Answer:
6 5/12
Step-by-step explanation:
2 2/3 as a mixed number is 8/3
3 3/4 as a mixed number is 15/4
To make the 2 fractions have a common denominator, they become:
32/12 and 45/12
to add them up it will be: 77/12 which can be written as 6 5/12
We have that
y = 2x²<span> + 2
step 1
</span><span>exchange the value of x for y and the value of y for x
</span>y = 2x² + 2------> x=2y²+2
step 2
clear y variable
x=2y²+2----> 2y²=x-2----> y²=[(x-2)/2]-----> y=(+/-)√[(x-2)/2]
step 3
the inverse is
f(x)-1=(+/-)√[(x-2)/2]
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
The manager already hired 9 people. Let 
be the number of people he still can hire. Since these people will add to the 9 he already hired, when the hiring campaign will be over he will have hired 
people. We know that he can't hire more than 14 people, so the number of people hired must be less than or equal to 14:
If we subtract 9 from both sides, we have
so, the manager can hire at most 5 other people
