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
It is this 100 60 40 30 800
-3x^2-2x (+-)6=0
ax^2+bx+c
a=-3
b=-2
c= either positive or negative 6 (there is no plus or minus in your question.)
What are the answer choices? It might help.
Answer:
GURL MY MATH TEACHER GAVE ME THE SAME SHEET A MONTH AGOO
YASS
1.) x is less than or equal to 3
2.) x greater than or equal to 2
3.) x < -1
4.) x greater than or equal to 3
5.) x>2
6.) x greater than or equal to 1
Answer:
Area of semicircle =( π r2)÷ 2
= 1/2 × 22/7 × 11 ×11
= 11 × 17.5
= 19.25 CM2
