Answer:
2 with a remainder of 10
Step-by-step explanation:
11 goes into 32 2 times which is 22. Then there is 10 leftover as a remainder.
Answer:
6c^2(-40 + 5c^3 + 3c^2)
Step-by-step explanation:
These three terms have the common factor c^2. Thus:
-240c^2+ 30c^5+ 18c^4 = c^2(-240+ 30c^3+ 18c^2).
Recognizing that 6 is common to all three terms inside the parentheses, we get:
6c^2(-40 + 5c^3 + 3c^2) This is as far as factoring of this expression can be taken.
Answer:
1/3
Step-by-step explanation:
3^2/5 x 3^-7/5
We know that a^b* a^c = a^(b+c)
3^(2/5 -7/5)
3^ (-5/5) = 3^ -1
We know that a^-b = 1/ a^b
3^-1 = 1 / 3
Jerry did the best at Spanish so the answer is D
If you divide the numerators by the denominators of each class you get his percentage
Math - 45/50 = .9 = 90%
Science- 19/25= .76 = 76%
Music- 8/10 = .8 = 80%
Spanish 19/20 = .95 = 95%
Answer:
D) 7
Step-by-step explanation:
Let y = number of yellow marbles in the bag
For the first marble
P(yellow) = number of yellow marbles over total
=y/12
For the second draw, there are y-1 yellow marbles in the bag, and only 11 marbles left since we keep the first marble
P(2nd marble yellow) = (y-1)/11
The probability that the first and second marble are yellow when we keep the first marble is
y/12 * (y-1)/11 = 5/33
y * (y-1) 5
---------- = ----------
121 33
We can use cross products to solve
33 y(y-1) = 660
Divide each side by 33
y(y-1) = 660/33
y(y-1) = 20
Distribute
y^2 -y =20
Subtract 20
y^2 -y -20 =0
Factor
(y-5)(y+4) =0
Using the zero product property
y = 5 or y = -4
Since we cant have negative marbles
We have 5 yellow marbles
Since there are 12 marbles
12-5 =7
There are 7 green marbles
