Answer:
No
Step-by-step explanation:
The answer is no because no pocket can be empty and there isn't enough money to satisfy the condition. At least, one dollar must be stored in each pocket but the number (integer) of dollars in each pocket is different.
Let's store the minimum amount of dollars in the pockets while satisfying the condition. Place 1 dollar in the first pocket. The second pocket must have 2 dollars (it can't be 1 dollar, it must be a different number of dollars). The third pocket must have third dollars.
Repeating this process, the ninth pocket must have 9 dollars. At this moment, we have arranged 1+2+3+4+5+6+7+8+9=45 dollars in our pockets. But we only had 44 dollars! Plus, the tenth pocket is still empty.
If you store more dollars on the first, second, nth pockets, you will just run out of money more quickly than in our process above. so it's impossible to arrange the money in such way.
The probability of drawing a blue card is equal to the quotient when the number of blue cards is divided by the total number of cards. This is, 16/38 which is also equal to 8/19. Thus, the probability of drawing a yellow card is 8/19. The answer is letter C.
F(x) = 1/(x+2) & g(x) = x/(x-3)
(f(x) + g(x) = 1/(x+2) + x/(x-3). Reduce to same denominator:
1/(x+2) + x/(x-3) =(x-3) + x(x-3)/(x+2).(x-3) ==> (x²+3x-3)/(x+2).(x-3)
L+S=8, L=8-S
500L+200S=2200, using L from above in this equation gives you:
500(8-S)+200S=2200
4000-500S+200S=2200
4000-300S=2200
-300S=-1800
S=6, and since L=8-S:
L=2
So Rita bought 2 large packets and 6 small packets.
