-4(x-8) + 3(2x+5)
-4x+32 + 6x+15
2x+47
2(10)+47
20+47
67
-3y+10=-5y+8 add 5y to both sides
2y+10=8 subtract 10 from both sides
2y=-2 divide both sides by 2
y=-1
Answer:
The probability that the two rats are from the first litter is 14.28%, and the probability that the two rats are from the second litter is 34.28%.
Step-by-step explanation:
Since a cage holds two litter of rats, and one litter comprises one female and five males, while the other litter comprises seven females and two males, and a random selection of two rats is done, to find the probability that the two rats are from the same litter the following calculation must be performed:
6/15 x 5/14 = 0.1428
9/15 x 8/14 = 0.3428
Therefore, the probability that the two rats are from the first litter is 14.28%, and the probability that the two rats are from the second litter is 34.28%.
90^6=531441000000
80^-4= 1/80^4= 1/4960000
For this, use negative power rule x^-a= 1/x^a
