There are two choices for each child: overweight (o) or underweight (u). So if the first child is o the next can be o or u. If the first is u the second can be o or u. This gives four possibilities. Here the first child is the letter noted first and the second is the one listed second:
OO
OU
UO
UU
There are 4 outcomes and if each is equally likely then the probability of each is 1/4. Thus the probability of UU is 1/4
The probability of one underweight and one over weight is 1/2 because in two of the outcomes listed above there is one O and one U (namely OU and UO). Since there are 4 outcomes the probability is 2/4 = 1/2
Week 24,assuming that both machines are cleaned at week 0, because multiples of 12 are 12, 24,36. And multiples of 8 are 8, 16, 24. 24 is also the LCM.
A=-.25+6.7t, s=.75+4.5t when Amir catches up a=s so:
-.25+6.7t=.75+4.5t add .25 to both sides
6.7t=1+4.5t subtract 4.5t from both sides
2.2t=1 divide both sides by 2.2
t=10/22 hr
t≈0.45 hr (to nearest hundredth)
