If you're asking for the model for the second problem (I didn't see any unsolved), here it is!
This is a binomial probability situation, since a dog either is adopted or is not adopted. The chances of a dog's being adopted in 0.20. Here we're speaking of 9 visits. Thus, n=9, p=0.20.
One way of doing this problem is to calculate the probability that ONE dog will be adopted, and then that that TWO dogs will be adopted, and so on, up to NINE dogs. Add together these nine probabilities to get your answer.
But a better (faster) approach would be to calculate the probability that ZERO dogs will be adopted, and then to subtract this from 1.000.
Using my TI-84Plus calculator, I figured that P(0 dogs will be adopted) is binompdf(9,0.20,0), or 0.134. Subtracting this from 1.000, we get 0.866 (answer to this problem).
15 electricians worked for 24 days to the whole job, now, there are 15 of them, so on any given day, each electrician worked one whole day, in 24 days, that one electrician worked 24 days total.
now, there were 15 electricians on any given day though, since each one of them worked the whole day that one day, so the "days work worth" on a day is 15, so the house gets 15days worth of work because of that.
so how many "days worth" did all 15 do on the 24 days, well, 15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15+15, namely 15 * 24, or 360 days worth of work.
since it takes 360 days worth of work to do the whole wiring, in how many days would 18 electricians do it? 360/18.
(4,-8)
Step-by-step explanation:
from -5 to -1 you would have to count from 1 to 4. and do the same for -8 to -16
