Interesting :0 did it rain that much-
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
Answer:
a) Binomial distribution B(n=12,p=0.01)
b) P=0.007
c) P=0.999924
d) P=0.366
Step-by-step explanation:
a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.
b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

Then,

c) In this case, the distribution is B(1200,0.01)

d) In this case, the distribution is B(100,0.01)
We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

Your question is not clear. I must guess what you meant. Did you mean "Find the two numbers?"
If so, let the two numbers be m and n. Then mn=60 and m=n-7.
Then (n-7)(n) = 60, or n^2 -7n - 60 = 0. This factors: (n+5)(n-12) = 0
Thus, n = -5 or n = 12.
check: does mn = 60? Does 12*5 = 60? Yes. Is m (which is 5) 7 less than n (which is 12)? Yes.
Note that you could also check to see whether n=-5 and m=n-7=-5-7=-12 is also a solution.