Answer: the actual answer it d. y=-1/3x-5
Step-by-step explanation:
The wording of the question is a little strange. The percentage of dog owners is already estimated at 52%, so no simulation seems useful for that. However, if you want to simulate dog ownership within any given household, you want to apply some algorithm to the given numbers so that about 52% of the time you will see the equivalent of "owns at least one dog."
We assume the numbers are uniformly distributed on 00000 .. 99999. You could, for example, take 4 of the 5-digit numbers (20 digits total), divide them into pairs of digits, and declare "owns at least one dog" if the pair of digits is 51 or less.
For example, the first set of 4 numbers so divided will be ...
95 91 15 52 41 74 05 34 10 02
and "owns at least one dog" would then be ...
no no yes no yes no yes yes yes yes . . . 6 of the 10 simulated households
_____
This sort of approach can work well if you're simulating something described by a percentage. If there is some other ratio involved, say 3 out of 248, then you could throw out any number that is 99944 or higher (403*248) and look at the remainder when dividing by 248. If it is 2 or less, your condition is satisfied.
Making use of random number tables is a bit of an art. The idea is to choose the algorithm for processing the numbers so that the desired distribution is obtained. If the desired distribution is non-uniform, then there are ways to apply functions to the numbers or simply put them in bins of different width so that you get the desired simulated result.
A. Y = 10+18
B. X*3=12
C. X/3+4=5
A. 10+18= 28
B. 12/3=4
C 5-4 = 1, 1*3= 3 X=3
-4, -1/5, 1/5, and 1 1/5?
