[needed very much 99 points ] what is the theoretical probability that the family has two dogs or two cats? describe how to use two coins to simulate which two pets the family has. flip both coins 50 times and record your data in a table like the one below. result frequency heads, heads heads, tails tails, heads tails, tails total 50 based on your data, what is the experimental probability that the family has two dogs or two cats? if the family has three pets, what is the theoretical probability that they have three dogs or three cats? how could you change the simulation to generate data for three pets?

let dogs be heads. Let cats be tails. A coin has two sides, in which you are flipping two of them. Note that there can be the possible outcomes

h-h, h-t, t-h, t-t.

How this affects the possibility of two dogs & two cats. Note that there are 1/2 a chance of getting those two (with the others being one of each), which means that out of 4 chances, 2 are allowed.

2/4 = 1/2

50% is your answer

hope this helps

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

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