The american veterinary association claims that the annual cost of medical care for dogs averages $100 with a standard deviation of $30. the cost for cats averages $120 with a standard deviation of $35. some basic algebraic and statistical steps show us that the average of the difference in the cost of medical care for dogs and cats is then $100 – $120 = –$20. the standard deviation of that same difference equals $46. if the difference in costs follows a normal distribution, what is the probability that the cost for someone's dog is higher than for the cat? group of answer choices

The probability that the cost for someone's dog is higher than for the cat is 33.2%.

Step-by-step explanation:

With the data given, we know that the difference in cost of medical care of dogs and cats have a normal distribution with μ=-20 and σ=46.

To know the probability of the cost for the dog is higher than the cat, we have to calculate the probability of P(d>0).

Then we have to calculate z, and look up in a standarized normal distribution table.

Calculate z:

The probability of the difference being higher than 0, we have:

We can say that the probability that the cost for someone's dog is higher than for the cat is 33.2%.

im with everythinglos so yeahhhhh

step-by-step explanation:

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !