Answer:
51,111%
Step-by-step explanation:
We know that 45% have a dog but no cat. While 23% have both a dog and a cat.
Then Pr(X has cat|X has dog) = Pr(X has dog and cat)/Pr(X has dog)
= Pr(X has dog and cat)/Pr(X has dog) = 0.23/0.45 = 0.51 ≈ 51,111%
The probability that a student’s family owns a cat if the family owns a dog is 51,111%
I’ll help if u attach the pic of the question:)
Answer:20
Step-by-step explanation:4:9
9x5=45
4x5=20
20:45
Bc
Answer:
24
Step-by-step explanation:
Answer:
D. We are 95% confident that between about 22% and 28% of ESPN Viewers thought the Eagles would still win the division.
Step-by-step explanation:
We have to calculate a 95% confidence interval for the population proportion. The population is made up of ESPN viewers.
The sample proportion is p=0.25.
The standard error of the proportion is:

The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the proportion of ESPN viewers is (0.222, 0.278) or (22.2%, 27.8%).