A study found that in a random sample of 50 burglaries, 16% of the criminals were arrested. In a random sample of 50 car thefts, 12% of the criminals were
arrested. The researchers wants to find a 90% confidence interval for the
difference between the percentages of people who committed burglaries and
were arrested and the percentages of people who committed car thefts and
were arrested. What is the point estimate in this situation? (Assume the
sampling distribution is normal).

won't that be plagrisim thought?

step-by-step explanation:

answer: rolling a number cube and flipping a coin.

step-by-step explanation:

this is a direct variation proportion. y = kx

y is the total amount due, k is the constant of variation, and x is the number of texts sent.

plug in the 3.20 and the 16 to solve for the "k". this is the constant of proportionality. it is the value that will not change.

3.20 = k(16). solving you get $.20 per text.

the equation would then be y = .20x

i would say c

step-by-step explanation: