When 258 college students are randomly selected and surveyed, it is found that 106 own a car. find a 99% confidence interval for the true proportion of all college students who own a car. 0.351 < p < 0.471 0.332 < p < 0.490 0.339 < p < 0.482 0.360 < p < 0.461

0.332 < p < 0.490

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

, where n= sample size

= sample proportion

z* = critical z-value.

As per given , we have

n= 258

Sample proportion of college students who own a car =

Critical z-value for 99% confidence interval is 2.576. (By z-table)

Therefore , the  99% confidence interval for the true proportion(p) of all college students who own a car will be :

Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

your question is incomplete, write all the numbers in the sequence that it tells you.  

step-by-step explanation: