In 2012, about 24% of high-school seniors reported binge drinking (defined as five or more drinks in a row in the past two weeks), a substantial drop since the late 1990s. a simple random sample of 500 high-school seniors is to be taken. the standard error is 0.019. what is the 95% confidence interval for the proportion of high-school seniors in the sample who would report binge drinking? (0.24 - 1.96*0.019, 0.24 + 1.96*0.019) o (0.24 - 1.645*0.019,0.24 + 1.645*0.019) (0.019 - 1.96*0.24, 0.019 +1.96*0.24) (0.019 - 1.645*0.24, 0.019 +1.645*0.24)

(0.24 - 1.96*0.019, 0.24 + 1.96*0.019)

Step-by-step explanation:

We know that the confidence interval population proportion (p) is given by :-

, where = sample proportion.

z* = Critical value (Two -tailed)

SE = standard error

Given : In 2012, about 24% of high-school seniors reported binge drinking (defined as five or more drinks in a row in the past two weeks), a substantial drop since the late 1990s.

SE = 0.019

Significance level =

Two-tailed value corresponds for :

z*=1.96   [Using z-table]

Now, the 95% confidence interval for the proportion of high-school seniors in the sample who would report binge drinking will be :-

Hence, the correct answer = (0.24 - 1.96*0.019, 0.24 + 1.96*0.019)

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic

2 apples

step-by-step explanation:

taking away equals subtraction. 6-4=2