A random sample of 1700 workers in a particular city found 578 workers who had full health insurance coverage. Find a 95% confidence interval for the true percent of workers in this city who have full health insurance coverage.

(31.75%, 36.25%)

Step-by-step explanation:

Let p be the proportion of workers in this city who have full health insurance coverage.

As per given,

Sample size : n= 170

Number of workers who had full health insurance coverage.=578

i.e. sample proportion:

Also, z-score for 95% confidence level : 1.96

Formula to find the confidence interval for p :

Hence, a 95% confidence interval for the true percent of workers in this city who have full health insurance coverage= (31.75%, 36.25%)

susie has 7 apples

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,