Apoll showed that 48 out of 120 randomly chosen graduates of california medical schools last year intended to specialize in family practice. what is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice? select one: a. ± .0736b. ± .0447c. ± .0876d. ± .0894

± 0.0736

Step-by-step explanation:

Data provided in the question:

randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = = 0.4

Confidence level = 90%

sample size, n = 120

Now,

For 90% confidence level , z-value = 1.645

Width of the confidence interval = ± Margin of error

= ±

= ±

= ± 0.07356 ≈ ± 0.0736

Hence,

The correct answer is option  ± 0.0736

good for u

step-by-step explanation:

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !