You measure 22 watermelons' weights, and find they have a mean weight of 36 ounces. Assume the population standard deviation is 6.9 ounces. Based on this, construct a 95% confidence interval for the true population mean watermelon weight. Give your answers as decimals, to two places

Step-by-step explanation:

Confidence interval formula is expressed as;

CI = xbar ±(z*s/√n)

xbar is the mean = 36

z is the z score at 95% CI = 1.96

s is the standard deviation = 6.9

n is the sample size = 22

Substitute

CI = 36  ±(1.96*6.9/√22)

CI = 36  ±(1.96*6.9/4.6904)

CI = 36  ±(1.96*1.4711)

CI = 36  ± 2.8833

CI = (36 - 2.8833, 36 + 2.8833)

CI = (33.1167, 38.8833)

Hence the 95% CI for the true population mean is 33.12<x<38.88

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.