a sample of 49 observations is taken from a normal population with a standard deviation of 10. the sample mean is 55. Determine the 99% confidence interval for the population mean.

55+/-3.69

= (51.31, 58.69)

Therefore, the 99% confidence interval (a,b)= (51.31, 58.69)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x =55

Standard deviation r = 10

Number of samples n = 49

Confidence interval = 99%

z-value (at 99% confidence) = 2.58

Substituting the values we have;

55+/-2.58(10/√49)

55+/-2.58(1.428571428571)

55+/-3.685714285714

55+/-3.69

= (51.31, 58.69)

Therefore, the 99% confidence interval (a,b)= (51.31, 58.69)

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: