In a sample of 57 temperature readings taken from the freezer of a restaurant, the mean is 29.6 degrees and the population standard deviation is 2.7 degrees. What would be the 80% confidence interval for the temperatures in the freezer?

(29.14 ; 30.06)

Step-by-step explanation:

Given that'

Sample size (n) = 57

Mean (m) = 29.6

Population standard deviation (σ) = 2.7

Confidence interval = 80%

= (1 - 0.8) / 2 = 0.1

Mean ± z * σ/√n

Using the Z probability calculator : Z0. 1 = 1.28

Hence,

29.6 ± 1.28 * (2.7 / √57)

29.6 - (1.28 * 0.3576237) ; 29.6 + (1.28 * 0.3576237)

29.142241664 ; 30.057758336

(29.14 ; 30.06)

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."