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)

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

answer: the anser is 3

step-by-step explanation:

c- 2o is equal to 9