In a sample of 57 temperature readings taken from the freezer of a restaurant, the mean is 29.6 degrees and the population stand
ard deviation is 2.7 degrees. What would be the 80% confidence interval for the temperatures in the freezer?
1 answer:
Answer:
(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)
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Answer:
B. 25.40 cm
You're welcome
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So, what you would have to do is figure out how old he is not which is 30 yrs old because it is half of 60.
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HOPES THIS HELPS!!!!
Answer:
(x + 1/2)^2 = 289/4
Step-by-step explanation:
x^2+x-72=0
Write like this:
(x + ___)^2 = ___
put 72 on other side
x^2 + x = 72
x^ + x + (1/4) = 72 + (1/4)
(x + 1/2)^2 = 289/4
x + (1/2) = 17/2
Answer:
The answer is -7
Step-by-step explanation:
7 – (-4) + 3(-6)
7 + 4 – 18 = -7
Thus, The answer is -7
<u>-TheUnknownScientist</u><u> 72</u>
