Answer:
Step-by-step explanation:
8
Answer:
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher
Step-by-step explanation:
Data provided in the question:
sample size, n = 103
Mean temperature, μ = 98.3
°
Standard deviation, σ = 0.73
Degrees of freedom, df = n - 1 = 102
Now,
For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]
Therefore,
CI = 
Thus,
Lower limit of CI = 
or
Lower limit of CI = 
or
Lower limit of CI = 98.11
and,
Upper limit of CI = 
or
Upper limit of CI = 
or
Upper limit of CI = 98.49
Hence,
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F
The value f x when y= 5 is 2
<h3>Variations </h3>
Let the given equation be y = k/x
where k is the constant
if y = 2 when x= 5, then;
k = xy
k= 2(5)
k = 10
In order to determine the value of x when y = 5
x = k/y
x = 10/5
x = 2
Hence the value f x when y= 5 is 2
Learn more on variation here: brainly.com/question/6499629
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Answer:
184.26 i dont exacaly know but i think it is just let me know if im right or wrong
Step-by-step explanation: