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PolarNik [594]
3 years ago
11

Evaluate the limits, if they exist. If not, write DNE.

Mathematics
2 answers:
GrogVix [38]3 years ago
6 0
2
DNE
-2 not sure about the answer
Blizzard [7]3 years ago
4 0
A) 2
B) DNE
C) -2

I hope this answers your question but I’m not so sure of my answer!
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This kind of question should not be placed under math. Secondly, this type of question is not something to be answered by yes or no, or logical-exact answer because the idea here, is to use your thoughts about "Trees are helpful" This might be a very long, long, long answer - and only you can answer.
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3 years ago
What is equivalent to this equation 4^x+3=64
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he average movie admission price for a recent year was $7.18. The population variance was 3.81. A random sample of 15 theater ad
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Answer:

Since the calculated value of  z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.

Step-by-step explanation:

The data given is

Population mean μ= $ 7.18

Population variance= σ²= 3.81

Population Standard Deviation = √σ²= √3.81= 1.952

Sample Mean= x`= $ 8.02

Sample Standard Deviation =s = $ 2.08

Sample Size = 15

Significance Level = ∝= 0.05

The null and alternate hypotheses are

H0: σ1=σ2 against the claim that Ha: σ1≠ σ2

where  σ1 is the population variance and

σ2 is the sample variance

The rejection region is Z ≥ ±1.96 for two tailed test  at ∝= 0.05

The test statistic z is used

z= x`-  μ/ σ/√n

Putting the values

Z= 8.02-7.18/1.952/√15

z= 0.84/0.51599

z= 1.6279

Since the calculated value of  z= 1.6279 does not fall in the critical region Z ≥ ±1.96 we conclude that there is no difference between the population variance and the sample variance.

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Put the answer as 20.8 unless specified otherwise.
.16 is 16%
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