Answer:
Kindly check explanation
Step-by-step explanation:
Given :
Sample size, n = 30
Tcritical value = 2.045
Null hypothesis :
H0: μ = 9.08
Alternative hypothesis :
H1: μ≠ 9.08
Sample mean, m = 8.25
Samole standard deviation, s = 1.67
Test statistic : (m - μ) ÷ s/sqrt(n)
Test statistic : (8.25 - 9.08) ÷ 1.67/sqrt(30)
Test statistic : - 0.83 ÷ 0.3048988
Test statistic : - 2.722
Tstatistic = - 2.722
Decision region :
Reject Null ; if
Tstatistic < Tcritical
Tcritical : - 2.045
-2.722 < - 2.045 ; We reject the Null
Using the α - level (confidence interval) 0.05
The Pvalue for the data from Tstatistic calculator:
df = n - 1 =. 30 - 1 = 29
Pvalue = 0.0108
Reject H0 if :
Pvalue < α
0.0108 < 0.05 ; Hence, we reject the Null
Answer:
Step-by-step explanation:
<u>Functions have graphs or slopes.</u>
<u>This is not a function.</u>
<u>This is not on a graph or on a slope.</u>
Your answer is not a function.
Answer:
- <em><u>1. There are infinite 8's</u></em>
- <em><u>2. The estimate is 130.89</u></em>
- <em><u>3. The difference is 0.01</u></em>
<em><u /></em>
Explanation:
There are countless, or infinite, 8's in the answer because the result is a periodic decimal:
The notation used for the periodic decimal numbers is an arc over the number that is repeated. In this case, it would be an arc over the first 8.
I will use a bar just to show how it is, but instead of a bar it should be an arc:
The <em>estimation</em> or approximation to <em>two decimal places</em> requires to drop the 8's since the third one and increase the second 8 to 9 (because the first decimal dropped is greater than or equal to 5).
Thus the estimated value would be 13.89.
The difference is:
- 130.89 - 130.888. . . = 0.0111. . .
- Rounded to 2 decimal places: 0.01
The logs rules work "backwards", so you can condense ("compress"?) strings of log expressions into one log with a complicated argument. When they tell you to "simplify" a log expression, this usually means they will have given you lots of log terms, each containing a simple argument, and they want you to combine everything into one log with a complicated argument. "Simplifying" in this context usually means the opposite of "expanding".
There is no standard definition, in this context, for "simplifying". You have to use your own good sense. If they give you a big complicated thing and ask you to "simplify", then they almost certainly mean "expand". If they give you a string of log terms and ask you to "simplify", then they almost certainly mean "condense".
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Answer:
Brainliest if correct!
Step-by-step explanation:
6561
9^4
the first one!!!
