Which best describes the effect of the narration in the excerpt? It suggests that the man being executed feels tranquil and at p
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P = 2x+2(450-x)=900
A = x(450 - x)
A = x(450 - x)
-x² + 450x = A
-x² +450x = A
A is going to be maximum, when x is a vertex.
-(x² - 450x) = A
-(x² - 2*225x+225²) - 225² =A
-(x-225)² -225² = A
Vertex when x=225
Largest are 225(450-225) = 225² = 50625 ft²
The square has a largest area.
A = x(450 - x)
A = x(450 - x)
-x² + 450x = A
-x² +450x = A
A is going to be maximum, when x is a vertex.
-(x² - 450x) = A
-(x² - 2*225x+225²) - 225² =A
-(x-225)² -225² = A
Vertex when x=225
Largest are 225(450-225) = 225² = 50625 ft²
The square has a largest area.
The standard error of the difference of sample means is 0.444
From the complete question, we have the following parameters
<u>Canadians</u>
- Sample size = 50
- Mean = 4.6
- Standard deviation = 2.9
<u>Americans</u>
- Sample size = 60
- Mean = 5.2
- Standard deviation = 1.3
The standard error of a sample is the quotient of the standard deviation and the square root of the sample size.
This is represented as:
The standard error of the Canadian sample is:
So, we have:
The standard error of the American sample is:
So, we have:
The standard error of the difference of sample means is then calculated as:
This gives
Take square roots
Hence, the standard error of the difference of sample means is 0.444
Read more about standard errors at:
Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).
<h3>What is a t-distribution confidence interval?</h3>The confidence interval is:
In which:
is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed <em>95% confidence interval</em>, with 49 - 1 = <em>48 df</em>, is t = 2.0106.
Hence:
The 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).
More can be learned about the t-distribution at brainly.com/question/16162795