This histogram is skewed to the left and its distribution's center would be three (3).
<h3>What is a histogram?</h3>
A histogram is used to graphically represent a set of data points into user-specified ranges, especially through the use of rectangular bars.
In this scenario, we can infer the following about the histogram distribution's center, spread, and overall shape:
- This histogram is skewed to the left.
- Its distribution's center would be three (3).
- The spread implies that more students are between 58-59 and 60-61 inches.
- The shape shows that many of the values are at the lower end.
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Answer:
C
Step-by-step explanation:
calculate like terms, then it is pretty simple from there.
Answer:
The answer to your question is Russia is 50 times greater than Finland.
Step-by-step explanation:
Data
Finland area = 3.4 x 10⁵ km²
Russia area = 1.7 x 10⁷ km²
Proportion = ?
Process
1.- Just divide the Russia's area by the Finland's area.
Proportion =
Proportion =
Porportion = 50
Midpoint Formula =
We are given the coordinates: A(-5,6) and B(3,2)
Plug the values into the equation to get:
M = ((3 - 5)/2, (2 + 6)/2)
Now, just solve it down to get M = (-1,4).
The midpoint of the segment with endpoints A(-5,6) and B (3,2) is (-1,4). Hope this helps ALot and have a wonderful day!
Part A
Answers:
Mean = 5.7
Standard Deviation = 0.046
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The mean is given to us, which was 5.7, so there's no need to do any work there.
To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046
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Part B
The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67
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At 95% confidence, the critical value is z = 1.96 approximately
ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949
The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67
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Part C
Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.
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At 95% confidence, the critical value is z = 1.96 approximately
ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657
L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99
U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21