If f(x) = ln(2), then limx--->2 (f(2)-f(x))/x-2
1 answer:
Answer:
- as written, -2
- with denominator parentheses, 0
- with f(x)=ln(x) and denominator parentheses, -1/2
Step-by-step explanation:
The problem as stated asks for the limit as x approaches 2 of (0/x) -2.
As written, the limit is (0/2) -2 = -2.
<u>Explanation</u>: f(x) is a constant, so the numerator is 0. The ratio 0/x -2 is defined as -2 everywhere except x=0. So, the value at x=2 is 0/2 -2 = -2.
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If you mean (f(2) -f(x))/(x -2), that limit is the limit of 0/(x-2) = 0 as x approaches 2.
<u>Explanation</u>: f(x) is a constant, so the numerator is 0. The ratio 0/(x-2) is zero everywhere except at x=2. The left limit and right limit are both 0 as x approaches 2. Since these limits agree, the limit is said to be 0.
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If you mean f(x) = ln(x) and you want the limit of (f(2) -f(x))/(x -2), that value will be -1/2.
<u>Explanation</u>: The value of the ratio is 0/0 at x=2, so we can find the limit using L'Hôpital's rule. Differentiating numerator and denominator, we have ...
lim = (-1/x)/(1)
The value is -1/2 at x=2.
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Answer:
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And using the z score we got:
So then we can conclude that 31 represent approximately the percentile 81 on the distribution given
He scored better than about 80.98 % of all MCAT takers.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
Where and
And for this case we have a score of 31
The percentile is calculated with the following probability:
And using the z score we got:
So then we can conclude that 31 represent approximately the percentile 81 on the distribution given
He scored better than about 80.98 % of all MCAT takers.