Assume that the probability of a driver getting into an accident is 7.6%, the average cost of an accident is $16,412.05, and the
2 answers:
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Answer:
(a) B. G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.
(b) Every function of the form is an antiderivative of 8x
Step-by-step explanation:
A function <em>F </em>is an antiderivative of the function <em>f</em> if
for all x in the domain of <em>f.</em>
(a) If , then
is an antiderivative of <em>f </em>because
Therefore, G(x) is an antiderivative of f(x) because G'(x) = f(x) for all x.
Let F be an antiderivative of f. Then, for each constant C, the function F(x) + C is also an antiderivative of <em>f</em>.
(b) Because
then is an antiderivative of
. Therefore, every antiderivative of 8x is of the form
for some constant C, and every function of the form
is an antiderivative of 8x.
Answer:
This measure is just 0.17 standard deviations from the mean, so we should not be surprised.
Step-by-step explanation:
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
If the z-score is lower than -2, or higher than 2.5, the score of X is considered unusual.
Subtraction of normal variables:
When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.
Let xS – xL represent the sampling distribution.
Mean s 6, means L 4. So
Standard deviation s is 2.5, for L is 1.5. So
Should we be surprised if the sample mean housebroken age for the small breed dogs is at least 2.5 months more than the sample mean housebroken age for the large breed dogs? Explain your answer.
We have to find the z-score for X = 2.5. So
This measure is just 0.17 standard deviations from the mean, so we should not be surprised.