How can I prove that
1 answer:
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The solution to the above factorization problem is given as f′(x)=4x³−3x²−10x−1. See steps below.
<h3>What are the steps to the above answer?</h3>Step 1 - Take the derivative of both sides
f′(x)=d/dx(x^4−x^3−5x^2−x−6)
Step 2 - Use differentiation rule d/dx(f(x)±g(x))=d/dx(f(x))±d/dx(g(x))
f′(x)=d/dx(x4)−d/dx(x^3)−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−d/dx(x3)−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x2−d/dx(5x^2)−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x^2−10x−d/dx(x)−d/dx(6)
f′(x)=4x^3−3x^2−10x−1−dxd(6)
f′(x)=4x^3−3x^2−10x−1−0
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Answer:
Yes, it would be unusual.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using 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 or
, the outcome X is considered unusual.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
Would it be unusual for this sample mean to be less than 19 days?
We have to find Z when X = 19. So
By the Central Limit Theorem
, so yes, the sample mean being less than 19 days would be considered an unusual outcome.