For what values of m does the equation mx = 5 have a single solution? can m have a value for which the equation does not have so
1 answer:
Answer:
Given equation: mx =5 ......[1]
To find for which values of m does the given equation have a single solution.
Divide equation [1] both sides by x,
Simplify:
The value of m have a single solution for all real values of x except x≠0.
No solution states that there is no answer to the equation
Infinite solutions states that any value for the variable would make the equation true.
we can write equation [1] as;
For the value of m = 0 , the given equation does not have solutions.
For any constant parameter m ,
, the given equation does not have any infinite number of solutions.
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Answer:
a) 0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.
b) 0.4129 = 41.29% probability that the mean return will be less than 8%
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 normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
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.
Mean 8.7% and standard deviation 20.2%.
This means that
40 years:
This means that
(a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 13%?
This is 1 subtracted by the pvalue of Z when X = 13. So
By the Central Limit Theorem
has a pvalue of 0.9115
1 - 0.9115 = 0.0885
0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.
(b) What is the probability that the mean return will be less than 8%?
This is the pvalue of Z when X = 8. So
has a pvalue of 0.4129
0.4129 = 41.29% probability that the mean return will be less than 8%