Question 1 demonstrates the Commutative Property.
Answer:
No solution exists
Step-by-step explanation:
Answer:
option d -1 is the answer
Et's give this a go:h(x) = cos(x) / f(x)
derivative (recall the quotient rule)h'(x) = [ f(x) * (-sin(x)) - cos(x)*f'(x) ] / [ f(x) ]^2
simplifyh'(x) = [ -sin(x)*f(x) - cox(x)*f '(x) ] / [ f(x) ]^2h'(π/3) = [ -sin(π/3)*f(π/3) - cox(π/3)*f '(π/3) ] / [ f(π/3) ]^2h'(π/3) = −(3–√/2)∗(3)−(1/2)∗(−7)/(3)2
h'(π/3) = (−33–√/2+7/2)/9
And you can further simplify if you want, I'll stop there.
Answer:
The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
A consumer organization finds that the mass of the chocolate bar is normally distributed with a mean of 200.4 g and a standard deviation of 0.05 g.
This means that
a) Find the proportion of the chocolate bars produced that have mass within one standard deviation of the mean.
pvalue of Z when X = 200.4 + 0.05 = 200.9 subtracted by the pvalue of Z when X = 200.4 - 0.05 = 199.9.
X = 200.9
has a pvalue of 0.8413
X = 199.9
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.