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3 years ago

For f(x)=0.01(2)power of x, find the average rate of change from x=3 to x=8

1 answer:

4 0

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)= 0.01(2)^x \qquad 
\begin{cases}
x_1=3\\
x_2=8
\end{cases}\implies \cfrac{f(8)-f(3)}{8-3}
\\\\\\
\cfrac{0.01(2)^8-0.01(2)^3}{8-3}$

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3 years ago

F(x) = x^2-3<br> What is the average rate of change on the interval [-5, 2] ?

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$(-3)$ .

Step-by-step explanation:

Over this interval, the change in the value of this function is:

$f(-5) - f(2) = 22 - 1 = 21$ .

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$(-5) - 2 = -7$ .

The average rate of change of function $f$ over this interval is equal to the change in the function value divided by the corresponding change in $x$ :

$\begin{aligned}& (\text{average rate of change}) \\ =\; & \frac{(\text{change in function value})}{(\text{change in $x$})} \\ =\; & \frac{f(-5) - f(2)}{(-5) - 2} \\ =\; & \frac{22 - 1}{-7} \\ =\; & \frac{21}{-7} \\ =\; & -3\end{aligned}$ .

Thus, the average rate of change of $f(x) = x^{2} - 3$ over the interval $[-5,\, 2]$ would be $(-3)$ .

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2 years ago

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2 years ago

IQ scores are measured with a test designed so that the mean is 116 and the standard deviation is 16. Consider the group of IQ s

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So the z-scores that separate the unusual IQ scores from those that are usual are Z = -2 and Z = 2.

The IQ scores that separate the unusual IQ scores from those that are usual are 84 and 148.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

In this question, we have that:

$\mu = 116, \sigma = 16$