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Andru [333]
3 years ago
5

Given the function g(x)= -x^2 + x + 5, determine the average rate of change of the function over the interval -4 ≤ x ≤ 3.​

Mathematics
1 answer:
OleMash [197]3 years ago
8 0

Answer:

The average rate of change is 2.

Step-by-step explanation:

We are given the function:

g(x)=-x^2+x+5

And we want to find the average rate of change over the interval:

-4\leq x\leq 3

The average rate of change is synonymous with the slope. So, we will evaluate the function at its endpoints and find the slope between them.

Our first endpoint is given by:

g(-4)=-(-4)^2+(-4)+5=-15

And our second endpoint is given by:

g(3)=-(3)^2+(3)+5=-1

This gives us two points (-4, -15) and (3, -1). The average rate of change will be the slope between them. Thus:

\displaystyle ARC=\frac{-1-(-15)}{3-(-4)}=\frac{14}{7}=2

The average rate of change is 2.

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A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced.
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Answer:

CI=(66.54,78.46)                                  

Step-by-step explanation:

Given : A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced. The mean yield of the 10 batches is 72.5% and the standard deviation is 5.8%. Assume the yields are independent and approximately normally distributed.

To find : A 99% confidence interval for the mean yield when the new catalyst is used ?

Solution :

Let X be the yield of the batches.

We have given, n=10 , \bar{X}=72.5\% , s=5.8%

Since the size of the sample is small.

We will use the student's t statistic to construct a 995 confidence interval.

\bar X\pm t_{n-1,\frac{\alpha}{2}}\frac{s}{\sqrt n}

From the t-table with 9 degree of freedom for \frac{\alpha}{2}=0.005

t_{n-1,\frac{\alpha}{2}}=t_{9,0.005}

t_{n-1,\frac{\alpha}{2}}=3.250

The 99% confidence interval is given by,

CI=72.5 \pm 3.25\frac{5.8}{\sqrt{10}}

CI=72.5 \pm 5.96

CI=(72.5+5.96),(72.5-5.96)

CI=(66.54,78.46)

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