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Assoli18 [71]
3 years ago
5

Plsss help ;)

Mathematics
2 answers:
Degger [83]3 years ago
8 0

Answer:

ave rate of change = (-6)^(2+2) +3 -  (-6)^(-1+2) +3

                                  ---------------------------------------------

                                      2+1

Step-by-step explanation:

To find the average rate of change

ave rate of change = f(x2) - f(x1)

                                  ----------------

                                   x2-x1

We know that x2 = 2 and x1 = -1

ave rate of change = f(2) - f(-1)

                                  ----------------

                                   2--1

ave rate of change = (-6)^(2+2) +3 -  (-6)^(-1+2) +3

                                  ---------------------------------------------

                                      2+1

Vilka [71]3 years ago
8 0

Answer:

ave rate of change = (-6)^(2+2) +3 -  (-6)^(-1+2) +3

Step-by-step explanation:


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5+15=20
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I'm probably wrong, but thats how I understood the question.
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What's the equation for circumference
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8 0
3 years ago
Read 2 more answers
In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen
Lesechka [4]

Step-by-step explanation:

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Type            ni                xi                  si

Graded       42           0.486            0.187

No-fines      42           0.359           0.158

To find - a. Formulate the above in terms of a hypothesis testing problem.

b. Give the test statistic and its reference distribution (under the null hypothesis).

c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.

Proof -

a.)

Hypothesis testing problem :

H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

b)

Test statistic :

Z = \frac{x_{1} - x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1}}  + \frac{s_{2}^{2} }{n_{2}} } }

Z = \frac{0.486 - 0.359 }{\sqrt{\frac{0.03496 }{42}  + \frac{0.02496 }{42} } }

Z = \frac{0.127 }{\sqrt{0.001468}}

Z = \frac{0.127 }{0.0377}

⇒Z(cal) = 3.3687

Z(tab) = 1.96

As Z (cal) > Z(tab)

So, we reject H0 at 5% Level of significance

p-value = 0.99962

Hence

There is significant difference in mean conductivity at the two materials.

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Answer:

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The function passes though -6

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