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suter [353]
4 years ago
13

What is the multiplicative rate of change for the exponential function graphed to the left?

Mathematics
1 answer:
schepotkina [342]4 years ago
7 0

Answer:

The multiplicative rate of change is 3

Step-by-step explanation:

The given graph passes through the points:

(-2,\frac{2}{9}),(-1,\frac{2}{3}),(0,2),(1,6),(2,18),(3,54)

The multiplicative rate of change is the factor that multiplies each y-value to give the subsequent y-values.

We can find that factor by dividing a y-value by the previous y-value of any two consecutive y-value.

The multiplicative rate of change is \frac{6}{2}=3 or \frac{18}{6}=3 or \frac{54}{18}=3  or \frac{2}{\frac{2}{3} } =3.

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Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expendi
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Answer:

a) Null hypothesis: \mu_d= 0

Alternative hypothesis: \mu_d \neq 0

b) t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{850 -0}{\frac{1123}{\sqrt{42}}}=4.905

The next step is calculate the degrees of freedom given by:

df=n-1=42-1=41

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v =2*P(t_{(41)}>4.905) =0.000015

So the p value is lower than any significance level given, so then we can conclude that we can reject the null hypothesis that the difference between he two groups are equal.

Step-by-step explanation:

Assuming the following questions:

We assume the following data: n = 42 ,\bar d= 850 , s_d = 1123

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

The system of hypothesis for this case are:

Null hypothesis: \mu_2- \mu_1 = 0

Alternative hypothesis: \mu_2 -\mu_1 \neq 0

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Null hypothesis: \mu_d= 0

Alternative hypothesis: \mu_d \neq 0

b. Use a .05 level of significance. Can you conclude that the population means differ? What is the p-value?

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{850 -0}{\frac{1123}{\sqrt{42}}}=4.905

The next step is calculate the degrees of freedom given by:

df=n-1=42-1=41

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v =2*P(t_{(41)}>4.905) =0.000015

So the p value is lower than any significance level given, so then we can conclude that we can reject the null hypothesis that the difference between he two groups are equal.

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