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Ede4ka [16]
2 years ago
3

Two students in your class, Hunter and Maggie, are disputing a function. Hunter says that for the function, between x = -2 and x

= 2, the average rate of change is 0. Maggie says that for the function, between x = -2 and x = 2, the graph goes up through a turning point, and then back down. Explain how Hunter and Maggie can both be correct, using complete sentences.

Mathematics
1 answer:
Troyanec [42]2 years ago
3 0

Answer:

consider an example  f(x)=-x^{2}+4.

Step-by-step explanation:

hunter says that for a function between x=  -2 and x=2, the average rate of change is 0.

the average rate of a change of a function between the points a and b is given by:

\dfrac{f(b)-f(a)}{b-a}

for it to be equal to zero we have f(a)=f(b).

hence according to Hunter f(-2)=f(2).

Now maggie says that for the function between x=-2 and x=2 the graph goes up through a turning point and back down.

for both of them to be correct we can consider the example f(x)=-x^{2}+4.




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Answer:

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Step-by-step explanation:

Let the radius of the sphere be r.  Then the volume of the sphere is

V = (4/3)(π)(r³).  

Next, recognize that the side length of the cube is 2r, and that the volume of the cube is thus

V = (2r)³, or 8r³.

Then the volume of the foam is equal to the volume of the cube less the volume of the sphere:

V-foam = 8r³ - (4/3)(π)(r³).  This could be factored into

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Jeff has recorded the cost of insurance premiums on his car for a particular insurance company.
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According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me
Reil [10]

Answer:

a) \bar X=10.65

Median =\frac{10.7+10.7}{2}=10.7

Mode= 10.7

b) Range = 11.8-8.3=3.5

s= 0.948

c)IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got: \bar X=10.65

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

Median =\frac{10.7+10.7}{2}=10.7

The mode would be the most repeated value on this case:

Mode= 10.7

Part b

The range is defined as Range =Max-Min and if we replace we got:

Range = 11.8-8.3=3.5

We can calculate the standard deviation with the following formula:

s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s= 0.948

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and Q_1= \frac{10.6+10.7}{2}=10.55

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got Q_3 = \frac{10.9+11.2}{2}=11.05

Then we can find the IQR like this:

IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0
2 years ago
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