Find the average value of f(x) = 7x^2(x^3+1)6 over the interval [0, 2].

Answers

Answer 1
Answer:

The average value is

\displaystyle\frac1{2-0}\int_0^27x^2(x^3+1)^6\,\mathrm dx

Let u=x^3+1, so that \mathrm du=3x^2\,\mathrm dx:

\displaystyle\frac12\int_1^9\frac73u^6\,\mathrm du=\frac16u^7\bigg|_1^9=\frac{9^7-1}6=\frac{2,391,484}3


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Problem 8-19 Because of high tuition costs at state and private universities, enrollments at community colleges have increased dramatically in recent years. The following data show the enrollment (in thousands) for Jefferson Community College for the nine most recent years. Click on the datafile logo to reference the data. Year Period (t) Enrollment (1,000s) 1 1 6.5 2 2 8.1 3 3 8.4 4 4 10.2 5 5 12.5 6 6 13.3 7 7 13.7 8 8 17.2 9 9 18.1 Use simple linear regression analysis to find the parameters for the line that minimizes MSE for this time series.If required, round your answers to two decimal places.
y-intercept, b0 = 4.7.17
Slope, b1 = 1.46
MSE = ???????? NEED THIS
What is the forecast for year 10? 19.283
Round your interim computations and final answer to two decimal places.

Answers

Answer:

a) find the attached graph

b) find the attachment no 4 and 5

c)T_(10)= 4.72+1.46(10) = 19.28

Step-by-step explanation:

a) A trend pattern exist if the time series plot gradually shifts to higher or lower values over a long period of time

find the attached graph

b) Liner Trend Equation

T_(1) =b_(0) +b_(1)t

Where T_(1) is the linear trend forecast in period t , b_(0) is the intercept of the linear trend time, b_(1) is the slope of the linear trend line, t is the time period

now computing the slope and intercept

formula is attached ( 3 no attachment)

Y_(t)is the value of the time series in period t, n is the number of time periods

Y(bar) is the average value and t(bar) is the average value of t

due to unavailability of equation in math-script i attached the calculation part of this question( 4th and 5th no attachment)

thus the linear trend equation is T_(t)= 4.72+1.46t                         (1)

T_(10)= 4.72+1.46(10) = 19.28

Final answer:

To find the Mean Squared Error (MSE), you can calculate the difference between the actual and predicted values, square these differences, and find their average. To forecast for a specific year, you can insert the year as the 'x' value into the simple linear regression equation.

Explanation:

The question is asking for the Mean Squared Error (MSE) for a simple linear regression model based on the enrollment data of Jefferson Community College. This involves using the y-intercept (b0) and slope (b1) values provided, and the given data points. You can calculate the MSE by taking the difference between the actual and predicted values (errors), squaring these differences, and then finding the average of these squared differences for the entire dataset.

Then, to forecast for year 10, you use the simple linear regression model equation, y = b0 + b1*x, where y represents the predicted enrollment. So, for year 10, you would insert 10 as your 'x' value into the equation, which results in the forecast value provided which is 19.283.

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Write the standard form of the quadratic function whose graph is a parabola with the given vertex and that passes through the given point. (Let x be the independent variable and y be the dependent variable.)Vertex: (−3, 4); point: (0, 13)

Answers

The standard form of the quadratic function whose graph is a parabola with the given vertex and that passes through the given point is;

y = x² + 6x + 13

We are given;

Vertex coordinate; (-3, 4)

A point on the graph; (0, 13)

The vertex form of a quadratic equation is given by;

y = a(x - h)² + k

Where h, k are the coordinates of the vertex.

a is the letter in general form of quadratic equation which is;

y = ax² + bx + c

Thus, at point (0, 13) at the vertex of (-3, 4), we have;

13 = a(0 - (-3))² + 4

⇒ 13 - 4 = 9a

9a = 9  

a = 9/9

a = 1  

Since y = a(x - h)² + k is the vertex form, let us put the vertex values for h and k as well as the value of a to get the quadratic equation;

y = 1(x - (-3))² + 4

y = x² + 6x + 9 + 4

y = x² + 6x + 13

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Answer:

The formula for this quadratic function is x*2 +6x+13

Step-by-step explanation:

If we have the vertex and one point of a parabola it is possible to find the quadratic function by the use of this

y= a (x-h)*2 + K

Quadratic function looks like this

y= ax*2 + bx + c

So let's find the a

y= a (x-h)*2 + K where

y is 13, x is 0, h is -3 and K is 4

13= a (0-(-3))*2 +4

13=9a +4

9=9a

9/9=a

1=a

The quadratic function will be

y= 1(x+3)*2 + 4

Let's get the classic form

(x+3)*2 = (x+3)(x+3)

(x*2+3x+3x+9)

x*2 +6x+13

f(0) = 13

Based on your observations from Question 1, what is the relationship between AE and BF when DFB and CEA measure something otherthan 90°? In this situation, what is the relationship between
AB
and CD ? Explain.

Answers

Final answer:

Without more context or details, we cannot determine any definitive relationships between AE and BF or AB and CD when DFB and CEA are not 90°. Generally, these lines or line segments could be skew, meaning they do not intersect and are not parallel.

Explanation:

Without additional context, specifics about the diagram, or an understanding of how AE, BF, AB, and CD are oriented towards each other, we can't determine any definitive relationships between AE and BF, or AB and CD when angles DFB and CEA don't equal 90°. However, in terms of general geometry, if you have two lines or line segments, and you know that the angles forming by intersecting lines are not 90°, then those two lines or segments are not perpendicular. In most cases, without knowing that they are parallel, AE and BF can be stated as skew lines. Similarly, unless specified, AB and CD could also be skew, not having any particular relationship. The term 'skew' is used to describe lines that do not intersect and are not parallel.

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Answer:

if angle dfb and angle cea measure something other than 90degrees, then line we is not equal to line bf. in this case, line and and line cd intersect at a single point.

Simplify 5/6i please help

Answers

Answer:

5/6 can´t be simplified

Step-by-step explanation:

Answer:

5i/6

Step-by-step explanation:

(5)/(6)i\n\n\mathrm{Multiply\:fractions}:\quad \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c)\n=(5i)/(6)

Which of the following rules best describes the matrix belowa. dilation of scale factor 2
b. reflection over the y-axis
c. reflection over the line y = x
d. translation 1 unit left and 1 unit up

Answers

Answer: D

Step-by-step explanation:

d. translation 1 unit left & 1 unit up

right on edg

Give two examples of addition of two mixed numbers with different denominators
SHOW ALL STEPS

Answers

Answer:

First Example: 3 1/2 + 4 3/4, Second Example: 6 3/8 + 7 9/15

Extra Example: 8 4/20 + 3 5/10

Step-by-step explanation:

First Example:

1/2 + 3/4

1/2 is equal to 2/4 so it is now compatible to be added to 3/4.

2/4 + 3/4

= 5/4

Now for the mixed numbers since its 3 and 4, 3 + 4 = 7.

Final answer is 7 5/4.

Second Example:

3/8 + 9/15

9/15 can be reduced to 3/5

Now the equation is 3/8 + 3/5

= 15/40 + 24/40 is an equivalent equation

15/40 + 24/40  

= 39/40

Now for the mixed numbers since its 6 and 7, 6 + 7 = 13

Final answer is 13 39/40.

I am going to include one last example just in case you need one:

Third Example:

4/20 + 5/10

We can reduce these to

1/5 + 1/2

= 2/10 + 5/10 is the equivalent equation

2/10 + 5/10

= 7/10

Now for the mixed numbers since its 8 and 3, 8 + 3 = 11.

Final answer is 11 7/10.

I Hope this helps!