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egoroff_w [7]
3 years ago
5

What is the effect on the graph of the function f(x) = x^2 when f(x) is changed to f(x − 6)

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

Answer:

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

Step-by-step explanation:

When we Add or subtract a positive constant, let say c, to input x, it would be a horizontal shift.  

For example:

Type of change               Effect on y = f(x)

y = f(x - c)                      horizontal shift: c units to right

So

Considering the function

f(x) = x^2

The graph is shown below. The first figure is representing f(x) = x^2.

Now, considering the function

\:f\left(x\right)=\left(x-6\right)^2

According to the rule, as we have discussed above, as a positive constant 6 is added to the input, so there is a horizontal shift, 6 units to the right.

The graph of \:f\left(x-6\right)=\left(x-6\right)^2 is shown below in second figure. It is clear that the graph of  \:f\left(x-6\right)=\left(x-6\right)^2  is shifted 6 units to the right as compare to the function f(x) = x^2.

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

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Given the sample −4, −10, −16, 8, −12, add one more sample value that will make the mean equal to 3.
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Answer: The value to add to the set is 52

=======================================================

For a moment, ignore the extra value we'll add in later.

The given set of values add to -34.

Let x be the extra value we add in. It adds to the -34 to get -34+x, then we divide by 6 because the set {-4,-10,-16,8,-12,x} has 6 items.

So we get the mean (-34+x)/6

Set this equal to the mean we want (3) and solve for x

(-34+x)/6 = 3

-34+x = 6*3

-34+x = 18

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x = 52

The set {-4,-10,-16,8,-12,x} updates to {-4,-10,-16,8,-12,52} and this set has a mean of 3.

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3 years ago
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A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decre
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Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

At the null hypothesis, it is <u>tested if the consumption is not different</u>, that is, if the subtraction of the means is 0, hence:

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At the alternative hypothesis, it is <u>tested if the consumption is different</u>, that is, if the subtraction of the means is not 0, hence:

H_1: \mu_1 - \mu_2 \neq 0

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s_1 = \frac{45.1}{\sqrt{22}} = 9.6154

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The distribution of the differences is has:

\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25

s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14

The test statistic is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 0 is the value tested at the null hypothesis.

Hence:

t = \frac{25 - 0}{11.14}

t = 2.2438

The p-value of the test is found using a <u>two-tailed test</u>, as we are testing if the mean is different of a value, with <u>t = 2.2438</u> and 22 + 22 - 2 = <u>42 df.</u>

  • Using a t-distribution calculator, this p-value is of 0.0302.

Since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

A similar problem is given at brainly.com/question/25600813

4 0
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So the mutiples are 1.3, 2.2, 2.6, 3.1, 3.5, 4.4, 5.3, 6.2, and 6.6

Greather than 6 is 1.6, 2.5, 3.4, 3.6, 4.3, 4.5 4.6, 5.2, 5.4, 5.5, 5.6, 6.1, 6.3, 6.4, and 6.5.

So the prop is 24/36 or 2/3=66%

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A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 505π ft3,
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Answer:

radius   x  = 3 ft

height   h = 23,8  ft

Step-by-step explanation:

From problem statement

V(t)  = V(cylinder) + V(hemisphere)

let x be radius of base of cylinder (at the same time radius of the hemisphere)

and h the height of the cylinder, then:

V(c)  = π*x²*h       area of cylinder = area of base + lateral area

                                              A(c)  = π*x²  +   2*π*x*h

V(h) = (2/3)*π*x³   area of hemisphere   A(h)  =   (2/3)*π*x²

A(t)  =  π*x²  +   2*π*x*h +   (2/3)*π*x²

Now A as fuction of x    

total volume   505  = π*x²*h  +  (2/3)*π*x³

h = [505 - (2/3)* π*x³ ]  /2* π*x    

Now we have the expression for A as function of x

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Taking derivatives both sides

A´(x) =  6πx -  2πx²              A´(x) =  0         6x  - 2x²  = 0

x₁  =  0  we dismiss

6 - 2x = 0

x = 3     and   h = [505 - (2/3)* π*x³]/2* π*x

h  =  (505 - 18.84) / 6.28*3

h  = 23,8  ft

7 0
3 years ago
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