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Paha777 [63]
3 years ago
9

Find the derivative of f(x) = (1 + 6x2)(x − x2) in two ways.

Mathematics
1 answer:
ipn [44]3 years ago
3 0
Using product rule;

f(x)=(1+6x²)(x-x²)

f'(x)=(12x)(x-x²) + (1-2x)(1+6x²) = 12x² -12x³ +1 +6x² -2x -12x³ = -24x³ +18x² -2x +1

Solving the bracket first;

f(x)=(1+6x²)(x-x²) = x -x² +6x³ -6x^4

f'(x)= 1 -2x +18x² -24x³ = -24x³ +18x² -2x +1

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A recipe submitted to a magazine by one of its subscribers’ states that the mean baking time for a cheesecake is 55 minutes. A t
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Answer:

t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =P(t_{(9)}>4.296)=0.001  

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55.

Step-by-step explanation:

Information given

We have the following data: 54 55 58 59 59 60 61 61 62 65

The sample mean and deviation can be calculated with the following formulas:

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

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

\bar X=59.4 represent the sample mean

s=3.239 represent the sample standard deviation

n=10 sample size  

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to test if the true mean is higher than 55, the system of hypothesis would be:  

Null hypothesis:\mu \leq 55  

Alternative hypothesis:\mu > 55  

Replacing the info given we got:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

And replacing the info given we got:

t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =P(t_{(9)}>4.296)=0.001  

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55

8 0
3 years ago
Plot the x-intercept(s), y-intercept, vertex, and axis of symmetry of this function:
Maslowich

Answer:

Part 1) The vertex is the point (1,-9)

Part 2) The equation of the axis of symmetry is x=1

Part 3) The y-intercept is the point (0,-8)

Part 4) The x-intercepts are (-2,0) and (4,0)

Part 5) The graph in the attached figure

Step-by-step explanation:

we have

h(x)=(x-1)^2-9

This is a vertical parabola open upward

The vertex represent a minimum

Part 1) Find the vertex

The quadratic equation is written in vertex form

y=a(x-h)^2+k

where

(h,k) is the vertex of the parabola

so

The vertex is the point (1,-9)

Part 2) Find the axis of symmetry

The equation of the axis of symmetry of a vertical parabola is equal the the x-coordinate of the vertex

so

The equation of the axis of symmetry is

x=1

Part 3) Find the y-intercept

we know that

The y-intercept is the value of the function when the value of x is equal to zero

so

For x=0

h(x)=(0-1)^2-9

h(0)=-8

therefore

The y-intercept is the point (0,-8)

Part 4) Find the x-intercepts

we know that

The x-intercepts are the values of x when the value of the function is equal to zero

so

For h(x)=0

(x-1)^2-9=0

solve for x

(x-1)^2=9

square root both sides

x-1=\pm3

x=1\pm3

x=1+3=4\\x=1-3=-2

therefore

The x-intercepts are (-2,0) and (4,0)

Part 5) Plot the graph of the quadratic function

Plot the following points to graph the function

vertex (1,-9)

axis of symmetry x=1

y-intercept (0,-8)

x-intercepts (-2,0) and (4,0)

The graph in the attached figure

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