find the average rate of change of the function in the interval 6,13

The assumption that is necessary before conducting a test of hypothesis is that there is a significant amount of complaints for use of e-tickets.

The assumption is that there is a significant amount of complaints for use of e-tickets.

H0: there is a significant amount of complaints for use of e-tickets.

H1: Number of complaints for use of e-tickets is not significant.

H1: parameter > value

Notice the inequality points to the right

Decision Rule: Reject H0 if test statistic > critical value

Decision Rule: Reject H0 if p -value = 0.05

The hypothetical mean is 15.00 and The actual mean is 13.50

The difference between these two values is -1.50.

The sample size is 20

Mean 13.50

SD 1.50

The test statistic will be -4.459.

Lastly z it's important to reject the null hypothesis since the test statistic is greater than the critical value of 1.7291.

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Answer:

Okay, so:

sinA = 22.5/25.5

<A = 62* (correct)

cos C = 22.5/25.5 (correct)

tanA = 22.5/12.0 = 1.875 (correct)

sinC = 12.0/25.5

tan C = 12.0/22.5 = 0.53 (correct)

Step-by-step explanation:

I'll explain the wrong ones in the comments hold on-

The answer is 48π units³ or 150.72 units³.

To find the volume of the cone, use the formula : 1/3 × πr²h

We are given that r = 6 and h = 4.

Solving :

a) The coordinates of point A that meet the linear equation y = 2 · x + 7 are A(x, y) = (0, 7).

b) The equation of the line parallel to the line y = 2 · x + 7 is the line y = 2 · x + 3.

c) The coordinates of the midpoint of line segment BC is M(x, y) = (1, 7).

Herein we find three cases related to the linear equation y = 2 · x + 7. a) We need to determine the coordinates of point A, the y-coordinate can be determined by evaluating the function:

y = 2 · 0 + 7

y = 0 + 7

y = 7

The coordinates of point A that meet the linear equation y = 2 · x + 7 are A(x, y) = (0, 7).

b) Two lines are parallel to each other whether both have the same slope but different intercept. Then, the new line is of the form:

y = 2 · x + b

Now we proceed to find the intercept of the line:

b = y - 2 · x

b = 3 - 2 · 0

b = 3

The equation of the line parallel to the line y = 2 · x + 7 is the line y = 2 · x + 3.

c) The coordinates of point C is:

y = 2 · 2 + 7

y = 11

If we know that B(x, y) = (0, 3) and C(x, y) = (2, 11), then the coordinates of the midpoint of BC is:

M(x, y) = 0.5 · B(x, y) + 0.5 · C(x, y)

M(x, y) = 0.5 · (0, 3) + 0.5 · (2, 11)

M(x, y) = (1, 7)

The coordinates of the midpoint of line segment BC is M(x, y) = (1, 7).

The statement presents typing mistakes and omissions, complete form is shown below:

a) The line of y = 2x + 7 meets the y-axis at A, where x = 0. Write down the co-ordinates of A.

b) A line parralel to y = 2x + 7 passes through B(x, y) = (0,3) Find the equation of this line.

c) C is the point on the line y = 2x + 7 where x = 2. Find the the co-ordinates of the midpoint of BC.

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Answer:

C

Step-by-step explanation:

|-80 - 15| = 95

The distance between -80 and 15 is 95

80 to 0 plus 15.

The measure of the side SQ from the given diagram is 19/6

Similar shapes are shapes that has equal length and equal side measures and angles.

From the given diagram, the measure of the sides TSis congruent to SP and the measure of MS is congruent to SQ.

Using the expression below to determine the value of x

MS/ST = SQ/SP

Given the following parameters

MS =30

ST = 10

SQ = 6x-1

SP =  3 - 2x

Substitute the given parameters into the formula to have:

30/10 = 6x-1/3-2x

Cross multiply

10(6x-1) = 30(3-2x)

Expand

60x - 10 = 90 - 60x

60x + 60x = 90 + 10

120x = 100

x = 10/12

x = 5/6

Determine the measure of SQ

SQ = 6x - 1

SQ = 5(5/6) - 1

SQ = 25/6  - 1

SQ = 19/6

Hence the measure of the side SQ from the given diagram is 19/6

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Angles are said to be produced whenever two or more straight lines intersect. Some types of angles are right angle, obtuse angle, acute angle, etc. Thus, the required proof is stated below:

Prove: <A > <C

           <B > <D

An obtuse angle is a given angle that is greater than [tex]90^{o}[/tex] but less than [tex]180^{o}[/tex]. While an acute angle is a given angle that is less than [tex]90^{o}[/tex].

Thus extend AB to F, such that BF = CD.

Draw a perpendicular from a to intersect CD at point E.

So that;

<BAD = <BAE + <DAE (addition property of an angle)

Such that;

<A is an obtuse angle, while <C is an acute angle. Thu,

<A > <C (different types of angles)

Also,

<ADC = <CDF - <ADF (subtraction property of angles)

But both angles B and D are acute angles.

So that;

<B > <C (comparison of measures of two angles)

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The perimeter of the triangle is 56.4 units.

The perimeter of the triangle can be found as follows:

The perimeter of a figure is the sum of the whole sides. Therefore, the perimeter of a triangle is the sum of the three sides.

perimeter of a triangle = x + y + z

where

Therefore,

perimeter of the triangle = 19.8 + (9.8 + 8.4) + (8.4 + (19.8 - 9.8))

perimeter of the triangle = 19.8 + 18.2 + 18.4

perimeter of the triangle = 56.4

perimeter of the triangle = 56.4 units

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The number of solutions of a+b+c+d+e+f = 2006 is 9.12 * 10^16

The equation is given as:

a+b+c+d+e+f = 2006

In the above equation, we have:

Result = 2006

Variables = 6

This means that

n = 2006

r = 6

The number of solutions is then calculated as:

(n + r - 1)Cr

This gives

(2006 + 6 - 1)C6

Evaluate the sum and difference

2011C6

Apply the combination formula:

2011C6 = 2011!/((2011-6)! * 6!)

Evaluate the difference

2011C6 = 2011!/(2005! * 6!)

Expand the expression

2011C6 = 2011 * 2010 * 2009 * 2008 * 2007 * 2006 * 2005!/(2005! * 6!)

Cancel out the common factors

2011C6 = 2011 * 2010 * 2009 * 2008 * 2007 * 2006/6!

Expand the denominator

2011C6 = 2011 * 2010 * 2009 * 2008 * 2007 * 2006/720

Evaluate the quotient

2011C6 = 9.12 * 10^16

Hence, the number of solutions of a+b+c+d+e+f = 2006 is 9.12 * 10^16

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Answer: 3.2

Step-by-step explanation: 3.2 is the answer because the number 7 is higher than 5. so the number 1 goes up by one, which that equals 3.2

Answer:

RT = 16[tex]\sqrt{3}[/tex]

Step-by-step explanation:

using the sine ratio in right triangle QRT and the exact value

sin60° = [tex]\frac{\sqrt{3} }{2}[/tex] , then

sin60° = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{RT}{QR}[/tex] = [tex]\frac{RT}{32}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )

2 RT = 32[tex]\sqrt{3}[/tex] ( divide both sides by 2 )

RT = 16[tex]\sqrt{3}[/tex]

Answer:

y-6= -5/3(x+2)

Step-by-step explanation:

the general way to write it is y-y1 = m(x-x1), so i plugged in the values

y1 = 6, x1 = -2

m = (y2-y1)/(x2-x1) = -5/3

Answer:

nope

Step-by-step explanation:

simplify the first equation

Answer:

  yes

Step-by-step explanation:

The associative and commutative properties of multiplication allow you to rearrange the product to the form shown in the question.

Division by 2 is effectively multiplication by 1/2.

  [tex](n\times(n+1))/2\qquad\text{given}\\\\=\dfrac{n(n+1)}{2}=\dfrac{n}{2}(n+1)\\\\=\boxed{(n/2)(n+1)}[/tex]

__

Additional comment

Evaluation of the expressions according to the order of operations will proceed differently for the two expressions.

For the first expression, evaluation steps are ...

For the second expression, evaluation steps are ...

As we said above, the properties of multiplication ensure the results are the same either way.

As a practical matter, for integer values of n, one of n and (n+1) will be even. It is usually convenient to divide the even number by 2. This means the evaluation might be ...

For certain computer representations of the numbers, results may differ depending on the specific numbers and the order of evaluation.

The periodic time, t, of a pendulum range directly as the square root of its length, l. t = 6 when l = 9. If l = 25 then the periodic time, T exists at 10.

The period, or periodic time, of a periodic variation of a quantity, exists described as the time interval between two consecutive repetitions.

Given: The periodic time, t, of a pendulum goes directly as the square root of its length, l if t = 6 when l = 9.

If [tex]T \alpha\ l^2[/tex] then [tex]T^2[/tex] α [tex]\sqrt{l}[/tex]

[tex]T^2 = l[/tex]

Let, [tex]T^2 = kl,[/tex] where k exists constant

t = 6 when l = 9.

So, [tex]6^2 = k*9[/tex]

[tex]k = 6^2/9 = 4[/tex]

[tex]T^2 = 4*l[/tex]

If l = 25

[tex]T^2 = 4 * 25 = 100[/tex]

[tex]T = \sqrt{100}[/tex]

T = 10

Therefore, the periodic time, T exists 10.

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The behavior sees the leading coefficient is already positive, then increasing the size of the inputs will simply result in the leading term being even more positively skewed. The line on the graph will start to slope to the right.

The behavior of the graph of the polynomial function f(x) as the variable x approaches either positive or negative infinity represents the end behavior of the function. The final behavior of a polynomial function's graph is determined by the degree of the function as well as the leading coefficient.

Generally, the equation for is zeros  mathematically given as

y=x^4−2x^3−11x^2+12x+36

Therefore

(x+2)^2(x-3)^2=0

x=-2

x=3

since the graph attached has a positive leading coefficient and a positive degree

In conclusion, If the leading coefficient is already positive, then increasing the size of the inputs will simply result in the leading term being even more positively skewed. The line on the graph will start to slope to the right.

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The sum of each of the 4 polynomials and their answers are respectively;

12x² + 3x + 6 + (-7x²) – 4x – 2  = 5x² - x + 4

(2x² - x) + (-x – 2x² – 2) = -2x - 2

(x³ + x² + 2) + (x² – 2 - x³) = 2x²

(x² + x) + (x² + 8x - 2) = 2x²+ 9x - 2

1) We want to find the sum of the polynomials (12x² + 3x + 6) and (-7x² – 4x – 2). Thus, we have;

12x² + 3x + 6 + (-7x²) – 4x – 2

= 5x² - x + 4

2)  We want to find the sum of the polynomials (2x² - x) and (-x – 2x² – 2). Thus, we have;

(2x² - x) + (-x – 2x² – 2)

= -2x - 2

3) We want to find the sum of the polynomials (x³ + x² + 2) and (x² – 2 - x³). Thus, we have;

(x³ + x² + 2) + (x² – 2 - x³)

= 2x²

4) We want to find the sum of the polynomials (x² + x) and (x² + 8x - 2). Thus, we have;

(x² + x) + (x² + 8x - 2) = 2x²+ 9x - 2

The sum of each of the 4 polynomials and their answers are respectively;

12x² + 3x + 6 + (-7x²) – 4x – 2  = 5x² - x + 4

(2x² - x) + (-x – 2x² – 2) = -2x - 2

(x³ + x² + 2) + (x² – 2 - x³) = 2x²

(x² + x) + (x² + 8x - 2) = 2x²+ 9x - 2

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Answer: 42 and 31

Step-by-step explanation:

       Let x be one number and y be the other.

       We will write mathematical equations for these two numbers.

The sum of 2 numbers is 73.

x + y = 73

Their difference is 11.

x - y = 11

       See attached for the graph.

       The point of intersection is our solution.

7.2 + c = 19

c = 19 - 7.2

c = 11.8

Answer:

Your answer is 5.

Step-by-step explanation:

7.2 + c = 19

or, 14 + c = 19

or, c = 19 - 14

or, c = 5       ans.

Hope its helpful :-)

The value of x from the given diagram is 20

According to theorem, the right triangle altitude theorem is a result in elementary geometry that describes a relation between the altitude on the hypotenuse in a right triangle and the two line segments it creates on the hypotenuse.

Using the theorem above;

RT^2 = 9 * 16

RT^2 = 144

R = 12 units

Determine the value of x using the Pythagoras theorem;

x² =12² + 16²

x² = 144 + 256

x² = 400

Take the square root of both sides

x = √400

x = 20

Hence the value of x from the given diagram is 20

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Answer:

Step-by-step explanation:

Convert 2y=-3x+2 into y=mx+b form

[tex]\frac{2y}{2} =\frac{-3x+2}{2}[/tex]

[tex]y=\frac{-3x}{2} +1[/tex]

Y intercept is 1, slope is -3/2

Graph looks like this:

Answer:

2.1 litres

Step-by-step explanation:

the 5 part of the ratio refers to 3.5 litres of red paint

divide this amount by 5 to find the value of one part of the ratio

3.5 litres ÷ 5 = 0.7 litres ← value of 1 part of the ratio , then

3 parts = 3 × 0.7 litres = 2.1 litres ← amount of blue paint used

With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

So, According to this theorem the equation becomes the

[tex]\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).[/tex]

Then after solving, it become and with theorem it says that the

[tex]\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .[/tex]

Hence by this way the given equation with convolution theorem is proved.

So, With convolution theorem the equation is proved.

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Answer:

b/a

Step-by-step explanation:

Perpendicular lines have slopes that are opposite sign and reciprocals (flipped over).

In the equation

ax + by = c,

the slope of the line is

-a/b

If you haven't memorized this pattern yet, you can calculate it by solving ax+by=c for y.

by = -ax +c

y = -a/b x + c/b

The slope is -a/b

So a perpendicular line would be opposite sign and flipped, b/a

The distance between the bird (B) and plane (P) exists at 10,779 feet.

Given the figure attached,

PT = OB = 7000 ft

PB = OT = x feet

and angle of elevation (∠POT) = 30°

By applying the Tan rule in the triangle POT,

tan 33° = Opposite side/Adjacent side = PT/OT

tan 33° = 7000/x

x = 7000/tan33°

x = 10779 feet

The distance between the bird (B) and plane (P) exists at 10,779 feet.

Therefore, the correct answer is option D. 10779 feet.

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Given

[tex]7 - px - x^2 = 16 - (q + x)^2[/tex]

expanding the right side gives

[tex]7 - px - x^2 = 16 - (q^2 + 2qx + x^2)[/tex]

[tex]7 - px - x^2 = 16 - q^2 - 2qx - x^2[/tex]

Two polynomials of equal degree are the same if their coefficients are identical. This means

[tex]\begin{cases}16 - q^2 = 7 \\ -2q = -p\end{cases}[/tex]

[tex]16 - q^2 = 7 \implies q^2 = 9 \implies q=\pm3[/tex]

[tex]-2q = -p \implies p = 2q = \pm6[/tex]

Both [tex]p[/tex] and [tex]q[/tex] are positive, so [tex]\boxed{p=6}[/tex] and [tex]\boxed{q=3}[/tex].

Answer:

23 customers on payment plan B

13 customers on plan text B

Probability: 13/23

Step-by-step explanation:

The correctly matched quadratic functions are given as follows:

Quadratics are polynomial equations of the second degree, which means that they contain at least one squared term.

Quadratic equations are another name for it. The quadratic equation has the following general form: ax2 + bx + c = 0.

Recall that:
f(x) - a (x -c) (x - d)

where

Part A) First graph

We are given to know

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive, the function therefore, is equal to

The vertex is equal to the point (1, -4)

Substitute and solve for a

-4 = a(1-3)(1+1)

-4 = a(-2)(2)

a = 1

Hence, the function is equal to f(x) = (x-3) (x + 1)

Part B) Second Graph

We are also given to know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to f(x) = a (x-1)(x+3)

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

to solve for a, we must substitute:

8 = a(-1-1) (-1+3)

8 = a(-2)(2)

a = -2

Hence, the function is equal to: f(x) = -2 (x - 1) ( x + 3)

Part C)

We are given to know that the solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to f(x)  = a(x-6) (x+2).

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

We substitute and solve for a:

-8 = a(2-6) (2+2)

-8 = a(-4)(4)

a = 0.5

Hence, the function is equal to f(x) = 0.5(x - 6) (x + 2)

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Answer:

  3

Step-by-step explanation:

The quadratic formula is used to solve a quadratic equation in standard form, based on the values of the coefficients.

The standard-form quadratic ax² +bx +c = 0 has solutions given by the quadratic formula:

  [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

For the quadratic 3x² +4x -2 = 0, we have a=3, b=4, c=-2 and the formula gives ...

  [tex]x=\dfrac{-4\pm\sqrt{4^2-4(3)(-2)}}{2(3)}=\dfrac{-4\pm\sqrt{16+24}}{6}\\\\x=\dfrac{-4\pm 2\sqrt{10}}{6}=\dfrac{-2\pm\sqrt{10}}{3}[/tex]

The denominator in the solution is 3.

Answer:

The answer is x=(-2±√10)/3

Step-by-step explanation:

The first integer is 5.

The second integer  is 10.

If an integer is higher than zero, it is positive; if it is lower than zero, it is negative. Zero can be either positive or negative. Since a b and c d, then a + c b + d, the ordering of integers is consistent with algebraic operations.

One integer is twice the other

so,

If one integer is x then

The other will be 2x.

Their product :

2x * x = 50

2x² = 50

x² = 50/2

x² = 25

x = √25

x = 5

the first integer is 5.

the second integer is 2x = 2 x 5

                                        = 10.

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