please help.

A report by the NCAA states that 57.5% of football injuries occur during practices. A head trainer claims that this is too high for this conference, so he randomly selects 36 injuries and founds that 17 occurred during practices. Is his claims correct at a=0.05?​

The nth term of the sequence is 2n - 8

The nth term of an arithmetic progression is expressed as;

Tn = a + (n - 1)d

where

a is the first term

d is the common difference

n is the number of terms

Given the following parameters

a = f(1)=−6

f(2) = −4

Determine the common difference

d = f(2) - f(1)

d = -4 - (-6)
d = -4 + 6

d = 2

Determine the nth term of the sequence

Tn = -6 + (n -1)(2)

Tn = -6+2n-2
Tn = 2n - 8

Hence the nth term of the sequence is 2n - 8

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By definition, we have

[tex]f(n) = f(n - 1) + f(n - 2)[/tex]

so that by substitution,

[tex]f(n-1) = f(n-2) + f(n-3) \implies f(n) = 2f(n-2) + f(n-3)[/tex]

[tex]f(n-2) = f(n-3) + f(n-4) \implies f(n) = 3f(n-3) + 2f(n-4)[/tex]

[tex]f(n-3) = f(n-4) + f(n-5) \implies f(n) = 5f(n-4) + 3f(n-5)[/tex]

[tex]f(n-4) = f(n-5) + f(n-6) \implies f(n) = 8f(n-5) + 5f(n-6)[/tex]

and so on.

Recall the Fibonacci sequence [tex]F(n)[/tex], whose first several terms for [tex]n\ge1[/tex] are

[tex]\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}[/tex]

Let [tex]F_n[/tex] denote the [tex]n[/tex]-th Fibonacci number. Notice that the coefficients in each successive equation form at least a part of this sequence.

[tex]f(n) = f(n-1) + f(n-2) = F_2f(n-1) + F_1 f(n-2)[/tex]

[tex]f(n) = 2f(n-2) + f(n-3) = F_3 f(n-2) + F_2 f(n-3)[/tex]

[tex]f(n) = 3f(n-3) + 2f(n-4) = F_4 f(n-3) + F_3 f(n-4)[/tex]

[tex]f(n) = 5f(n-4) + 3f(n-5) = F_5 f(n-4) + F_4 f(n-5)[/tex]

[tex]f(n) = 8f(n-5) + 5f(n-6) = F_6 f(n-5) + F_5 f(n-6)[/tex]

and so on. After [tex]k[/tex] iterations of substituting, we would end up with

[tex]f(n) = F_{k+1} f(n - k) + F_k f(n - (k+1))[/tex]

so that after [tex]k=n-2[/tex] iterations,

[tex]f(n) = F_{(n-2)+1} f(n - (n-2)) + F_{n-2} f(n - ((n-2)+1)) \\\\ f(n) = f(2) F_{n-1} + f(1) F_{n-2} \\\\ \boxed{f(n) = -4 F_{n-1} - 6 F_{n-2}}[/tex]

Answer:

g(0)⁻¹ = 6

Step-by-step explanation:

First, you must find the inverse of the function. Remember, another way of representing g(x) is with "y".  To find the inverse, you must swap the positions of the "x" and "y" variables in the equation. Then, you must rearrange the equation and isolate "y".

g(x) = 18 - 3x                                    <----- Original function

y = 18 - 3x                                        <----- Plug "y" in for g(x)

x = 18 - 3y                                        <----- Swap the positions of "x" and "y"

x + 3y = 18                                        <----- Add 3y to both sides

3y = 18 - x                                         <----- Subtract "x" from both sides

y = (18 - x) / 3                                    <----- Divide both sides by 3

y = 6 - (1/3)x                                     <----- Divide both terms by 3

Now that we have the inverse function, we need to plug x = 0 into the equation and solve for the output. In the inverse function, "y" is represented by the symbol g(x)⁻¹.

g(x)⁻¹ = 6 - (1/3)x                                  <----- Inverse function

g(0)⁻¹ = 6 - (1/3)(0)                               <----- Plug 0 in for "x"

g(0)⁻¹ = 6 - 0                                       <----- Multiply 1/3 and 0

g(0)⁻¹ = 6                                             <----- Subtract

The total cost to run Val’s AC costs for a summer month of 30 days is $94.5.

Quantity of electricity used in kilowatts

1000 watts = 1 kilowatts

2,500 watts = 2.5 kilowatts

Cost of Val’s AC to run for a summer month of 30 days = Quantity of electricity used × Number of hours Val runs AC per day × Number of days × Cost of electricity

= 2.5 × 7 × 30 × 0.18

= $94.5

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Answer:

1.

Step-by-step explanation:

10C0 = 1.

No the parents do not feel differently as they used to feel twenty years ago

Given :-

Po: p=0.46

Pa: p NE 0.46

alpha=0.01

Test statistic is one sample proportion [tex]z=(phat-p)/\sqrt{{ {0.46)(0.54)/800}}[/tex]

Critical value z> |2.576|

z=0.025/0.0176

=1.418

This is a p-value of 0.078

Fail to reject Po  and insufficient evidence to support the claim that parents feel differently.

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Answer:

-40x

Step-by-step explanation:

f(x) =-3x-5

g(x) =4x-2

(f+g)(x) =?

now,

f(g(x))

=f(4x-2)

=-3×4x-2-5

=-10×4x

=-40x

See below for the distance between the points and the lines

Question 2

The line and the points are given as:

x = y

P = (4, -2)

Rewrite the equation as:

y = x

The slope of the above equation is

m = 1

The slope of a line perpendicular to it is

m = -1

A linear equation is represented as:

y = mx + b

Substitute m = -1

y = -x + b

Substitute (4, -2) in y = -x + b

-2 = -4 + b

Solve for b

b = 2

Substitute b = 2 in y = -x + b

y = -x + 2

So, we have:

x = y and y = -x + 2

Substitute x for y

x = -x + 2

Solve for x

x = 1

Substitute x = 1 in y = x

y = 1

So, we have the following points

(1, 1) and (4, -2)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(1 - 4)² + (1 + 2)²

Evaluate

d = 3√2

Hence, the distance between x = y and P = (4, -2) is 3√2 units

Question 3

The line and the points are given as:

y = 2x + 1

Q = (2, 10)

The slope of the above equation is

m = 2

The slope of a line perpendicular to it is

m = -1/2

A linear equation is represented as:

y = mx + b

Substitute m = -1/2

y = -1/2x + b

Substitute (2, 10) in y = -1/2x + b

10 = -1/2 * 2 + b

Solve for b

b = 11

Substitute b = 11 in y = -1/2x + b

y = -1/2x + 11

So, we have:

y = 2x + 1 and y = -1/2x + 11

Substitute 2x + 1 for y

2x + 1 = -1/2x + 11

Solve for x

x = 4

Substitute x = 4 in y = 2x + 1

y = 9

So, we have the following points

(4, 9) and (2, 10)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(4 - 2)² + (9 - 10)²

Evaluate

d = √5

Hence, the distance between the line and the point is √5 units

Question 4

The line and the points are given as:

y = -x + 3

R = (-5, 0)

The slope of the above equation is

m = -1

The slope of a line perpendicular to it is

m = 1

A linear equation is represented as:

y = mx + b

Substitute m = 1

y = x + b

Substitute (-5, 0) in y = x + b

0 = 5 + b

Solve for b

b = -5

Substitute b = 5 in y = x + b

y = x + 5

So, we have:

y = x + 5 and y = -x + 3

Substitute x + 5 for y

x + 5 = -x + 3

Solve for x

x = -1

Substitute x = -1 in y = x + 3

y = 2

So, we have the following points

(-1, 2) and (-5, 0)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(-1 + 5)² + (2 - 0)²

Evaluate

d = 2√5

Hence, the distance between the line and the point is 2√5 units

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Considering it's discriminant, it is found that:

A. The classmate is wrong, as the discriminant is of zero, hence the equation has one solution.

B. The quadratic equation has 1 x-intercept.

A quadratic equation is modeled by:

y = ax^2 + bx + c

The discriminant is:

[tex]\Delta = b^2 - 4ac[/tex]

The solutions are as follows:

In this problem, the equation is:

y = 9x² - 6x + 1.

The coefficients are a = 9, b = -6 and c = 1, hence the discriminant is:

[tex]\Delta =(-6)^2 - 4(9)(1) = 36 - 36 = 0[/tex]

Since the discriminant is zero, the classmate is wrong, as it means that the equation has one solution = one x-intercept.

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well, with the assumption that a year has 365 days, that means one day is really just 1/365th of a year, so then 54 days will be 54/365 of a year.

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$17100\\ r=rate\to 9.65\%\to \frac{9.65}{100}\dotfill &0.0965\\ t=years\dotfill &\frac{54}{365} \end{cases} \\\\\\ I = (17100)(0.0965)(\frac{54}{365})\implies \stackrel{\textit{interest owed}}{I\approx 244.13}~\hfill \underset{amount~owed}{\stackrel{17100~~ + ~~244.13}{\approx 17344.13}}[/tex]

Question 13

Part (a)

[tex]\frac{3840-3905}{110}=\boxed{0.5\overline{90}}[/tex]

By similar logic, the other answers are

(b) 2.6818181....

(c) 3.5909090...

(d) 1.1363636...

Question 14

Part (a)

[tex]\frac{46800-41400}{4700} \approx \boxed{1.15}[/tex]

Part (b)

[tex]-2.57=\frac{x-41400}{4700} \\ \\ -2.57(4700)=x-41400 \\ \\ x=41400-2.57(4700) \\ \\ x \approx \boxed{29300}[/tex]

Answer:

a=4.40

Step-by-step explanation:

To find value of a use pythagoras theorem:

c^2=b^2+a^2

Rearrange the equation:

a^2=c^2-b^2

Substitute the values:

a^2=(4.58)^2-(1.26)^2

After calculation:

a=4.40

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$132.90\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ t=years \end{cases} \\\\\\ 132.90 = (6000)(0.055)(t)\implies \cfrac{132.90}{(6000)(0.055)}=t\implies \cfrac{443}{1100}=t \\\\\\ \stackrel{\textit{converting that to days}}{\cfrac{443}{1100}\cdot 365} ~~ \approx ~~ 147~days[/tex]

now, if we move back from August 14th by 147 days backwards, that'd put us on March 20th.

Answer: 19/40

Step-by-step explanation:

First Write 0.475 as 0.4751Multiply both numerator and denominator by 10 for every number after the decimal point0.475 × 10001 × 1000 = 4751000. Reducing the fraction gives The answer

Answer:

47°

Step-by-step explanation:

If angles are complementary, that means those angles are added up to 90°.

Therefore,

43 + x = 90

Take 43 to the right side to make x as the subject.

x = 90 - 43

x = 47°

Note that a complementary angle adds up to 90 degrees.

In this problem, we get the equation:

43° + x = 90°

So, we need to find the value of x by doing:

90° – 43° = 47°

x = 47°

43° + 47° = 90°

We get the final result as:

x = 47°

Hope this helps :)

Answer: B

Step-by-step explanation:

[tex]\frac{4+5i}{4i}\\ \\ =\frac{1}{4} \left(\frac{4+5i}{i} \right)\\\\=\frac{1}{4}((-i)(4+5i))\\\\=\frac{1}{4}(-4i-5i^2)\\\\=\frac{5-4i}{4}[/tex]

Answer:

So g(6) comes out to be 2.

Step-by-step explanation:

So f(6)=5 means when f acts on 6, the result is 5....so the inverse would take 5 back to 6...or g(5)=6

So f(2)=6 means that f reassigns 2 to 6....so the inverse would take 6 back to 2....or g(6)=2

hope this helps :)

Answer: 3

Step-by-step explanation:

If f(x) = y, then invf(y) = x
So, f(5) = invg(5) = 3

a) The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function g(x) in factored form is g(x) = [(x - 6) · (x + 3)] / (x - 2). there is no slant asymptotes.

c) There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

a) In the first part of this question we need to determine the equation of a polynomial in expanded form, derived from its factor form defined below:

f(x) = Π (x - rₐ), for a ∈ {1, 2, 3, 4, ..., n}         (1)

Where rₐ is the a-th root of the polynomial.

If we know that r₁ = 6, r₂ = - 1 and r₃ = - 3, then the polynomial in factor form is:

f(x) = (x - 6) · (x + 1) · (x + 3)

f(x) = (x - 6) · (x² + 4 · x + 4)

f(x) = (x - 6) · x² + (x - 6) · (4 · x) + (x - 6) · 4

f(x) = x³ - 6 · x² + 4 · x² - 24 · x + 4 · x - 24

f(x) = x³ + 10 · x² - 20 · x - 24

The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function is introduced below:

g(x) = (x³ + 10 · x² - 20 · x - 24) / (x² - x - 2)

g(x) = [(x - 6) · (x + 1) · (x + 3)] / [(x - 2) · (x + 1)]

g(x) = [(x - 6) · (x + 3)] / (x - 2)

The slope of the slant asymptote is:

m = lim [g(x) / x] for x → ± ∞

m = [(x - 6) · (x + 3)] / [x · (x - 2)]

m = 1

And the intercept of the slant asymptote is:

n = lim [g(x) - m · x] for x → ± ∞

n = Non-existent

Hence, there is no slant asymptotes.

c) There is vertical asymptote at a x-point if the denominator is equal to zero. There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

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A 0.13 sec and a 86 kVp technical changes are the changes that would have to produce the radiographic density and the contrast that are more similar to the original.

The radio graphic density can de defined to be the total amount or the overall darkening that can be found in a particular radiograph. This is usually known to have a certain type of density range that lies between  0.3 to 2.0 density.

When there is a density that is less than 0.3, the problem is basically because there is the issue of the density which would be found at the base and the fact that the film that is being used has fog in it.

It is known that based on the values that we have here the density and contrast would have to be of the given kvp of 86 and 0.13 seconds in order to be said to be similar to the original.

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The particle will stop at D.

If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

In pentagon ABCDE shown above, each side is 1 cm. If a particle starts at point A and travels clockwise 723 cm along ABCDE, then

Number of sides = 5

Let x be the one complete revolution.

5 x = 725

x = 145

Now, 145 - 2 = 143

Hence, it will stop at D which is 2 less than A.

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Answer:

the perpendicular bisector

Explanation:

The perpendicular bisector will intersect the segment at its midpoint.

Answer:

below

Step-by-step explanation:

1) slope = rise / run

2 coordinates are (-4, 0), (0, 2).

2 - 0 = 2

0 -- 4 = 4

2 / 4 = 1/2 so the slope is 0.5 or ½

2) it crosses the y axis at the average of the origin and 4.

4 + 0 = 4 / 2 = 2 so y intercept is 2.

3) in y= mx + b form

f(x) = ½x + 2, or, f(x) = 0.5x + 2

The limit of j (x) as x approaches 3 is 4.

According to the question, A line begins at an open circle (2, 6) on a coordinate plane and descends through ( -2, 2). At, a complete circle is (3, 6). From a solid circle (2, 3), a curve leads to an open circle (3, 4). From the open circle to the closed circle, a line runs (6, 5).

From the graph, it can be seen that the limit of the function j(x) as the value of x approaches 3 is 4.

A diagram or pictorial representation that organizes the depiction of data or values is known as a graph.

The relationships between two or more items are frequently represented by the points on a graph.

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Answer:

4

Step-by-step explanation:

egg 2023

The measure of angle a is 70 degrees

The angle in a semicircle is a right angle.

This means that:

a + d = 90

Where O is the center of the circle

From the figure, we have:

Angle d and the angle with a measure of 20 degrees are corresponding angles

This means that

d = 20

Substitute d = 20 in a + d = 90

a + 20 = 90

Subtract 20 from both sides of the equation

a = 70

Hence, the measure of angle a is 70 degrees

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46) True

47) False

These are true by definition.

Answer:

4, 6, 4

6, 12, 8

8, 12, 6

Step-by-step explanation:

Tetrahedron:

4 faces, 6 edges, 4 vertices

Cube:

6 faces, 12 edges, 8 vertices

Octahedron:

8 faces, 12 edges, 6 vertices

you can also use the v - e + f = 2 to check

Answer:

22

Step-by-step explanation:

9. The curve passes through the point (-1, -3), which means

[tex]-3 = a(-1) + \dfrac b{-1} \implies a + b = 3[/tex]

Compute the derivative.

[tex]y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}[/tex]

At the given point, the gradient is -7 so that

[tex]-7 = a - \dfrac b{(-1)^2} \implies a-b = -7[/tex]

Eliminating [tex]b[/tex], we find

[tex](a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}[/tex]

Solve for [tex]b[/tex].

[tex]a+b=3 \implies b=3-a \implies \boxed{b = 5}[/tex]

10. Compute the derivative.

[tex]y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6[/tex]

Solve for [tex]x[/tex] when the gradient is 2.

[tex]x^2 - 5x + 6 = 2[/tex]

[tex]x^2 - 5x + 4 = 0[/tex]

[tex](x - 1) (x - 4) = 0[/tex]

[tex]\implies x=1 \text{ or } x=4[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}[/tex]

[tex]\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}[/tex]

11. a. Solve for [tex]x[/tex] where both curves meet.

[tex]\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5[/tex]

[tex]\dfrac{x^3}3 - 2x^2 - 9x = 0[/tex]

[tex]\dfrac x3 (x^2 - 6x - 27) = 0[/tex]

[tex]\dfrac x3 (x - 9) (x + 3) = 0[/tex]

[tex]\implies x = 0 \text{ or }x = 9 \text{ or } x = -3[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}[/tex]

[tex]B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}[/tex]

[tex]C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}[/tex]

11. b. Compute the derivative for the curve.

[tex]y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8[/tex]

Evaluate the derivative at the [tex]x[/tex]-coordinates of A, B, and C.

[tex]A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}[/tex]

[tex]B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}[/tex]

[tex]C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}[/tex]

12. a. Compute the derivative.

[tex]y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}[/tex]

12. b. By completing the square, we have

[tex]12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4[/tex]

so that

[tex]\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}[/tex]

13. a. Compute the derivative.

[tex]y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}[/tex]

13. b. Complete the square.

[tex]3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3[/tex]

Then

[tex]\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}[/tex]

5 and 6 thousandths in standard form is 5.006

The number is given as:

5 and 6 thousandths

This can be rewritten as

5 +  6 thousandths

Express as numbers

5 +  0.006

Evaluate the sum

5.006

Hence, 5 and 6 thousandths in standard form is 5.006

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The length of the diagonal that he cut to the nearest whole inch is 36 inches

A red dotted line crosses the rectangle from the upper left corner to the lower right corner to form a triangle.

Length of the diagonal of a rectangle;

Hypotenuse² = adjacent² + opposite²

= 18² + 36²

= 324 + 1296

hyp² = 1620

hyp = √1620

hyp = 35.4964786985976

Approximately,

hypotenuse = 36 inches

Therefore, the length of the diagonal that he cut to the nearest whole inch is 36 inches.

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