Your professor selects three students from your Stats class to form a study team. How many study teams are possible from a class of 30 students

1. The time taken for the water to fall that much is 0.2 s

2. The initial speed of the water is 4 m/s

h = ½gt²

0.2 = ½ × 9.8 × t²

0.2 = 4.9 × t²

Divide both side by 4.9

t² = 0.2 / 4.9

Take the square root of both side

t = √(0.2 / 4.9)

t = 0.2 s

s = ut

Divide both sides by t

u = s / t

u = 0.8 / 0.2

u = 4 m/s

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Question 4

[tex]\sin 30^{\circ}=\frac{y}{14\sqrt{3}}\\\\\frac{1}{2}=\frac{y}{14\sqrt{3}}\\\\\boxed{y=7\sqrt{3}}\\\\\\\\\cos 30^{\circ}=\frac{x}{14\sqrt{3}}\\\\\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\\\\\boxed{x=21}[/tex]

Question 5

[tex]\cos 45^{\circ}=\frac{x}{28}\\\\\frac{1}{\sqrt{2}}=\frac{x}{28}\\\\x=\frac{28}{\sqrt{2}}\\\\\boxed{x=14\sqrt{2}}\\\\\\\\\sin 45^{\circ}=\frac{y}{28}\\\\\frac{1}{\sqrt{2}}=\frac{y}{28}\\\\y=\frac{28}{\sqrt{2}}\\\\\boxed{y=14\sqrt{2}}[/tex]

The average increase in the number of flowers pollinated per day between days 4 and 10 is 39, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].

In the question, we are asked for the average increase in the number of flowers pollinated per day between days 4 and 10, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].

To find the average increase in the number of flowers pollinated per day between days 4 and 10, we use the formula {f(10) - f(4)}/{10 - 4}.

First, we find the value of the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex], for f(10) and f(4).

[tex]f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(10) = (3)^{\frac{10}{2} }\\\Rightarrow f(10) = 3^5 = 243[/tex]

[tex]f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(4) = (3)^{\frac{4}{2} }\\\Rightarrow f(10) = 3^2 = 9[/tex]

Thus, the average increase

= {f(10) - f(4)}/{10 - 4},

= (243 - 9)/(10 - 4),

= 234/6

= 39.

Thus, the average increase in the number of flowers pollinated per day between days 4 and 10 is 39, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].

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For complete question, refer to the attachment.

The cross-section would be the same shape as the base. A cross-section cannot be a sphere because a cross-section must be two-dimensional, but a sphere is three-dimensional.

Si triplicamos la cantidad de cada artículo, el subtotal será de 2 940 soles, el impuesto al consumo será de 529,20 soles y el total será de 3 469,20 soles.

En esta pregunta tenemos un recibo con una lista de productos, cantidades, precios unitarios y precios de compra, además de un impuesto que se aplica al consumo por concepto de los productos comprados y el total, el cual refleja tanto el subtotal como el concepto por impuestos.

En cuanto a la compra de una determinada cantidad para un artículo, el coste total respectivo es igual a producto del número de artículos por el precio unitario.

El subtotal se determina mediante la suma de los costes totales asociados a cada artículo y luego se le recarga con el impuesto al consumo. Por último, el total de compra es la suma del subtotal y los pagos tributarios.

Si triplicamos el número de cada artículo comprado, tenemos los siguientes costes para cada artículo:

Chompas de dralom

12 / 4 × 280 = 840

Chompas de Orlon

6 / 2 × 100

300

Chompas de lana de alpaca

15 / 5 × 600

1 800

El subtotal es 2 940 soles y el impuesto al consumo es 18 / 100 × 2 940 = 529,20 soles. Por tanto, el total es 3 469,20 soles.

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Answer:

1st option

Step-by-step explanation:

let y = h(x) and rearrange making x the subject , that is

y = 6x - 6 ( add 6 to both sides )

y + 6 = 6x ( divide both sides by 6 )

[tex]\frac{y+6}{6}[/tex] = x

change y back into terms of x with x = [tex]h^{-1}[/tex] (x)

[tex]h^{-1}[/tex] (x) = [tex]\frac{x+6}{6}[/tex]

a

jfjoydx9tdzrzfzixitxxytxixutx

The test that represents the best choice if you wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr is one sample t-test

The purpose of the one sample t-test is to determine if a sample observations could follow a specific parameter like the mean.

This test is used to determine whether an unknown population mean is different from a specific value.

one sample t-test is used to compare the mean of a single group against that of a known mean.

We wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr

So this test (one sample t-test ) would be the best choice to compare mttr.

Therefore, the test that represents the best choice if you wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr is one sample t-test

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The two equivalent forms of [tex]sec^2x[/tex] are [tex]\frac{1}{cos^2x}[/tex] and [tex]1+tan^2x[/tex]

The given trigonometric expression is:

[tex]sec^2x[/tex]

Note that:

[tex]sec(x)=\frac{1}{cos(x)}[/tex]

Substitute this equivalence to the given expression

[tex]sec^2(x)=\frac{1}{cos^2(x)}[/tex]

Also from [tex]cos^2(x)+sin^2(x)=1[/tex]

Divide through by [tex]cos^2x[/tex]

[tex]\frac{cos^2x}{cos^2x} +\frac{sin^2x}{cos^2x}=\frac{1}{sin^2x}[/tex]

Simplifying the resulting expression:

[tex]1+tan^2x=sec^2x\\\\sec^2x=1+tan^2x[/tex]

Therefore, the two equivalent forms of [tex]sec^2x[/tex] are [tex]\frac{1}{cos^2x}[/tex] and [tex]1+tan^2x[/tex]

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Answer:

$270

Step-by-step explanation:

18x15=$270

HOPE THAT HELPED:)

Answer:

a.1

b. 17

Step-by-step explanation:

So the | | around an equation makes it an absolute value which can be defined as the "distance from zero" or in other words, if a value is positive, it stays positive, and if a value is negative, it becomes positive.

|-8 + 9| = |1| = 1

|-8| + 9 = 8 + 9 = 17

A irrational number is a number that can't be expressed as a ratio of two whole numbers. That's it.

For examples (in increasing order of difficulty)

1 is a rational number because it is 1/1

0.75 is a rational number because it is equal to 3/4

2.333... (infinite number of digits, all equal to three) is rational because it is equal to 7/3.

sqrt(2) is not a rational number. This is not completely trivial to show but there are some relatively simple proofs of this fact. It's been known since the greek.

pi is irrational. This is much more complicated and is a result from 19th century.

As you see, there is absolutely no mention of the digits in the definition or in the proofs I presented.

Now the result that you probably hear about and wanted to remember (slightly incorrectly) is that a number is rational if and only if its decimal expansion is eventually periodic. What does it mean ?

Take, 5/700 and write it in decimal expansion. It is 0.0057142857142857.. As you can see the pattern "571428" is repeating in the the digits. That's what it means to have an eventually periodic decimal expansion. The length of the pattern can be anything, but as long as there is a repeating pattern, the number is rational and vice versa.

As a consequence, sqrt(2) does not have a periodic decimal expansion. So it has an infinite number of digits but moreover, the digits do not form any easy repeating pattern.

Answer:

Step-by-step explanation:

From the right triangle with cot theta = 4/7  = (adjacent side / opposite side), Fatima can find the length of the hypotenuse using Pythagoras. This gives  a value of 8.0823. So cos theta = 4 / 8.0823 = 0,49614,

Then she could find the values of sin theta  using

sine squared theta + cos squared theta = 1.

Answer: its D or 1 + cotangent squared theta = cosecant squared theta

Step-by-step explanation:

Answer:

[tex]\sf A) \quad \begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]

B)  see below

C)  points C, E and F

Step-by-step explanation:

Given points:

A system of inequalities is a set of two or more inequalities in one or more variables.

To create a system of inequalities that only contains C and F in the overlapping shaded region, create a linear equation where points C, F and E are to the right of the line and a linear equation where points C, F, A, B and D are to the left of the line.

The easiest way to do this is to find the slope of the line that passes through points C and F, then add values to move the lines either side of the points.

[tex]\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{y_F-y_C}{x_F-x_C}=\dfrac{-4-1}{3-2}=-5[/tex]

Therefore:

[tex]\sf y = -5x + 5[/tex]  →  points C, F and E are to the right of the line.

[tex]\sf y=-5x+12[/tex]  →  points C, F, A, B and D are the left of the line.

Therefore, the system of inequalities that only contains points C and F in the overlapping shaded regions is:

[tex]\begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]

To graph the system of inequalities:

To verify that the points C and F are solutions to the system of inequalities created in Part A, substitute the x-values of both points into the system of inequalities.  If the y-values satisfy both inequalities, then the points are solutions to the system.

Point C (2, 1)

[tex]\implies \sf x=2 \implies 1 > -5(2)+5 \implies 1 > -5\quad verified[/tex]

[tex]\implies \sf x=2 \implies 1 < -5(2)+12\implies 1 < 2 \quad verified[/tex]

Point F (3, -4)

[tex]\implies \sf x=3 \implies -4 > -5(3)+5 \implies -4 > -10\quad verified[/tex]

[tex]\implies \sf x=3 \implies -4 < -5(3)+12\implies -4 < -3 \quad verified[/tex]

Method 1

Graph the line y = 7x - 4 (making the line dashed since it is y < 7x - 4).

Shade below the dashed line.

Points that are contained in the shaded region are the houses in which Erica is interested in living:  points C, E and F.

Method 2

Substitute the x-value of each point into the given inequality y < 7x - 4.

Any point where the y-value satisfies the inequality is a house that Erica is interested in living.

[tex]\sf Point\: A: \quad x=-5 \implies 5 < 7(-5)-4 \implies -5 < -39 \implies no[/tex]

[tex]\sf Point\: B: \quad x=-4 \implies -2 < 7(-4)-4 \implies -2 < -32 \implies no[/tex]

[tex]\sf Point\: C: \quad x=2 \implies 1 < 7(2)-4 \implies 1 < 10 \implies yes[/tex]

[tex]\sf Point\: D: \quad x=-2 \implies 4 < 7(-2)-4 \implies 4 < -18 \implies no[/tex]

[tex]\sf Point\: E: \quad x=2 \implies 4 < 7(2)-4 \implies 4 < 10 \implies yes[/tex]

[tex]\sf Point\: F: \quad x=3 \implies -4 < 7(3)-4 \implies -4 < 17 \implies yes[/tex]

This follows from the identity

[tex]\cot(x) = \dfrac1{\tan(x)} = \tan(90^\circ - x)[/tex]

The overall product is 1 because

[tex]\tan(10^\circ) \tan(80^\circ) = \cot(80^\circ) \tan(80^\circ) = 1[/tex]

[tex]\tan(20^\circ) \tan(70^\circ) = \cot(70^\circ) \tan(70^\circ) = 1[/tex]

[tex]\tan(30^\circ) \tan(60^\circ) = \cot(60^\circ) \tan(60^\circ) = 1[/tex]

[tex]\tan(40^\circ) \tan(50^\circ) = \cot(50^\circ) \tan(50^\circ) = 1[/tex]

Answer:

Step-by-step explanation:

The high school graduation is the dependent variable.

In this question,

Jordan is studying whether or not race affects high school graduation in utah.

A dependent variable is a variable in an expression that depends on the value of another variable. It is a variable that represents a quantity that changes based on other quantities being manipulated in an experiment. It is the variable being tested, and therefore, it is called the dependent variable.

In the above statement, the high school graduation gets affected by race. So, the the high school graduation is the dependent variable.

Hence we can conclude that the high school graduation is the dependent variable.

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Answer:

35449 people

Explanation:

Given equation: y = 1100.74x -1976.47

Here year '2005' represents x = 005 = 5

So, the year '2034' will represent x = 034 = 34

Insert this into equation:

y = 1100.74(34) -1976.47

y = 35448.69

y ≈ 35449 (rounded to nearest whole number)

The Polynomial Roots are given as follows:
 

x₁=−1.2085

x₂=0.2709+1.2576i

x₃=0.2709−1.2576i

It is to be noted that the Polynomial 3x³+ 2x2 + 3x + 6 has no real rational roots that can be found using the rational root text. Hence, this was solved using the Qubic Formulas.

The roots (also known as zeroes or solutions) of a polynomial P (x) P(x) P(x) are the x values for which P (x) P(x) P(x) equals zero.

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​

Answer: 12

Step-by-step explanation:

I got it right

Quadratic expressions are mathematical expressions that have the power of one of its alphabets being equal to and not more than two. The given expression in the question can be simplified as 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 112

Algebraic expressions are mathematical expressions that consist of given number(s) and alphabet(s). When the power (exponential) of one of the alphabets is given as two, then the expression can be classified as a quadratic expression.

Thus the given expansion of the quadratic expression can be simplified as follows:

([tex]9x^{2}[/tex] + 9x - 4)([tex]9x^{2}[/tex] + 9x - 10) - 72

But,

([tex]9x^{2}[/tex] + 9x - 4)([tex]9x^{2}[/tex] + 9x - 10) = [tex]9x^{2}[/tex]([tex]9x^{2}[/tex] + 9x - 10) + 9x([tex]9x^{2}[/tex] + 9x - 10) - 4([tex]9x^{2}[/tex] + 9x - 10)

               = 81[tex]x^{4}[/tex] + 81[tex]x^{3}[/tex] - 90[tex]x^{2}[/tex] + 81[tex]x^{3}[/tex] + 81 [tex]x^{2}[/tex] - 90x - 36 [tex]x^{2}[/tex] + 36x - 40

               = 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 40

So that we have;

(81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 40) - 72  = 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 40 -72

                  = 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 112

Thus the simplification of the given expression is 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 112

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It is an example of Permutation without repetition.

To find the number of pictures can create.

An arrangement of things in a certain sequence is what we refer to as a permutation. The constituents or components of sets are presented in this section in a sequential or chronological fashion. A coming together or merging of several parts or characteristics in which each of the constituent parts or traits retains its unique identity.

Given that:

50 unique pieces of puzzle

Use them all to solve puzzle

No repetition is allowed therefore

Since no repetition is allowed because all puzzle pieces are required to be used.

Therefore each of the 50 pieces can be used max one time

Therefore if we start with piece like this " 1 2 3 .. up to ... 50 " it is one combination

Another could be "2 1 3 ..  up to .. 50" or "3 1 2 .. up to 50" and so on

So we see that we can see that we can select the first piece in 50 ways 49 left which could be selected for second piece then 48, 47 respectively

Therefore 50X49X48X47X46X....X1 = 50!

Hence, It is an example of Permutation without repetition.

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Option is missing

A. Combination with repetition

B. Permutation with repetition

C. Combination without repetition

D. Permutation without repetition

Answer:

[tex]h(x)=-0.11(3)^{-x}[/tex]

Step-by-step explanation:

Given function:

[tex]f(x)=0.11(3)^x[/tex]

Reflection in the x-axis:

When reflecting a function in the x-axis, make the whole function negative:

[tex]\implies g(x)=-f(x)[/tex]

[tex]\implies g(x)=-0.11(3)^x[/tex]

Reflection in the y-axis:

When reflecting a function in the y-axis, make the x variable negative:

[tex]\implies h(x)=g(-x)[/tex]

[tex]\implies h(x)=-0.11(3)^{-x}[/tex]

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Answer:

the answer to this question is d

Step-by-step explanation:

h(x)=-0.11(3)^-x

The speed of light is 300,000 kilometers per hour. hence option c is correct.

According to the statement

we have to explain that the by how much speed the light travels.

So, Firstly

Light is the natural agent that stimulates sight and makes things visible.

and speed of light means that the distance covered by the light with respect to time.

So, it means those distance which is covered by the light with respect to time is called the speed of light.

The speed of light is measures in km, meters and in other units also.

And we can calculate the speed of light per hour or per minute or in other time measurements.

So, The speed of light is 300,000 kilometers per hour. hence option c is correct.

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a.

b.

A quadratic function is a function of the form y = ax² + bx + c

To write y = 5(x - 2)² + 7, we expand the bracket and simplify.

So, y = 5(x - 2)² + 7

y = 5(x² - 4x + 4) + 7

y = 5x² - 20x + 20 + 7

y = 5x² - 20x + 27

So,  the quadratic function y = 5(x - 2)² + 7 in standard form is y = 5x² - 20x + 27

The y intercept is gotten when x = 0.

So, y = 5x² - 20x + 27

y(0) = 5(0)² - 20(0) + 27

y(0) = 0 - 0 + 27

y(0) = 27

The y intercept is (0, 27)

Find the graph in the attachment

To write y = 3(x + 1)² + 12, we expand the bracket and simplify.

So, y = 3(x + 1)² + 12

y = 3(x² + 2x + 1) + 12

y = 3x² + 6x + 3 + 12

y = 3x² + 6x + 15

So, the quadratic function y = 3(x + 1)² + 12 in standard form is y = 3x² + 6x + 15

The y intercept is gotten when x = 0.

So, y = 3x² + 6x + 15

y(0) = 3(0)² - 6(0) + 15

y(0) = 0 - 0 + 15

y(0) = 15

The y-intercept is (0, 15)

Find the graph in the attachment

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The answers to all the parts are given below.

The height equation is:

h(t) = - 16t^2 + 100t + 10

A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.

v(t) = - 2 * 16 * t + 100 = 0

t = 100/(2*16) = 3.125s

Now we put this time in our height equation:

h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.

B) The lowest height of the potato will be h = 0ft when the potato hits the ground.

C) the lowest time that works for that function is t = 0s when the potato is fired by the gun.

D) the maximum time can be found when the potato hits the ground, after that point the equation does not work anymore, let's find it,

h(t) = 0 = -16t^2 +100t+10

we can solve it using Bhaskara's equation:

[tex]t=\frac{-100+-\sqrt{100^{2} +4*16*10} }{-2*16} =\frac{-100+-103.2}{-32}[/tex]

So the two solutions are:

t = 6.3s

t = -0.1s

We need to choose the positive time, as we already discussed that the minimum time that works for the equation is t = 0s.

so here the answer is t = 6.3s

E) the domain is  0s ≤ t ≤ 6.3s

The range is 0ft ≤ h ≤ 166.25 ft.

Therefore, the answers to all the parts are shown.

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