Complete the two column proof.

The number of times Aarf the Dog bark for 15 seconds if the dog starts barking at midnight and stops at 1:01 AM is 91.5 times

Total time of barking and silence = 15 seconds + 25 seconds

= 40 seconds

Difference in time = 1:01 AM - 12:00 am

= 1 hour 1 minutes

Convert hours and minutes to seconds

1 hour 1 minutes = 61 minutes × 60 seconds

= 3,660 seconds

Number of times Aarf the Dog bark for 15 seconds = Total hours Aarf dog barks / Total time of barking and silence

= 3,660 / 40

= 91.5 times

So therefore, the number of Aarf the Dog bark for 15 seconds is 91.5 times

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Answer:

Step-by-step explanation:

[tex]\begin{array}{ccll} hours&gallons\\ \cline{1-2} 2.5 & 1625\\ h& 3185 \end{array} \implies \cfrac{2.5}{h}~~=~~\cfrac{1625}{3185}\implies \cfrac{2.5}{h}=\cfrac{25}{49} \\\\\\ (2.5)(49)=25h\implies \cfrac{(2.5)(49)}{25}=h\implies 4.9=h\qquad \textit{4 hours and 54 minutes}[/tex]

The required expression which is equivalent to -51 -(-60) is -51 + 60.

Given that,
Which expression is equivalent to -51 - (-60) is to be determined.

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
Given expression =  -51 - (-60)
Simplify by applying the property of distribution under the parenthesis,
= -51 + 60

Thus, The required expression which is equivalent to -51 -(-60) is -51 + 60.

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The ratio sinC is given by 4/5 and in decimal form is given by sinC = 0.8.

The trigonometric functions are real functions in mathematics that connect the angle of a right-angled triangle to the ratios of its two side lengths. They are also known as circular functions, angle functions, or goniometric functions. They are extensively employed in all geosciences, including navigation, solid mechanics, celestial mechanics, geodesy, and many more. As some of the most basic periodic functions, they are frequently employed in Fourier analysis to examine periodic events.

The sine, cosine, and tangent are the trigonometric functions that are most commonly utilized in contemporary mathematics. The cotangent, secant, and their less common reciprocals, the cosecant, are each of their reciprocals. These six trigonometric functions each have an analog in the hyperbolic functions, as well as an equivalent inverse function.

The value of sin C is given by

sinC = perpendicular/ hypotenuse

sinC = AB/AC

sinC = 4/5

In decimal form, it is written as

sinC = 0.8

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Answer:

B

Step-by-step explanation:

f(-1)=(-1)²-3(-1)-4

f(-1)=1-3(-1)-4

f(-1)=1+3-4

f(-1)=0

Do this same process with all the equations until you find one that doesn't equal 0.

m ∠5 and m ∠7 is  72 ° . All straight angles are 180 degrees, according to the straight angle theorem. A straight angle is formed when the angle's legs point precisely in the opposite directions. The symbol for a straight angle is 180 degrees, or  π (in radians).

An angle is a shape created by two rays that share a terminus and are referred to as the angle's sides and vertices, respectively. Angles created by two rays are in the plane where the rays are located. The meeting of two planes also creates angles. We refer to these as dihedral angles.

The names of basic angles are Acute angle, Obtuse angle, Right angle, Straight angle, reflex angle and full rotation. An angle is geometrical shape formed by joining two rays at their end-points. An angle is usually measured in degrees.

m ∠4 = 108°

m ∠5 = m ∠1

m ∠1 is in same line of m ∠4

So

m ∠1 = 180 - m ∠4

m ∠1 = 180 - 108

m ∠1 = 72 °

So m ∠5 = 72 °

m ∠7 is opposite to m ∠5 , opposite angles will be equal

Therefor m ∠7 = 72 °

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Step-by-step explanation:

as the graph shows :

4x + 12x = 320

16x = 320

x = 20

The probability that the sum of the numbers rolled on the dice is even =1/2.

For the given question,

A fair dice is rolled twice.

Six different results are available with a single roll of the dice (1,2,3,4,5,6)

Therefore, if two dice are rolled, there are a total of 36 possible results, or 6².

Sample space for the sum of even = [(1,1), (1,3), (1,5), (2,2), (2, 4),(2,6), (3, 1), (3,3),(3,5),(4,2),(4,4),(4,6), (5, 1), (5,3),(5,5),(6,2),(6,4),(6,6)]

Total sample space for sum of even = 18

For the probability that the sum of the numbers rolled is even is -

probability (sum is even) = 18 / 36 = 1/2

Thus, the probability that the sum of the numbers rolled is even is 1/2.

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If they are competing for 1st, 2nd, and 3rd place ribbons. The number of different ways can the winners be chosen is:  B. 60.

Given data:

Number of people =5

Number of places = 1st, 2nd, and 3rd

Hence,

Different combinations =5 ×4×3

Different combinations = 60 ways

Therefore the winners will be chosen by  60 different possible ways ,

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The graph of the coordinate points of the vertices of ABCD indicates;

Part A The perimeter of ABCD ia approximately 61.2 units

Part B. The area of ABCD is 199.5 square units

The perimeter of a figure is length of the line that goes round the figure.

A. The perimeter of the polygon is the sum of the length of the sides of the polygon, which is found as follows;

AB = √((21-11)²+(11-(-1))²) ≈ 15.6

BC = √((21-26)²+(11-(-1))²) = 13

CD = √((26-2)²+(-1-(-8))²) = 25

AD = √((2-5)²+(-8-(-1))²) ≈ 7.6

The perimeter of the polygon = 15.6 + 13 + 25 + 7.6 = 61.2

B. Plotting the figure indicates that the figure consists of two triangles, ΔABC and ΔACD

The base length of triangle ΔABC = 26 - 5 = 21

The height of triangle ΔABC = 11 - (-1) = 12

The area of triangle ΔABC = 0.5 × 21 × 12 = 126

Base length of triangle ΔADC = 21

Height of triangle ΔADC = -1 - (-8) = 7

Area of triangle ΔADC = 0.5 × 21 × 7 = 73.5

Area of the composite figure is therefore;

A = 126 + 73.5 = 199.5

The area of the figure is therefore; 199.5 square units

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The solution after divide by 8/5 instead of 4/5 will be 2.

What is mean by Fraction?

A fraction is a part of whole number, and a way to split up a number into equal parts.

Or, A number which is expressed as a quotient is called fraction.

It can be written as the form of p : q, which is equivalent to p / q.

Given that;

The model shows that 3 1/5 divided by 4/5 = 4.

Now,

Substitute 8/5 instead of 4/5, we get;

⇒ 3 1/5 ÷ 8/5

⇒ 16/5 ÷ 8/5

⇒ 16/5 × 5/8

⇒ 2

Thus, The solution after divide by 8/5 instead of 4/5 will be 2.

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Answer:

Step-by-step explanation:

The variation equation that connects x and y is y³ = 4.5x(x - 1) and the value of y is 4.48 when x equals  5

From the question, the variation statement is given as

The cube of y is directly proportional to x(x - 1)

Also, we have the initial values of x and y to be

x = 3 and y= 3

When the variation statement is represented as an equation, we have:

y³ = kx(x - 1)

Where k is the variation constant

Substitute the known values in the above equation So, we have the following equation

3³ = k x 3 x (3 - 1)

Evaluate

27 = 6k

Divide both sides by 6

k = 4.5

Substitute k = 4.5 in y³ = kx(x - 1)

y³ = 4.5x(x - 1)

Hence, the equation is y³ = 4.5x(x - 1)

Here, we are given that

x = 5

Substitute 5 for x in the equation y³ = 4.5x(x - 1)

So, we have

y³ = 4.5 x 5 x (5 - 1)

Evaluate

y³ = 90

Take the cube root of both sides

y = 4.48

Hence, the value of y is 4.48

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Answer: 0.4754

A poker hand is a combination of 5 cards from the pack of 52 cards.

One hand full is said when we have 2 cards of the same rank.

In this question, we have to find the probability of 2 hands full of different ranks.

So,

We use the formula : C(n,k)=\frac{n!}{k!(n-k)!}

From permutations and combinations, since order doesn't exist.

From a suit of 13 cards any two cards are chosen:

C(13,2)=\frac{13!}{2!(13-2)!}

C(13,2)=\frac{13×12×11!}{2!11!}

C(13,2)=\frac{13×12!}{2!}

C(13,2)=156/2

C(13,2)=78

78 ways are there to choose for each of the pairs .

Now,

If one of the ranks is of any particular suit or maybe a face card out of which two of them are chosen.

C(4,2)=\frac{4!}{2!(4-2)!}

C(4,2)=6

And same for the other sets of cards of another pair.

We have 78×6×6= 2808 ways to pick the first 4 cards.

Now,

For the fifth card which can be any card except for the paired suit i.e, 13-2=11

There are 4 suits to pick from, 4×11= 44 cards to pick for the final card.

For the first card = 2080

For the fifth card= 44

Total number of two pair hands possible= 2080×44= 123,552

Out of:

C(52,5)=\frac{53!}{5!(52-5)!}

C(52,5)= 2,598,960

Probability of event= 123,552/2,598,960

= 0.4753901560624

Final answer= 0.4754

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The probability of being dealt a two-pair is 0.4754.

Given;

What is the likelihood of receiving a two-pair when dealing a five-card poker hand from a 52-card deck that has been properly shuffled?

We use combinations to solve the given case;

From each suit, any two cards are chosen,

C(13,2) = [tex]\frac{13!}{2!(13-2)!}[/tex]

           = [tex]\frac{13*12*11!}{2!11!}[/tex]

           = 156/2

           = 78

Hence, there are 78 ways to choose for each of the pairs.

Now,

If one of the ranks is a face card, two of them are chosen, or if one of the ranks is a certain suit.

C(4,2) = [tex]\frac{4!}{2!(4-2)!}[/tex]

          =6

Similar to the other sets of cards of another pair.

We have 78×6×6= 2808 ways to pick the first 4 cards.

Now,

The fifth card, which can be any card other than the paired suit, is dealt as follows: 13 - 2 = 11.

There are 44 cards total to choose from among the 4 suits, or 4 x 11.

2080 for the primary card.

44 for the fifth card.

Possible two-pair hands total 2080 x 44 = 123,552.

Then,

C(52,5) = [tex]\frac{53!}{5!(52-5)!}[/tex]

            = 2,598,960

Probability of event = Possible two-pair hands / Total outcome

                                 = 123,552 / 2,598,960

                                 = 0.47539

Hence, the probability of being dealt a two-pair is 0.4754.

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The graph of the function  y=3 -1/2x is attached below.

Function:

Function refers the relationship between inputs where each input is related to exactly one output.

Given,

Here we have the function  y=3 -1/2x.

Now, we have to plot the graph of the function.

In order to plot the graph of the function, we have to take some sample values for x.

So, we will take the values of the x as -2, -1, 0, 1, 2.

Now, we have to apply the values on the given function then we get,

=> x = -2 => y = 3 - 1/2(-2) => y = 3 + 1 => y = 4

=> x = -1 => y = 3 - 1/2(-1) => y = 3 + 1/2 => y = 7/2 = 3.5

=> x = 0 => y = 3 - 1/2(0) => y = 3 - 0 => y = 3

=> x = 1 => y = 3 - 1/2(1) => y = 3 - 1/2 => y = 5/2 = 2.5

=> x = 2 => y = 3 - 1/2(2) => y = 3 - 1 => y = 2

When we enter these values into the table, then we get the table like the following.

By using the table we have to plot the graph and then we get the graph like the following.

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Answer:

x=22

Step-by-step explanation:

Let the unknown number be x

:- the power of x = x³

From the question

x³-15=7 and x²

x³=7+15

x³ * x²= 22

x= 22

[tex]T(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{3}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{ \cfrac{3}{5}}(x-\stackrel{x_1}{4}) \\\\\\ y-1=\cfrac{3}{5}x-\cfrac{12}{5}\implies y=\cfrac{3}{5}x-\cfrac{12}{5}+1\implies {\Large \begin{array}{llll} y=\cfrac{3}{5}x-\cfrac{7}{5} \end{array}} \\\\[-0.35em] ~\dotfill[/tex]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the line AB

[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-2}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}} \implies \cfrac{ -3 }{ 4 } \implies - \cfrac{3 }{ 4 } \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{4}} ~\hfill \stackrel{reciprocal}{\cfrac{4}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{4}{-3}\implies \cfrac{4}{3}}}[/tex]

so we're really looking for the equation of a line that has a slope of 4/3 and that it passes through (3 , 2)

[tex](\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{3}) \\\\\\ y-2=\cfrac{4}{3}x-4\implies {\Large \begin{array}{llll} y=\cfrac{4}{3}x-2 \end{array}}[/tex]

Answer:

A base

Step-by-step explanation:

its not an acid bc a ph level of 14 means is has a strong base.

Answer:

Step-by-step explanation:

The answer is 6.

I got this by subtracting 23 by 12 (the number of numbers shown). Then I got 11. So I followed the pattern until I got six.

23-12=11

2,4,6,8,2,4,6,8,2,4,6,8,......2,4,6,8,2,4,6,8,2,4,6

The dimensions of the box that minimize total cost are height h =3.17m, width, w and length, l=9.52m  are equal to  .

Let h is the height of the box.

Given Volume of the box is 216 m³  with no top.

And the cost of the bottom is 40USD/m².

For the sides, it is 30USD/m².

The bottom of the box is square, so l = w.

So Volume

[tex]V=hl^{2}[/tex]

[tex]216=hl^{2}[/tex]

[tex]h=\frac{216}{l^{2} }[/tex]

Now Surface Area of the box without a top is

S = 4hl + l²

So,

cost = 4 × 30hl +40l²

cost = 120 hl + 40 l²

Putting h ,

cost = 120 × [tex](\frac{216}{l^{2} })l + 40l^{2}[/tex]

cost = [tex]\frac{25960}{l^{2} }+40l^{2}[/tex]

To find minimum cost, derivate the  by using the calculator at

[tex]l=3\sqrt[3]{4}[/tex]

l=4.762203 m

[tex]h=\frac{216}{4.762203^{2} }[/tex]

h = 9.5244069354 m

Hence the dimensions of the box are

l=w

 =9.52 m

and height ,h=3.17 m  

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The probability that a card drawn randomly from a standard deck of 52 cards is a red queen is 1/26 or 0.038461. The result is obtained from the ratio of the number of red queen to 52 cards.

Probability of an event can be expressed as

P(A) = n(A) / n(S)

Where

In case a card drawn randomly from 52 cards, what is the probability that it is a red queen?

The four suits for a standard deck of 52 cards are hearts, diamonds, clubs, and spades of 13 cards each. The hearts and diamonds are red. While, the clubs and spades are black. The 13 cards are Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

From that information, the number of red queen in a standar deck of 52 cards is 2 cards. So,

The probability of drawing a red queen is

P(A) = n(A) / n(S)

P(A) = 2/52

P(A) = 1/26

P(A) = 0,038461

Hence, the probability that a card drawn is a red queen is 1/26 or 0,038461.

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The value of function is positive so the function have maximum value 30 at x=3.

In the given question we have to find the maximum or the minimum value of a function.

The given function is

f(x) = -3x^2+18x+3

To find the maximum or minimum value we firstly find the value of f'(x).

f'(x) = -6x+18

Now put f'(x)=0

-6x+18=0

Subtract 18 on both side we get

-6x=-18

Divide by -6 on both side we get

x = 3

Now finding the value of function at x=3

f(3)= -3*(3)^2+18*3+3

f(3)= -3*9+54+3

f(3)= -27+54+3

f(3)= 30

Since the value of function is positive so the function have maximum value 30 at x=3.

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The graph of the cosine function y = 3cos(4θ) - 1 is given by the image presented at the end of the answer.

The cosine function is defined by the rule presented as follows:

g(x) = acos(bx+c)+d.

The coefficients have the roles listed as follows:

In this problem, the definition of the function is:

y = 3cos(4θ) - 1

Considering the amplitude of 3 and the vertical shift of -1 the function will oscillate between:

-4 and 2.

(Without the vertical shift it would be between -3 and 3).

The period of the function will be as follows:

2π/b = 4

b = 2π/4

b = π/2.

Considering these features, the graph is given at the end of the answer.

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The value of their horizontal change is 2.7 feet

The equation of the function is given as

y = -8/3x + 40

From the question, the distance travelled by the rider is given as

Distance = 8 feet downward

This means that

Horizontal change = slope

A linear equation is represented as

y = mx + c

Where

Slope = m

By comparing y = mx + c and y = -8/3x + 40, we have

Slope = m = 8/3

This gives

Horizontal change = 8/3

Evaluate

Horizontal change = 2.7

Hence, the horizontal change is 2.7 feet

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The type of bias that can be present in stratified sampling is that the students cannot be selected to accurately reflect the population.

Samples may be classified according to the five options as follows:

In this problem, an equal amount of students from each class was surveyed, hence a stratified sample was used.

The type of bias present in stratified sampling is that the students cannot be selected to accurately reflect the population.

The problem asks for which type of bias is present in this sample.

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Answer:

x = -1

Step-by-step explanation:

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Answer:

-1

Step-by-step explanation:

Replace the variable x with 1 in the expression .

f(1)= 2(1) - 3

Remove the parenthasis

2(1) - 3

Multiply 2 by 1

2 - 3

subtract 2 from 3

-1

The number of batches of muffins Ms. Kirkland can bake with 7 ½ pounds of flour is 5 batches.

Given that, Ms. Kirkland is baking muffins. Each batch of muffins uses 1 ½ pounds of flour.

The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.

Here, [tex]1\frac{1}{2} = \frac{3}{2}[/tex] and [tex]7\frac{1}{2} = \frac{15}{2}[/tex]

The number of batches of muffins can she bake

= 15/2 ÷ 3/2

= 15/2 × 2/3

= 5 batches

Hence, the number of batches of muffins Ms. Kirkland can bake with 7 ½ pounds of flour is 5 batches.

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Answer: The answer is 5

Step-by-step explanation: I have my ways ;>

Answer:

true

Step-by-step explanation:

Answer:

True

Step-by-step explanation: