The probability of buying a movie ticket with a popcorn coupon is 0.629 and without a popcorn coupon is 0.371. If you buy 29 movie tickets, we want to know the probability that more than 16 of the tickets have popcorn coupons.

Consider tickets with popcorn coupons as successes in the binomial distribution. Do not include p= in your answer.

Based on the number of years that Peter and Blair have till retirement, the total amount that they need to save to completely fund their income shortfall is $65,545.72.

First, find the income shortfall:

= 75,017 - 54,000

= $21,017

The amount they need to save to cover this shortfall is the future value of this amount at retirement. The rate will be the real rate which is:

= 6% - 2% inflation

= 4%

The future value is:

= Income shortfall x future value interest factor, 4%, 29 years

= 21,017 x 3.1187

= $65,545.72

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Answer:

(x-5)(x+1)

Step-by-step explanation:

x² - 4x + 5 = (x-5)(x+1)

The probability that the student plays football is 20/33.

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

The more likely the event is to happen, the closer the probability value would be to 1. The less likely it is for the event not to happen, the closer the probability value would be to zero.

The probability that the student plays football = total number of students who play football / total number of students

The probability that the student plays football = 40/66 = 20/33

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Answer: [tex]4(x+2\sqrt{2})(x-2\sqrt{2})[/tex]

Step-by-step explanation:

We can first take out the common factor of 4, as both 4x² and -32 are divisible by 4.

[tex]4(x^2-8)[/tex]

From here, we can assume that x²-8 is a difference of two squares even though 8 is not a perfect square.

For review, a difference of two squares [tex]a^2-b^2[/tex] can be factored into [tex](a+b)(a-b)[/tex].

[tex]4(x^2-8)\\4(x+\sqrt{8})(x-\sqrt{8})\\4(x+2\sqrt{2})(x-2\sqrt{2})[/tex]

1. Solving linear equations:

(a) x = 4.

(b) r = 2/3.

2. Proportion:

(a) When x = 40, y = 100.

(b) When x = 40, y = 4.

3. System of equations:

(a) r = 8, s = 3.

(b) p = 5, q = -2.

4. Graphing lines:

(a) The slope of the line through (3, 4) and (-1, 3) is 1/4.

(b) The slope of the line 3x - 4y = 7 is 3/4.

(c) The slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.

5. Introductory quadratics:

(a) The solutions to the equation 4x² = 81 are -9/2, and 9/2.

(b) The solutions to the equation x² + 8x + 12 are -6, and -2.

(c) The solutions to the equation x² - 3x - 88 are -8, and 11.

6. The weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.

7. x = 3.

1. Solving linear equations:

(a) 3x - 7 = 9 - x,

or, 3x + x = 9 + 7,

or, 4x = 16,

or, x = 16/4 = 4.

Thus, x = 4.

(b) (7 - 2r)/3 = 4r,

or, 7 - 2r = 3*4r = 12r,

or, -2r - 12r = -7,

or, -14r = -7,

or, r = -7/-14 = 1/2.

Thus, r = 1/2.

2. Proportion:

(a) x and y directly proportion, means x/y = constant.

When x = 8, y = 20.

Thus, x/y = constant, or, 8/20 = constant, or, constant = 0.4.

When x = 40,

x/y = 0.4,

or, y = x/0.4 = 40/0.4 = 100.

Thus, when x = 40, y = 100.

(b) x and y indirectly proportion, means xy = constant.

When x = 8, y = 20.

Thus, xy = constant, or, 8*20 = constant, or, constant = 160.

When x = 40,

xy = 160,

or, 40y = 160,

or, y = 160/40 = 4.

Thus, when x = 40, y = 4.

3. System of equations:

(a) r - s = 5 ...(i)

3r - 5s = 9 ... (ii)

3*(i) - (ii) gives:

3r - 3s = 15

3r - 5s = 9

(-) (+)    (-)

_________

2s = 6,

or, s = 3.

Substituting in (i), we get

r - s = 5,

or, r - 3 = 5,

or, r = 8.

Thus, r = 8, s = 3.

(b) 3p + 7q = 1 ...(i).

5p = 14q + 53,

or, 5p - 14q = 53 ...(ii).

2*(i) + (ii) gives:

6p + 14q = 2

5p - 14q = 53

____________

11p = 55,

or, p = 5.

Substituting in (i), we get:

3p + 7q = 1,

or, 3*5 + 7q = 1,

or, 7q = 1 - 15 = -14,

or, q = -14/7 = -2.

Thus, p = 5, q = -2.

4. Graphing lines:

(a) Slope of the line through (3, 4) and (-1, 3) is,

m = (4 - 3)/(3 - (-1)),

or, m = 1/4.

Thus, the slope of the line through (3, 4) and (-1, 3) is 1/4.

(b) The graph given: 3x - 4y = 7.

Representing in the slope-intercept form, y = mx + b, gives:

3x - 4y = 7,

or, 4y = 3x - 7,

or, y = (3/4)x + (-7/4).

Thus, the slope of the line 3x - 4y = 7 is 3/4.

(c) Slope of the line through (5, -2) and (-1, 6) is,

m = (6 - (-2))/(-1 - 5),

or, m = 8/(-6) = -4/3.

Substituting m = -4/3 in the slope-intercept form, y = mx + b, gives:

y = (-4/3)x + b.

Substituting y = 6, and x = -1 gives:

6 = (-4/3)(-1) + b,

or, b = 6 - 4/3 = 14/3.

Thus, the slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.

5. Introductory quadratics:

(a) 4x² = 81,

or, 4x² - 81 = 0,

or (2x)² - 9² = 0,

or, (2x + 9)(2x - 9) = 0.

Either, 2x + 9 = 0 ⇒ x = -9/2,

or, 2x - 9 = 0 ⇒ x = 9/2.

Thus, the solutions to the equation 4x² = 81 are -9/2, and 9/2.

(b) x² + 8x + 12 = 0,

or, x² + 2x + 6x + 12 = 0,

or, x(x + 2) + 6(x + 2) = 0,

or, (x + 6)(x + 2) = 0.

Either, x + 6 = 0 ⇒ x = -6,

or, x + 2 = 0 ⇒ x = -2.

Thus, the solutions to the equation x² + 8x + 12 are -6, and -2.

(c) x² - 3x - 88 = 0,

or, x² - 11x + 8x - 88 = 0,

or, x(x - 11) + 8(x - 11) = 0,

or, (x + 8)(x - 11) = 0.

Either, x + 8 = 0 ⇒ x = -8,

or, x - 11 = 0 ⇒ x = 11.

Thus, the solutions to the equation x² - 3x - 88 are -8, and 11.

6. We assume the weight of one orange, one apple, and one banana to be x, y, and z units respectively.

Thus, we have:

3x + 2y + z = 15 ... (i)

5x + 7y + 2z = 44 ... (ii)

x + 3y + 5z = 26 ... (iii)

2(i) - (ii) gives:

6x + 4y + 2z = 30

5x + 7y + 2z = 44

(-)  (-)  (-)        (-)

______________

x - 3y = -14 ... (iv)

5(i) - (iii) gives:

15x + 10y + 5z = 75

x + 3y + 5z = 26

(-) (-) (-)          (-)

________________

14x + 7y = 49 ... (v)

14(iv) - (v) gives:

14x - 42y = -196

14x + 7y = 49

(-)    (-)     (-)

_____________

-49y = -245,

or, y = 5.

Substituting in (v), we get:

14x + 7y = 49,

or, 14x + 35 = 49,

or, x = 14/14 = 1.

Substituting x = 1 and y = 5 in (i), we get:

3x + 2y + z = 15,

or, 3 + 10 + z = 15,

or, z = 2.

Thus, the weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.

7. 3/(1 - (2/x)) = 3x,

or, 3/((x - 2)/x) = 3x,

or, 3x/(x - 2) = 3x,

or, 3x = 3x(x - 2),

or, 3x = 3x² - 6x,

or, 3x² - 9x = 0,

or, 3x(x - 3) = 0.

Either, 3x = 0 ⇒ x = 0,

or, x - 3 = 0 ⇒ x = 3.

Since we had a term 2/x, x cannot be 0.

Thus, x = 3.

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Answer:

TS=QV

Step-by-step explanation:

To prove SAS congruence, you need to prove that two lines and the angle between them are all respectively equal.

In the diagrams we already have RS=WV and ∠RST=∠WVQ, so it follows that we need to prove that TS=QV.

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

For triangles to be congruent by SAS congruency criteria, one angle and the two adjacent sides to the angle should be equal to corresponding angle and it's adjacent aides of another triangle.

And we have already been given the angle and one of the side. so the third side that need to be equal in both side to pe proved congruent are :

[tex] \qquad \large \sf {Conclusion} : [/tex]

Answer:

there are 8 million different phone numbers whose phone numbers whose digits aren't 0 or 1

The equation that represents the distance, in kilometers, Seb travels (d) based on how many hours he bikes (t) at this rate is d = 20t

The given parameter is

constant rate of 20 kilometers per hour.

This means that

Rate = 20km per hour

When Seb bikes for t hours, the distance is

d = rate * t

This gives

d = 20t

Hence, the equation that represents the distance, in kilometers, Seb travels (d) based on how many hours he bikes (t) at this rate is d = 20t

This means that

t = 5

So, we have:

d = 20 * 5

Evaluate

d = 100

Hence, Seb will travel 100 kilometers if he bikes at this rate for 5 hours

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Answer:

1. d=20t

2. 50

Using the normal distribution, there is a 0.1894 = 18.94% probability that the sample average will exceed $75.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The parameters for this problem are given as follows:

[tex]\mu = 70, \sigma = 40, n = 50, s = \frac{40}{\sqrt{50}} = 5.66[/tex]

The probability that the sample average will exceed $75 is one subtracted by the p-value of Z when X = 75, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

Z = (75 - 70)/5.66

Z = 0.88

Z = 0.88 has a p-value of 0.8106.

1 - 0.8106 = 0.1894.

0.1894 = 18.94% probability that the sample average will exceed $75.

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Answer:

Step-by-step explanation:

I think you are trying to use synthetic division for

[tex]x^{4} -2x^{3} -2x^{2} +3x-5[/tex] divided by x-3

First  

x -3 = 0 , make it equal to 0 and add 3 to both sides

x = 3

Then we write all the coefficients and continue with

a pattern of multiply by 3 and add to the next coefficient.

See attachment.

a) The preparation of a traditional income statement for Carbon Company is as follows:

Carbon Company

Sales revenue                         $600,000

Costs:

Materials               $ 101,300

Energy                       21,590

Setups                      24,360

Purchasing              26,200

Customer service   20,200

Long-term labor     29,400

Administrative        30,700     $253,750

Operating Income                  $346,250

b) The preparation of an activity-based income statement for Carbon Company is as follows:

Carbon Company

Sales revenue                                                                     $600,000

                   Resources supplied    Unused Resources   Used Resources

Materials               $ 101,300                     $ 2,300              $99,000

Energy                       21,590                         2,210                  19,380

Setups                      24,360                                0                 24,360

Purchasing              26,200                         7,800                  18,400

Customer service   20,200                         9,700                 10,500

Long-term labor     29,400                         4,650                 24,750

Administrative        30,700                         5,950                 24,750        

Total                   $253,750                      $32,610              $221,140

Operating Income                                                            $378,860

                          Unused Resources Capacity     Cost Driver Volume

Resources used

Materials                        $ 2,300                                 9,000 pounds

Energy                               2,210                        380 machine-hours

Setups                                     0                                         87 setups

Purchasing                      7,800                         80 purchase orders

Customer service           9,700                                         70 returns

Long-term labor              4,650                              330 labor-hours

Administrative                5,950                              450 labor-hours

Materials $ 101,300

Energy 21,590

Setups 24,360

Purchasing 26,200

Customer service 20,200

Long-term labor 29,400

Administrative 30,700

Sales revenue  $600,000

                   Resources supplied       Unused Resources   Used Resources

Materials               $ 101,300                         $ 2,300              $99,000

Energy                       21,590                             2,210                  19,380

Setups                      24,360                                    0                 24,360

Purchasing              26,200                             7,800                  18,400

Customer service   20,200                             9,700                 10,500

Long-term labor     29,400                             4,650                 24,750

Administrative        30,700                             5,950                 24,750        

Total                   $253,750                          $32,610              $221,140

Thus, the traditional income statement for Carbon Company shows an operating income of $346,250, while the activity-based income statement shows an operating income of $378,860.

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Answer: Choice 3

Step-by-step explanation:

Oof, this is a tough one. But anyways

Angle 19 + Angle 20 = 180

So Angle 20 = 180 - Angle 19

As l and m are parallel, Angle 20 = Angle 6 from the z rule

So Angle 6 = 180 - Angle 19

Substitute the information given at the start into the equation

2x - 4 = 180 - (5x - 8)

180 - 5x + 8 = 2x - 4

7x = 192

x = 27.42

From what we have found earlier, Angle 20 = Angle 6 so

Angle 20 = 2x - 4 = 2 * 27.42 - 4 = 50.85 rounding it to 50.9

So the answer is 50.9 or Choice 3

A. The pair of linear equations are: x + y = 55 and y = x + 25

B. Number of time she spend running everyday is: 15 minutes.

C. It is not possible.

A system of equations consist of linear equations that have unknown variables whose values make the linear equations in the system true.

Part A: To write a pair of linear equations, do the following,

Let x represent the time Jackie runs

Let y represent the time Jackie dances

We are given that, Jackie runs and dances for a total of 55 minutes every day, therefore, the first linear equation would be:

x + y = 55 --> equation 1

We are also given that Jackie dances for 25 minutes longer than she runs, therefore, the second linear equation would be:

y = x + 25 --> equation 2

Therefore, the two pair of linear equations that shows the relationship between the number of minutes Jackie runs and dances everyday are:

x + y = 55

y = x + 25

Part B: Substitute y = (x + 25) into equation 1

x + y = 55 --> equation 1

x + x + 25 = 55

2x + 25 = 55

2x = 55 - 25

2x = 30

x = 15

Therefore, the number of time she spend running everyday is: 15 minutes.

Part C:

If she spent 45 minutes dancing (y = 45), and x = 15, substitute the values into x + y = 55 to find out if it is possible.

15 + 45 = 55

60 = 55 (not true)

Therefore, it is not possible.

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Using the relation between velocity, distance and time, it is found that the rate of the current is of 3.33 mph.

Velocity is distance divided by time, hence:

v = d/t

A small motorboat travels 12mph in still water. With the current, upstream, 46 miles are traveled in t hours, hence:

12 + r = 46/t

r = 46/t - 12

Downstream, the time is of t + 2 hours, hence:

12 - r = 46/(t + 2)

r = 12 - 46/(t + 2)

Hence, equaling the values for r:

46/t - 12 = 12 - 46/(t + 2)

46/t + 46/(t + 2) = 24

[tex]\frac{46t + 92 + 46t}{t(t + 2)} = 24[/tex]

92t + 92 = 24t² + 48t

24t² - 44t - 92 = 0

Using a quadratic equation calculator, the solution is t = 3. Hence the rate is found as follows:

r = 46/t - 12 = 46/3 - 12 = 3.33 mph.

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Answer:

750 - y(y-5) = 0

Step-by-step explanation:

Area is Length times width.  In a rectangle we have two lengths and 2 widths.

We can use y to represent the length.  The width would be y-5.

Area = LW

750 = y(y-5) or another way to write this is

750 - Y(y-5) = 0

The dosage needed for this in an  order for morphine sulfate 1/4 gr IM stat is 1.5 ml.

What is an equation?

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depend on any other variable for its value while a dependent variable is a variable that depends on other variable.

One grain =  60mg, hence:

1/4 gr = 1/4 gr * 60 mg per gr = 15 mg

Hence, for 10 mg per ml:

Dosage = 15 mg /  10 mg per ml = 1.5 ml

The dosage needed for this in an  order for morphine sulfate 1/4 gr IM stat is 1.5 ml.

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[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

According to given figure,

Angle 11, Angle 20 and Angle 16 forms angles of a triangle,

so, by angle sum property of a triangle :

[tex]\qquad❖ \: \sf \: \angle11 + \angle20 + \angle16 = 180 \degree[/tex]

[tex]\qquad❖ \: \sf \: \angle11 +66 + 43 = 180 \degree[/tex]

[tex]\qquad❖ \: \sf \: \angle11 +109 = 180 \degree[/tex]

[tex]\qquad❖ \: \sf \: \angle11 = 180 \degree - 109[/tex]

[tex]\qquad❖ \: \sf \: \angle11 =71 \degree [/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

Answer:

d = [tex]\sqrt{65}[/tex]

Step-by-step explanation:

The distance between two points is found by

d = [tex]\sqrt{( x2-x1)^2 + ( y2-y1)^2}[/tex] where (x1,y1) and (x2,y2) are the two points

d = [tex]\sqrt{(-2 - -1)^2+(-6 - -14)^2 }[/tex]

d =[tex]\sqrt{( -2+1)^2 + (-6+14)^2}\\[/tex]

d = [tex]\sqrt{(-1)^2 +( 8)^2 }[/tex]

d = [tex]\sqrt{1+64}[/tex]

d = [tex]\sqrt{65}[/tex]

Taking squares on both sides leads to

[tex]\sqrt{-x + 6} = x - 6[/tex]

[tex]\left(\sqrt{-x + 6}\right)^2 = (x - 6)^2[/tex]

[tex]-x + 6 = x^2 - 12x + 36[/tex]

[tex]x^2 - 11x + 30 = 0[/tex]

[tex](x - 5) (x - 6) = 0[/tex]

Solving for [tex]x[/tex], we get

[tex]x - 5 = 0 \text{ or } x - 6 = 0[/tex]

or

[tex]x = 5 \text{ or } x = 6[/tex].

Evaluating both sides of the starting equation at these solutions, we have

[tex]\sqrt{-6 + 6} = 6 - 6 \implies 0 = 0[/tex]

which is true, so [tex]\boxed{x=6}[/tex] is a valid solution. However,

[tex]\sqrt{-5 + 6} = 5 - 6 \implies \sqrt1 = -1 \implies 1 = -1[/tex]

which is not true, so [tex]\boxed{x=5}[/tex] is an extraneous solution.

The fraction and decimal forms which match the long division problem is 4/9 and 0.4 respectively. option D

9√4.000

= 4.000/9

= 0.444

Approximately,

= 0.4

Fraction

This is a ratio of two numbers, the numerator and the denominator, usually written one above the other and separated by a horizontal bar.

Decimal

This is a number expressed in the decimal system (base 10).

Therefore, the fraction and decimal forms which match the long division problem is 4/9 and 0.4 respectively.

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Answer:

Step-by-step explanation:

D because the quadratic formula is basically [tex]\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]±

so if we place the equation D into the formula

a = 1

b = +3

c = +24

we can clearly say that D eq is correctly used and can be.

Answer:

See below

Step-by-step explanation:

[tex]A = \{2, 4, 6\}[/tex]

[tex]A^c =\{1, 3, 5\}\\B = \{4, 5, 6\}\\B^c = \{ 1,2, 3\}[/tex]

To prove 1 we have

[tex](A \bigcup B) = \{2, 4, 6\} \bigcup \{4, 5, 6\} = \{2, 4, 5, 6\}\\(A \bigcup B) ^c =\bold{\{1,3\}}\\A^c \bigcap B^c = {\{1, 3, 5\} \bigcap \{1, 2, 3\} =\bold{\{1,3\}}\\[/tex]

Hence proved

To prove 2

[tex](A \bigcap B) = \{2, 4, 6\} \bigcap \{4, 5, 6\} = \{4, 6\}\\\\(A \bigcap B)^c = \bold{\{1,2, 3, 5\}}\\A^c \bigcup B^c = \{1, 3, 5\} \bigcup \{1, 2, 3\} = \bold{\{1,2,3,5\}}\[/tex]

Hence proved

Notes

The complement of a set is the set of all elements not in the set

Here the set of all outcomes is {1, 2, 3,4,5, 6}

So if A = {2, 4, 6} then the complement of A s=is the set of outcomes not in A ie the set {1, 3, 5} which is the set of odd numbers on a die throw

The union of two sets is the set of all elements in both sets without duplication

The intersection of two sets is the set of all elements common to both sets

The number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

Note that the formula for combination is given as;

Combination = [tex]\frac{n!}{r!(n-r)!}[/tex]

From the information we have that;

There are 28 questions in the pool

The test should have a total of 10 questions;

6 difficult , 10 average and 12 easy questions

We are asked to determine the combination of;

3 difficult questions

4 average questions

3 easy questions

6C3 = [tex]\frac{6!}{3!(6-3)!}[/tex]

6C3 = [tex]\frac{720}{36}[/tex]

6C3 = 20

10C4 = [tex]\frac{10!}{4!(10-4)!}[/tex]

10C4 = [tex]\frac{3628800}{17280}[/tex]

10C4 = 210

12C4 = [tex]\frac{12!}{4!(12-4)!}[/tex]

12C4 = 220

The number of ways of constructing the questions is

= 20 × 210 × 220

= 924, 000 ways

Thus, the number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

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Answer:

1.

Step-by-step explanation:

10C0 = 1.

The simplified forms of the given expressions are

a. 32x

b. 0

c. -26t + 14x

d. 3t

From the question, we are to simplify the given expressions by adding or subtracting

a. (10x) + (22x)

= 32x

b. (-6x) - (-6x)

= -6x + 6x

= 0

c. (-13t) + (14x) - (+13t)

= -13t + 14x - 13t

Collect like terms

= -13t -13t + 14x

= -26t + 14x

d. (-4t) + (+4t) - (-3t)

= -4t + 4t + 3t

= 3t

Hence, the simplified forms of the given expressions are

a. 32x

b. 0

c. -26t + 14x

d. 3t

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According to the adjustable-rate mortgage The new monthly payment in year 8 of the loan is $1292.

According o the statement

we have given that the

Remy obtains a 30-year mortgage in the amount of $625,000 And

She secures a 7/1 ARM at an initial interest rate of 3%. Her initial monthly payment is $2,635.03.

And we have to find the new monthly payment when its interest rate increased to the 4.925%.

So, For this purpose we know that the

An adjustable-rate mortgage (ARM) is a loan with an interest rate that changes.

The amount in which she bought mortgage = 625,000

Interest rate = 3%

Initial amount = 2635.03

From this it clear that the

The amount she pays in 7 years = 2635.03*7

The amount she pays in 7 years = 18445.21

The left amount = 625,000 - 18445.21

The left amount = 606554

It means the rate of interest increased on the amount 606554$

So, According to the 4.925% interest the monthly payment will become

amount on interest rate 4.925% = 4.925% *606554/23 years

amount on interest rate 4.925% = $1292.

So, According to the adjustable-rate mortgage The new monthly payment in year 8 of the loan is $1292.

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The Area of the shaded region of the circle = 60.75π m²

The Length of arc ADB = ∅/360 × 2πr = 13.5π m.

The area of the shaded region in the circle = area of the sector of a circle = ∅/360 × πr², where r is the radius of the circle and the central angle is ∅.

The length of the arc on a circle = ∅/360 × 2πr, where r is the radius of the circle and the central angle is ∅.

Given the following:

Central angle (∅) = 360 - 90 = 270°

Radius (r) = 9 m.

Area of the shaded region of the circle = ∅/360 × πr² = 270/360 × π(9²)

Area of the shaded region of the circle = 270/360 × π81

Area of the shaded region of the circle = 60.75π m²

Length of arc ADB = ∅/360 × 2πr = 270/360 × 2π(9)

Length of arc ADB = ∅/360 × 2πr = 270/360 × 18π

Length of arc ADB = ∅/360 × 2πr = 13.5π m.

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The y- coordinate that divides the directed line segment from J to K into a ratio of 2:3 is 5. Option C

Let's the point that divides the line segment as point S.

We have that,

Point S divides the line segment into ratio 2:3

The ratio 2:3 means that we are to divide the line segment into;

= 2+3

= 5 equal parts.

We then have that the horizontal distance between the two coordinates is 5

The vertical distance between the two coordinates is 10

Now, let's divide both vertical and horizontal distance into five equal parts,

Horizontal distance = 5/5 = 1

Vertical distance = 10/ 5 = 2

The horizontal distance is 1

The vertical distance is 2

For every one unit move to the left from point J and two units up, we are dividing the line segment into five equal parts as shown in the picture.

The coordinate of point S that divides the line segment into 2 parts and 3 parts is (-5,5)

Thus, the y- coordinate that divides the directed line segment from J to K into a ratio of 2:3 is 5. Option C

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