Yogurtland charges $2.40 for 8 ounces of frozen yogurt. At this price, how frozen yogurt can you buy with a $15 gift card?

Answer:

A)  [tex]15x^2+22x+8[/tex]

B)  Degree 2, Quartic Expression

C)  The dimensions of the rectangle are polynomials. When multiplied together, the area of the rectangle is also a polynomial.

Step-by-step explanation:

A)  The formula for area of a rectangle is Area = L * W. The length and width are represented by (3x+2) and (5x+4). So we can say that

[tex]Area = (5x+4)(3x+2)[/tex]

Use FOIL to multiply the the polynomials. First, Outside, Inside, Last

[tex]Area = (5x)(3x) + (5x)(2) + (3x)(4) + (2)(4)\\Area = 15x^2 + 10x + 12x + 8\\Area = 15x^2 + 22x + 8[/tex]

The expression that represents the area of the rectangle is

[tex]15x^2 + 22x + 8[/tex].

B)  The degree of the expression is 2, because two is the highest power of x. The classification of the expression is quadratic because the graph of the expression is a parabola.

Degrees vs. Classification

Degree 0:  Zero Polynomial or Constant

Degree 1:  Linear (line)

Degree 2:  Quadratic (parabola)

Degree 3: Cubic

Degree 4: Quartic

...

C)  Closure property for polynomials applies to addition, subtraction, and multiplication. It means that the result of multiplying two polynomials will also be a polynomial. Part A demonstrates polynomial closure under multiplication because the dimensions of the rectangle are polynomials and so is the area.

Answer:

Step-by-step explanation:

[tex]D=\frac{\theta}{360} \times2\times\pi \times r[/tex]

[tex]D=\frac{60}{360} \times2\times\pi \times 42[/tex]

    [tex]=\frac{1}{6} \times84\times\pi[/tex]

    [tex]=12\pi cm^2[/tex]

Alexander would have to pay $15 to buy 2 pounds of walnuts with the coupon. The expression is cost = ($8/pound) x pp - $1.

The functional connection between cost and output is referred to as the cost function. It examines the cost behaviour at various output levels under the assumption of constant technology. An essential factor in determining how well a machine learning model performs for a certain dataset is the cost function. It determines and expresses as a single real number the difference between the projected value and expected value.

Given that, the price per pound of walnuts is $8.

2 pounds x $8/pound = $16

Alexander would get $1 off the final amount.

Thus,

$16 - $1 = $15

So Alexander would have to pay $15 to buy 2 pounds of walnuts with the coupon.

The expression for the cost can be written as:

cost = ($8/pound) x pp - $1

Hence, Alexander would have to pay $15 to buy 2 pounds of walnuts with the coupon. The expression is cost = ($8/pound) x pp - $1.

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value of variable x is 65 degree.

A straight line is a geometric object that extends infinitely in both directions and has a constant slope or gradient. It is the shortest distance between two points, and it can be described by an equation in the form of y = mx + c, where m is the slope or gradient of the line and c is the y-intercept.

The term "supplementary angles" refers to two angles whose sum is equal to 180 degrees. In other words, if angle A and angle B are supplementary angles, then:

∠A +∠B = 180 degrees

To find value of x

A straight line's angle total is 180°.

2x+50°=180°

2x=130°

x=65

Hence, value of variable x is 65 degree.

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Answer:

Step-by-step explanation:

You ask yourself, "600 is 30% of how many students?". In equation form this looks like:

600 = .30x

Divide both sides by .30 and you'll get that the number of students is 2000

Answer:

Let x be the number of hopts and y be the number of totts.

The first statement relates to the total amount of money expected to be made:

3x + 7y ≥ 84

This inequality states that the total revenue from selling hopts and totts should be at least $84.

The second statement relates to the total number of units expected to be sold:

x + y ≤ 24

This inequality states that the total number of hopts and totts sold should be at most 24 units. Therefore, the system of inequalities in standard form is:

3x + 7y ≥ 84

x + y ≤ 24

where x ≥ 0 and y ≥ 0 (since we cannot sell a negative number of hopts or totts).

Answer: Let "x" be the number of hopts that Farmer Naxvip plans to sell, and let "y" be the number of totts that she plans to sell. Then we can write the following system of inequalities to model the relationships between the amounts of hopts and totts:

3x + 7y >= 84

x + y <= 24

The first inequality represents the condition that Farmer Naxvip expects to make at least $84, while the second inequality represents the condition that she expects to sell at most 24 units.

Brainliest Appreciated! (:

Answer:

Step-by-step explanation:

45342

a) The probability of that happening is P(W1 = 1, W2 = 2) = P(W1 = 2, W2 = 1) = p1p2.

b) P(W1 = 1)P(W2 > 1) + P(W1 = 1)P(W2 > 2) + P(W1 = 2)P(W2 > 3) + ...= p1(1 - p2) + p1(1 - p2)p2 + p1p2(1 - p2) + ...= p1(1 - p2)(1 +p2 + p22 + ...) = p1(1 - p2)/(1 - p2)2= p1/(1 - p2)2

c) The probability that their minimum value is k is P(min(W1, W2) = k) = P(W1 = k, W2 > k) + P(W1 > k, W2 = k) = p1(1 - p2)k-1p2 + p2(1 - p1)k-1p1

d) The probability that their maximum value is k is P(max(W1, W2) = k) = P(W1 = k, W2 = k) + P(W1 = k, W2 > k) + P(W1 > k, W2 = k) = p1p2k-1 + p1(1 - p2)k-1 + p2(1 - p1)k-1

a) P(W1 = W2)The probability that W1 = W2 is 0. If W1 and W2 have different values, then W1 is equal to either 1 or 2. The probability of that happening is P(W1 = 1, W2 = 2) = P(W1 = 2, W2 = 1) = p1p2. The probability of that happening is P(W1 = 1, W2 = 2) = P(W1 = 2, W2 = 1) = p1p2.b) P(W1 < W2)The probability of W1 being less than W2 is P(W1 = 1, W2 = 2) + P(W1 = 1, W2 = 3) + P(W1 = 2, W2 = 3) + ... This may be written as P(W1 = 1)P(W2 > 1) + P(W1 = 1)P(W2 > 2) + P(W1 = 2)P(W2 > 3) +...= p1(1 - p2) + p1(1 - p2)p2 + p1p2(1 - p2) + ...= p1(1 - p2)(1 + p2 + p22 + ...) = p1(1 - p2)/(1 - p2)2= p1/(1 - p2)2

d) Distribution of min(W1, W2)If W1 and W2 are independent geometric random variables with parameters p1 and p2, respectively, then the probability that their minimum value is k is P(min(W1, W2) = k) = P(W1 = k, W2 > k) + P(W1 > k, W2 = k) = p1(1 - p2)k-1p2 + p2(1 - p1)k-1p1e) Distribution of max(W1, W2)If W1 and W2 are independent geometric random variables with parameters p1 and p2, respectively, then the probability that their maximum value is k is P(max(W1, W2) = k) = P(W1 = k, W2 = k) + P(W1 = k, W2 > k) + P(W1 > k, W2 = k) = p1p2k-1 + p1(1 - p2)k-1 + p2(1 - p1)k-1

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Answer:

The remainder when 5x3 + 2x2 - 7 is divided by x + 9 is -692.

Explanation:

We can use long division to divide 5x^3 + 2x^2 - 7 by x + 9:

-5x^2 + 43x - 385

x + 9 | 5x^3 + 2x^2 + 0x - 7

5x^3 + 45x^2

--------------

-43x^2 + 0x

-43x^2 - 387x

--------------

387x - 7

Therefore, the remainder when 5x^3 + 2x^2 - 7 is divided by x + 9 is 387x - 7.

Hope this helps, sorry if this is wrong! :]

Therefore , the solution of the given problem of equation comes out to be  y = 0.5x²- 3

The foundation of a model of linear regression is the equation y=mx+b. The inclination is B, and the y-intercept is m. Although it is true that y and y are separate parts, the the above sentence is frequently referred to as a "mathematics problem with two variables". Y=mx+b is the formula for a singular system of equations, where m denotes the gradient and b the y-intercept. The equation is Y=mx+b, where m denotes slopes and b denotes the y-intercept.

Here,

=> x  y

  -4  5

   -2  -1

   0  -3

    2  -1

To find : the equation which passes through the point

=> y = 0.5x²- 3

We can see,

=> y = 0.5(-4)² - 3

=> y = 8 -3

=> y =5

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Answer:

Step-by-step explanation:

You want the coordinates of triangle ABC after dilation by a factor of 1/2.

When the center of dilation is the origin, each of the coordinates is multiplied by the dilation factor:

  (x, y) ⇒ (x/2, y/2)

  A(-4, 0) ⇒ A'(-2, 0)

  B(0, 2) ⇒ B'(0, 1)

  C(2, -2) ⇒ C'(1, -1)

90% people entering the emergency department is within the interval of [0.1559, 0.2933].

The confidence interval of the percent of patients entering the emergency department who are admitted to the hospital is [0.1763, 0.3137]. It is not reasonable for the hospital to assume that 25% of people who enter the emergency department are admitted to the hospital. Here's why.How to calculate a 90% two-sided confidence interval for p, the percent of people entering the emergency department who are admitted to the hospital:$$CI_p =\bigg(\hat{p}-Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\bigg)$$ where $\hat{p} = \frac{x}{n}$, $\alpha = 0.10$, $Z_{\alpha/2} = 1.645$ (for a 90% confidence interval), and $n = 187$. The margin of error is given by $$ME = Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$Plugging in the values, we get $$\hat{p} = \frac{42}{187} = 0.2246$$$$ME = 1.645 \cdot \sqrt{\frac{0.2246\cdot 0.7754}{187}} \approx 0.0687$$Therefore, the confidence interval for $p$ is $$CI_p = (0.2246-0.0687, 0.2246+0.0687) = (0.1559, 0.2933)$$The 90% two-sided confidence interval for the percent of people entering the emergency department who are admitted to the hospital is [0.1559, 0.2933].Since the interval doesn't include 0.25, the hospital should not use the assumption that 25% of people who enter the emergency department are admitted to the hospital. This is because the interval does not overlap with the value of 0.25. As a result, we are 90% confident that the true proportion of people who are admitted to the hospital after entering the emergency department is within the interval of [0.1559, 0.2933].

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a)  approx 0.562$$

When playing backgammon, the doubling cube is an important feature that allows players to increase the stakes of the game. Imagine playing with a "tripling cube" instead, with faces of 3, 9, 27, 81, 243, and 729. Using the analysis given for the doubling cube, we can compute the probability of winning above which a triple should be accepted.To determine the probability of winning above which a triple should be accepted, we need to use the formula derived from the analysis of the doubling cube:$$P_w = \frac{q^2}{1-2q^2}$$where Pw is the probability of winning, and q is the probability of losing. We can substitute 1-q for p, the probability of winning, to get:$$P_w = \frac{(1-p)^2}{1-2(1-p)^2}$$Now, we need to modify this formula to account for the tripling cube. If we triple the current stakes, then we have effectively tripled the value of the doubling cube. In other words, if the current stakes are 1, then the value of the tripling cube is 3. If the current stakes are 2, then the value of the tripling cube is 9. More generally, if the current stakes are n, then the value of the tripling cube is 3^n. Using this information, we can modify the formula as follows:$$P_w = \frac{q^{3^n}}{1-2q^{3^n}}$$. This is the formula we need to use to compute the probability of winning above which a triple should be accepted. We can solve for q using the quadratic formula:$$q = \frac{1\pm\sqrt{1-4(1-2P_w)(-P_w^{3^n})}}{2}$$The value of q we want is the smaller one, because we want to compute the probability of losing. Once we have q, we can substitute it into the formula for Pw to get the probability of winning above which a triple should be accepted. Example Suppose we have a backgammon game with a current stake of 4, and we are considering accepting a triple from our opponent. Using the formula above, we can compute the probability of winning above which a triple should be accepted as follows:$$n = \log_3{3^2} = 2$$$$P_w = \frac{q^{3^2}}{1-2q^{3^2}}$$$$q = \frac{1-\sqrt{1-4(1-2P_w)(-P_w^{3^2})}}{2}$$$$q = \frac{1-\sqrt{1-8P_w^9}}{2}$$$$q = \frac{1-\sqrt{1-8(0.5)^9}}{2}$$$$q \approx 0.515$$$$P_w = \frac{q^{3^2}}{1-2q^{3^2}}$$$$P_w = \frac{(0.515)^9}{1-2(0.515)^9}$$$$P_w \approx 0.562$$. Therefore, if we have a probability of winning greater than 0.562, then we should accept the triple.

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If the volume of a right cone is 3,525, then the height of the cone is approximately 5 units.

We can use the formula for the volume of a cone to solve for the height of the cone:

V = (1/3)πr²2h

where V is the volume of the cone, r is the radius (half of the diameter), and h is the height of the cone.

Given that the diameter of the cone is 30 units, the radius is 15 units. Substituting this and the volume V = 3525, we get:

3525 = (1/3)π(15)²2h

3525 = 225πh/3

h = 3(3525)/(225π)

h ≈ 5

Therefore, the height of the cone is approximately 5 units.

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Answer:

34.28 meters

Step-by-step explanation:

We can use trigonometry to solve this problem. Let's call the horizontal distance between the boy and the pole "d". Then we can draw a right triangle with the boy's height (5m) as one leg, the distance "d" as the other leg, and the hypotenuse being the distance from the boy's eyes to the top of the pole (which we don't know yet).

We can use the angle of elevation to find the length of this hypotenuse. The angle of elevation is the angle between the horizontal and the line of sight from the boy's eyes to the top of the pole. Since the boy is looking up at the bird, this angle is also the same as the angle between the hypotenuse and the vertical (i.e. the angle at the top of the triangle). So we have:

tan(30°) = opposite/adjacent

where "opposite" is the height of the pole (20m) and "adjacent" is the hypotenuse. Solving for "adjacent", we get:

adjacent = opposite/tan(30°) = 20/tan(30°)

We can simplify tan(30°) to 1/√3, so:

adjacent = 20/(1/√3) = 20√3

Now we can use the Pythagorean theorem to find the horizontal distance "d":

d^2 + 5^2 = (20√3)^2

Simplifying and solving for "d", we get:

d = √[(20√3)^2 - 5^2] = √(1200 - 25) = √1175

So the horizontal distance between the boy and the pole is approximately 34.28 meters (rounded to two decimal places

Part A:

The equation Y = 10X + 5 is in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. Therefore, in this equation, the slope is 10 and the y-intercept is 5.

The slope represents the rate of change in the monthly rate of the gym membership. For every one unit increase in the number of months, the monthly rate will increase by $10. The y-intercept represents the initial cost of joining the gym, which is $5.

Part B:

To find out how much money Justin saves the first month by joining the gym at the discounted price, we need to calculate the difference between the regular price and the discounted price for the first month.

The regular price for the first month can be found by plugging in X = 1 into the equation Y = 15X + 20, which gives Y = 35.

The discounted price for the first month can be found by plugging in X = 1 into the equation Y = 10X + 5, which gives Y = 15.

Therefore, Justin saves $20 (35 - 15) the first month by joining the gym at the discounted price rather than the regular price.

Part C:

The system of equations is:

Y = 10X + 5 (discounted price)

Y = 15X + 20 (regular price)

The solution to the system of equations is (-3, -25), which means that if X = -3, then Y = -25 is a solution to both equations. However, this solution is not possible in this situation because X represents the number of months, which cannot be negative. Therefore, the point (-3, -25) is not a valid solution.

3.31 is an estimate for the radius of the circle.

The circle fοrm is a clοsed twο-dimensiοnal shape because every pοint in the plane that makes up a circle is evenly separated frοm the "centre" οf the fοrm.

Each line tracing the circle cοntributes tο the fοrmatiοn οf the line οf reflectiοn symmetry. In additiοn, it rοtates arοund the center in a symmetrical manner frοm every perspective.

he radius of the circle, we need to use the formula for the area of a sector, which is:

Area of sector = (θ/360) x π x r

the radius of the circle "x" for now, and set up an equation using the given information

51.3 = (120/360) x π x

Simplifying this equation, we get:

= (51.3 x 360)/(120 x π)

x² ≈ 10.95

x ≈ √10.95

x ≈ 3.31

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The simplified difference quotient for the function [tex]\(f(x)=6\) is \(\frac{x+h-1}{h/6}\).[/tex]

The difference quotient for the function \(f(x)=6\) is given by the formula \( \[tex]frac{f(x+h)-f(x)}{h}\)[/tex]. To simplify this difference quotient, substitute the given value of \(f(x)\) into the equation:

[tex]\( \frac{f(x+h)-f(x)}{h} = \frac{f(x+h)-6}{h} \).[/tex]  Next, substitute the value of \(f(x)\) into the numerator:

[tex]\frac{6(x+h-1)}{h} \\[/tex]  .

Finally, divide both sides of the equation by 6 to simplify the equation:

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

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Robert is currently 10 years old.Given that five years ago Sharon was twice as old as Robert was then, we can set up the following equation: y + 5 = 2(y)

We are given the information that Sharon is five years older than Robert, and five years ago Sharon was twice as old as Robert was then. This means we can create a system of equations to solve for Robert's age.

Let x = Robert's current age

Let y = Robert's age five years ago

Given that Sharon is five years older than Robert, we can set up the following equation:

x + 5 = Sharon's current age

Given that five years ago Sharon was twice as old as Robert was then, we can set up the following equation:

y + 5 = 2(y)

Solving the first equation for x, we get x = Sharon's current age - 5. Substituting this into the second equation, we get:

y + 5 = 2(Sharon's current age - 5)

Solving this equation for y, we get y = (Sharon's current age - 5)/2.

Since Sharon is five years older than Robert, Sharon's current age is x + 5. Substituting this into our equation for y, we get:

y = (x + 5 - 5)/2

Simplifying this equation, we get y = x/2. This means that Robert's age five years ago was half of his current age.

Since we know that Robert is currently 10 years old, Robert's age five years ago was 5. Therefore, Robert is currently 10 years old.

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Answer:

d. 27/4, or 6.75

Step-by-step explanation:

[tex]2( \frac{1}{8} + 4) - ( \frac{3}{16} \times 8)[/tex]

[tex]2( \frac{33}{8} ) - \frac{3}{2} [/tex]

[tex] \frac{33}{4} - \frac{3}{2} = \frac{33}{4} - \frac{6}{4} = \frac{27}{4} = 6.75[/tex]

Answer:

[tex]14in^2[/tex]

Step-by-step explanation:

Volume of a rectangluar prism = Length x Width x Height

Volume, Length, and Width is given now we can solve for Height, y

[tex]V = l * w * h[/tex]

[tex]3in^3 = 1in * 1in* h[/tex]

[tex]h = 3in[/tex]

Surface area of a rectangular prisim is A = 2(wl +hl +hw)

A = [tex]2(1in * 1in + 3in * 1in + 3in * 1in)\\2(1 in^2+ 3 in^2+ 3in^2)\\2(7in^2)\\14in^2[/tex]

Hope this helps!

Brainliest is much appreciated!

Answer:

Tip is $34

Round up bill amount $170

Step-by-step explanation:

20/ 100 x 170 = $34

Tip is $34

BILL plus tip is $204

Answer:

To evaluate this expression, we need to follow the order of operations, which is commonly remembered using the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction):

1. Start with any operations inside parentheses. There are no parentheses in this expression.

2. Evaluate any exponents. There are no exponents in this expression.

3. Evaluate multiplication and division, from left to right. In this expression, we have:

7 + (35/3) + 5 - (7 + (35/3) / (3 - 5))

= 7 + (35/3) + 5 - (7 + (11.67) / (-2))

= 7 + (35/3) + 5 - (7 - 5.835)

= 7 + (35/3) + 5 + 5.835 - 7

= 15.835 + (35/3) - 7

4. Finally, evaluate addition and subtraction, from left to right:

15.835 + (35/3) - 7

= 15.835 + 11.67

= 27.505

Therefore, the value of the expression is 27.505.

Step-by-step explanation:

The pass rate in the afternoon subgroup is greater than 1, we cannot draw a reliable conclusion without further investigation or clarification of the data.

To compare the probability that a student will pass the challenge in the morning with the probability that a student will pass the challenge in the afternoon, we need to calculate the proportion of students who passed in each subgroup.

The proportion of students who passed the challenge in the morning subgroup is:

Pass rate in the morning subgroup = Number of students who passed in the morning subgroup / Total number of students in the morning subgroup

Pass rate in the morning subgroup = 40 / 60

Pass rate in the morning subgroup = 0.67

The proportion of students who passed the challenge in the afternoon subgroup is:

Pass rate in the afternoon subgroup = Number of students who passed in the afternoon subgroup / Total number of students in the afternoon subgroup

Pass rate in the afternoon subgroup = 50 / 40

Pass rate in the afternoon subgroup = 1.25

that the pass rate in the afternoon subgroup is greater than 1, which is not possible as probabilities must be between 0 and 1. This implies that there might be a data error.

Assuming the data is correct, we can conclude that the probability of passing the challenge in the afternoon subgroup is higher than the probability of passing the challenge in the morning subgroup. However, since the pass rate in the afternoon subgroup is greater than 1, we cannot draw a reliable conclusion without further investigation or clarification of the data.

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(a) The solution of the equation is √6 + 2.

(b) a = 10 and b = 2, and the solution of the equation is √10 + 7.

((a) To solve this equation, we need to isolate the term with the variable x on one side and move all other terms to the other side.

Let's start by simplifying each term using the fact that;

√24 = √(4 × 6) = 2√6,

√96 = √(16 × 6) = 4√6,

√108 = √(36 × 3) = 6√3, and

√12 = √(4 × 3) = 2√3.

Then, we have:

x√24 + √96 = √108 + x√12

2x√6 + 4√6 = 6√3 + 2x√3

2(√6 + 2√3) = (x√3 + 2x√6)

2√6 + 4√3 = x(√3 + 2√6)

Now, we can equate the coefficients of √6 and √3 on both sides to get a system of equations:

2 = x

4 = 2x

Solving this system, we find that x = 2 and therefore a = 6 and b₁ = 2.

So, the solution of the equation is √6 + 2.

(b) To solve this equation, we also need to isolate the term with the variable x on one side and move all other terms to the other side.

Let's start by squaring both sides of the equation to eliminate the square roots:

(x√40)² = (x√5 + √10)²

40x² = 5x² + 10 + 2x√50 + 10

35x² - 20 = 2x√50

Now, we can square both sides again to eliminate the remaining square root:

(35x² - 20)² = (2x√50)²

1225x⁴ - 1400x² + 400 = 0

This is a quadratic equation in x². We can solve it using the quadratic formula:

x² = (1400 ± √(1400² - 4 × 1225 × 400)) / (2 × 1225)

x² = (1400 ± 200) / 245

x² = 2 or x² = 8/7

Since x² cannot be negative, we have x² = 2 and therefore x = √2.

Substituting this value of x back into the original equation, we have:

x√40 = x√5 + √10

√80 = √10 + √10

√80 = 2√10

So, a = 10 and b = 2, and the solution of the equation is √10 + 7.

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Answer:

a) x = 20

b) x = 9

Step-by-step explanation:

  a)     [tex]\boxed{\bf Mean =\dfrac{sum \ of \ all \ the \ data}{Number \ of \ data}}[/tex]

          [tex]\dfrac{16 +x + 3 + 14 +57}{5}=22\\[/tex]

                                 x + 90 = 22*5

                                 x + 90 = 110

                                         x = 110 - 90

                                        x = 20

b)

         [tex]\bf \dfrac{4 + x + 5 +6 + 1}{5} = 5[/tex]

                          x + 16   = 5 * 5

                            x + 16 = 25

                                   x = 25 - 16

                                   x = 9

Step-by-step explanation:

2.1 Factoring 3x2-2x-23

The first term is, 3x2 its coefficient is 3 .

The middle term is, -2x its coefficient is -2 .

The last term, "the constant", is -23

Step-1 : Multiply the coefficient of the first term by the constant 3 • -23 = -69

Step-2 : Find two factors of -69 whose sum equals the coefficient of the middle term, which is -2 .

-69 + 1 = -68

-23 + 3 = -20

-3 + 23 = 20

-1 + 69 = 68

Answer:

8.41

Step-by-step explanation:

8.14+3.45=11,59

20-11,59=8.41

Therefore, the fair price of one raffle ticket is $2.60, which is slightly less than the amount Raul paid ($3).

a) The fair price of one raffle ticket is $3. This is because 500 tickets were sold at this price and one prize of $200 is to be awarded. Therefore, the 500 tickets collected add up to $1,500, while the prize to be awarded is $200. The net amount to be divided among all the ticket holders is $1,300.

b) The fair price of one raffle ticket is $2.60. This is calculated by dividing the total prize money of $200 by the total number of tickets sold (500). Therefore, 500 tickets multiplied by $2.60 gives a total prize money of $1,300 which is equal to the total ticket sales of $1,500 less the prize money of $200.

For more such ticket related questions

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