Describe in practical terms the meanings of f^(-1)(0.7), f^(-1)(0.5),f^(-1)(0.2)

Answer:

900/π square centimeters or  286.62 square centimeters

Step-by-step explanation:

The circumference of a circle is given by the formula:

C = 2πr

where C is the circumference and r is the radius of the circle.

We are given that the circumference is 60 cm, so we can set up the equation:

60 = 2πr

Solving for r, we get:

r = 60/(2π) = 30/π

The area of a circle is given by the formula:

A = πr^2

Substituting the value of r, we get:

A = π(30/π)^2 = 900/π square centimeters

Therefore, the area of the circle is approximately 286.62 square centimeters, rounded to two decimal places.

After 45 blue counters are removed, the remaining number of blue counters is 63 more than yellow counters. Setting up an equation, we find that there were 162 blue counters to start with.

Let's start by assigning variables to the unknowns:

Let x be the number of yellow counters.

Then, the number of blue counters is 3x.

After 45 blue counters are removed, the number of blue counters is 3x - 45.

Finally, there are 63 more blue counters than yellow, so we can set up the equation:

3x - 45 = x + 63

Solving for x, we get:

2x = 108

x = 54

Therefore, there were 3x = 3(54) = 162 blue counters to start with.

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Answer:

Geometric sequences and exponential functions are alike in that they both involve repeated multiplication by a constant factor. In a geometric sequence, each term is found by multiplying the previous term by the same constant factor. In an exponential function, the value of the function is found by raising a constant base to an exponent that increases by a constant amount.

However, they differ in the way they are expressed. In a geometric sequence, each term is written as a discrete value in the sequence, while in an exponential function, the value of the function is written as a continuous function of the input variable. Additionally, geometric sequences are often used to model discrete phenomena, while exponential functions are often used to model continuous phenomena.

Step-by-step explanation:

The journal entry to close the Income Summary will include a C) credit to Emile, Capital for $26,667.

To close the Income Summary account, the net income of $40,000 needs to be allocated to the partners' capital accounts based on their profit and loss sharing ratio.

Dana is allocated 2/3 of the net income, which is $26,667 (2/3 x $40,000). Emile is allocated 1/3 of the net income, which is $13,333 (1/3 x $40,000).

Therefore, the journal entry to close the Income Summary account would be:

Credit Income Summary for $26,667 (to close the account)

Debit Dana, Capital for $26,667 (to allocate Dana's share of net income)

Debit Emile, Capital for $13,333 (to allocate Emile's share of net income)

This is calculated by multiplying the net income of $40,000 by the ratio of Emile's share (1/3), which equals $26,667.

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The amplitude οf the functiοn used tο mοdel the height οf a rider οn a Ferris wheel shοuld be equal tο the radius οf the Ferris wheel.

212 feet is the average height. A 550-fοοt ferris wheel is the tallest οne. This may be fοund in Las Vegas, Nevada, and is knοwn as the high rοller.

The amplitude οf a periοdic functiοn, such as the height οf a rider οn a Ferris wheel as a functiοn οf time, is the maximum distance frοm the average οr equilibrium pοsitiοn. In this case, the average οr equilibrium pοsitiοn wοuld be the height οf the rider when the Ferris wheel is at its lοwest pοint. The distance frοm the lοwest pοint tο the highest pοint οf the Ferris wheel is equal tο the diameter, and half οf the diameter is equal tο the radius. Therefοre, the amplitude οf the functiοn shοuld be equal tο the radius οf the Ferris wheel.

The area and circumference οf the Ferris wheel are nοt directly related tο the height οf a rider as a functiοn οf time, sο they are nοt relevant tο the amplitude οf the functiοn.

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Answer: D. the radius of the Ferris wheel

Step-by-step explanation:

Answer:

There are four kings in a standard deck of 52 cards, so the probability of drawing a king is 4/52 or 1/13.

Since we know that the card drawn is a king, there are now only four possible cards that could have been drawn: the king of spades, the king of hearts, the king of clubs, and the king of diamonds.

Out of these four possibilities, only one of them is the king of diamonds. Therefore, the probability that the card is a king of diamonds, given that the card drawn is a king, is 1/4.

In other words, the conditional probability of drawing the king of diamonds given that a king has been drawn is:

P(king of diamonds | king) = P(king of diamonds and king) / P(king) = 1/52 / 4/52 = 1/4

Hope this helps! If not, I'm sorry. Feel free to ask for help from me! :]

Answer:

problem 1. 53/20 problem 2. 24/27. Problem 3. 22/48

Step-by-step explanation:

to solve the first two, the process is similar. For the first one, they can be simplifed back into fraction form and not mixed form, so 1 2/5 can be simplified by multiplying 5 by one then adding it to 2, which gives you 7/5, and the same process for 2 1/2, multiply 2 by two, get 4 add it to one to get 5/2. Now we have 7/5 + 5/2, when we have fractions that we need to add or subtract, but down have the same bottom number, the easiest way to get a answer is to multiply both bottom numbers together then multiply the top number by the opposite bottom number, so 5x4 = 20, 4x7=28, 5x5=25, and we get 28/20 + 25/20 = 53/20. The process for problem two is the same, but just with subtraction and different numbers.

for problem three, the process is almost the same, but the steps differ. The bag is made up of R G and P counters, 3/8 of them are red, 1/6 are green, and the rest are purple because nothing else is in the bag. So another way of saying this is 3/8 + 1/6 + purple = Full Amount in bag. But now we need purple, so we know 1/2 one half and so does 2/4, and since we have fractions we can find the amount in the bag by adding them, 3/8 + 1/6 = 26/48, and we know the full bag is 1/1 or 48/48, and fhe only one left is purple, so to get purple, we subtract 48/48 from 26/48 and get 22/48 which is the amount of purple.

Decimals are used to write a number that is not an integer. Decimals are numbers between integers.

The supermarket chain of Product are :

Bakery, Beverages, Non-food and pharmaceutical products (such as cigarettes, lottery tickets and over-the-counter medicines), rental of DVDs, books and magazines, including supermarket tabloids, greeting cards, toys, a small number of household items such as light bulbs, Personal care such as cosmetics, soaps, shampoos, Produce (fresh fruits and vegetables), etc.

Decimals are used together to represent whole numbers and fractions. Here we will separate whole numbers and fractions by inserting a “.”, called a decimal point. In decimal form, we write this as 1.5 pizzas.

Decimals are used to write a number that is not an integer. Decimals are numbers between integers. An example is 12.5, which is a decimal number between 12 and 13. Greater than 12, but less than 13.

Continuing with the previous example, 12.5 is the same number as the fractional 12½. This is true regardless of the complexity of the decimal point. For example, the number 0. 75 equals ¾. If you want, you can go further and say that 0.75 equals 75%.

Complete Question:

Investigate supermarket chain products, observe which products have decimals in their values, list 10 products and use them to perform different decimal operations, build a word problem and demonstrate your understanding of decimals.

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Answer:

We can use the expression 2x = y , where 'x' is the amount she saves yesterday and y being the amount she saves this day.

Table = if x = 0 , y = 0
            if x = 1 , y = 2
            if x = 2 , y =4
            if x = 4 , y= 8

Therefοre , the sοlutiοn οf the given prοblem οf lοgarithm cοmes οut tο be  ln(4) is the value οf x at which f(x) equals 0, and chοice is the right respοnse (2).

The inverse οf an integer is represented mathematically by the lοgarithm. Therefοre, the expοnential that bc needs tο be multiplied by tο get a specific integer, x, is identical tο its base b expοnent. As an illustratiοn, 1000 = 103, sο lοg10 = 3 is 1000's base-10 lοgarithm, which is 3. As an example, the cube οf twenty is actually οne hundred while the base-10 derivative οf ten appears tο be twο. lοg  100 = 2.

Here,

The fοrmula is prοvided as f(x) = 2e(2x) - 32. and we must determine the x-value at which f(x) equals a particular number. Let's wοrk thrοugh each sοlutiοn:

(1) ln(5/2):

[tex]\Rightarrow f(ln(5/2)) = 2e^{(2*ln(5/2))} - 32[/tex]

[tex]\Rightarrow2e^{(ln(25/4))} - 32[/tex]

[tex]\Rightarrow 2(25/4) - 32[/tex]

[tex]\Rightarrow -19/2[/tex]

As a result, f(ln(5/2)) does not equal 0.

(2)  ln(4):

[tex]\Rightarrow f(ln(4)) = 2e^{(2*ln(4))}- 32[/tex]

[tex]\Rightarrow 2e^{(ln(16))}- 32[/tex]

[tex]\Rightarrow 2(16) - 32[/tex]

[tex]\Rightarrow 0[/tex]

As a result, choice (2) is the appropriate response because f(ln(4)) equals 0.

(3) ln(8):

[tex]\Rightarrow f(ln(8)) = 2e^{(2*ln(8))} - 32[/tex]

[tex]\Rightarrow2e^{(ln(64))} - 32[/tex]

[tex]\Rightarrow 2(64) - 32[/tex]

[tex]\Rightarrow 96[/tex]

As a result, f(ln(8)) does not equal 0.

(4 y = ln(2/5):

This choice is illogical because we must determine the value of x, not y.

As a result, ln(4) is the value of x at which f(x) equals 0, and choice is the right response (2).

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The correct statements about the transformation are the shape of the parallelogram is preserved and the size of the parallelogram is changed.

The rule (x,y) ↛ ((2x+2y-4) represents a linear transformation that involves scaling and translating the coordinates of each point. Specifically, the transformation scales the x-coordinate by a factor of 2 and the y-coordinate by a factor of 2, and then translates the resulting coordinates 4 units to the right and 4 units down.

Based on this, we can make the following observations:

The shape of the parallelogram L'm'n'p is preserved under the transformation. That is, the transformed parallelogram LMNP is also a parallelogram.

The size of the parallelogram L'm'n'p is changed under the transformation. Specifically, the dimensions of the transformed parallelogram LMNP are twice as large as those of the original parallelogram L'm'n'p.

Therefore, the correct statements about the transformation are:

The shape of the parallelogram is preserved.

The size of the parallelogram is changed.

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The solution to x/-3-2<-4 is x>18.

In mathematics, inequality is a statement that expresses the relationship between two values, where one value is greater than, less than, or equal to the other. Inequalities can be expressed using symbols such as > (greater than), < (less than), and = (equal to). Inequality can be used to compare two values or expressions, or to determine if a given number is within a certain range. Inequalities can be used to solve problems, such as finding the area of a triangle, or to model real-world situations, such as determining the maximum number of people that can fit in a room.

To solve this inequality, first we subtract 2 from both sides to isolate x:

x/-3<-6

Then, we divide both sides by -3 to solve for x:

x>18

Therefore, the solution to x/-3-2<-4 is x>18.

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The equation for the function g(x) is g(x) = (x + 4)² - 5.

In Mathematics, the translation of a geometric figure to the left simply means subtracting a digit from the value on the x-coordinate (x-axis) of the pre-image of a function while a geometric figure that is translated downward simply means subtracting a digit from the value on the y-coordinate (y-axis) of the pre-image.

In Mathematics, a vertical translation to the negative y-direction (downward) is modeled by this mathematical expression g(x) = f(x) - N.

Where:

N represents an integer.

g(x) and f(x) represent a function.

Based on the graph of the parent function f(x) = x², an equation for g(x) in vertex form after a translation of 5 units downward and 4 units to the left is given by:

g(x) = (x + 4)² - 5.

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Answer:

186

Step-by-step explanation:

97+89=186

The equation for line L in slope-intercept form is y = -1/4x + x₁/4 + y₁

The exponential equation is 53−x=251. Therefore, we need to express each side as a power of the same base and then equate exponents. Let's express both sides with the same base 5. We know that 251 is the same as 5². Thus, the equation can be rewritten as:

53−x=251=5².

We need to express 53−x as a power of 5. To do that, we can write it as:

53−x = 5³/x.

Now, we can substitute this expression into our equation to get:

5³/x=5²

Let's multiply both sides of the equation by x to eliminate the fraction:

5³=5²x.

Divide both sides by 5² to get:x = 5. We can check our solution by plugging it back into the original equation:

53−5=25, which is true.

Thus, the solution set is {5}.The equation for line L in the point-slope form: We know that L is perpendicular to y=4x. Thus, the slope of L is -1/4. We also know that L passes through a point (x₁, y₁). Let's write the point-slope form of the equation:

y - y₁ = -1/4(x - x₁).

The equation for line L in slope-intercept form: Let's solve this equation for y to get it in slope-intercept form:

y - y₁ = -1/4(x - x₁).

Multiply both sides by 4 to eliminate the fraction:

4y - 4y₁ = -x + x₁.

Simplify by moving -x + x₁ to the right side:

4y = x₁ - x + 4y₁.

Add x to both sides:

4y + x = x₁ + 4y₁

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The given sequence is a recursive sequence. Hence -2, 9, -24, 75 is the first four terms of the sequence.

The given sequence is a recursive sequence, where each term is defined in terms of the previous term. Here, each term is obtained by multiplying the previous term by -3 and adding 3. The first term, a1, is given. Using this, we can find the second term, a2, and then using a2, we can find a3, and so on.

Given: an = -3an-1 + 3 and a1 = -2

To find: the first four terms of the sequence

a1 = -2

a2 = -3a1 + 3 = -3(-2) + 3 = 9

a3 = -3a2 + 3 = -3(9) + 3 = -24

a4 = -3a3 + 3 = -3(-24) + 3 = 75

Therefore, the first four terms of the sequence are: -2, 9, -24, 75.

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Answer:

Step-by-step explanation:

If h is the height of the ball off the ground, if we want to find the time, t, when the ball hits the ground again, we set h equal to 0, because the height of something when it is on the ground is 0.

[tex]0=-t^2+4t[/tex] and solve for the values of t. Begin by factoring out a -t:

[tex]0=-t(t-4)[/tex]

By the Zero Product Property, either -t = 0 or t - 4 = 0. We already know that at time 0 the ball hit the ground for the first time because that was given in the problem. That means that the ball hit the ground for the second time 4 seconds later.

Twο pοints (8,8) and (-2,6) are the set οf the equatiοn  [tex]y < \frac{3}{4} x+6\\[/tex] οf the graph [tex]y = \frac{3}{4} x+6[/tex].

Graph, a diagram that shοws hοw a variable varies in relatiοn tο οne οr mοre οther variables (such as a cοllectiοn οf pοints, lines, line segments, curves, οr regiοns).

Nοw we have tο find the value οf the equatiοn i.e. [tex]y < \frac{3}{4} x+6\\[/tex],

(x, y) = (8,8) , we get 8 < 12 , First pοint

(x, y) = (-4,3) , we get 3 = 3 , Nο match

(x, y) = (6,-2) , we get -2 < 10.5 , Secοnd Pοint

(x, y) = (0, 8) , we get 8 > 6 , Nο Match

(x, y) = (-9,2) , we get 2>-0.75 , Nο Match

Sο, we get οur pοints, which are pοints (8,8) and (-2,6) fοr the equatiοn [tex]y < \frac{3}{4} x+6\\[/tex].

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Answer:

Let's use a Venn diagram to solve this problem.

First, let's label the three regions of the Venn diagram: Computer only (C), Optional Maths only (M), and both Computer and Optional Maths (C ∩ M). We can also label the region outside the circles as neither Computer nor Optional Maths (N).

We know that 30% of the students like Computer only, which means that the percentage of students in region C is 30%. Similarly, 25% of the students like both Computer and Optional Maths, so the percentage of students in region C ∩ M is 25%.

We are also given that 5% of the students don't like either subject, so the percentage of students in region N is 5%.

Finally, we are told that 390 students like Optional Maths, which includes the students in regions M and C ∩ M. We don't know the percentage of students in region M, but we do know that the percentage of students in region C ∩ M is 25%.

Using this information, we can set up an equation to solve for the total number of students:

C + M + C ∩ M + N = 100%

Substituting the percentages we know, we get:

30% + M + 25% + 5% = 100%

Simplifying the equation, we get:

M = 40%

This means that 40% of the students like Optional Maths only, which is the percentage of students in region M.

Now we can use the fact that 390 students like Optional Maths to solve for the total number of students:

M + C ∩ M = 390

0.4T + 0.25T = 390

0.65T = 390

T = 600

Therefore, the total number of students is 600.

The statement that correctly explains the association in the scatter plot is; A. Since the y-values decrease as the x-values increase, the scatter plot shows a negative association.

When interpreting scatterplots, it is possible to claim that two variables have a negative connection when the values of one variable tend to fall as the values of the other variable rise.

Similarly to this, we can say that two variables have a positive connection when one variable's values tend to rise along with those of the other.

In the scatterplot shown, we can see that there is a negative correlation between the two variables because the y-value is falling while the x-axis is rising.

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Answer:

(5, -1) or x = 5, y = -1

Step-by-step explanation:

[tex]\mathrm{Substitute\:}y=2x-11[/tex]

[tex]\begin{bmatrix}2x+3\left(2x-11\right)=7\end{bmatrix}[/tex]

[tex]8x-33=7[/tex]

[tex]\mathrm{Isolate}\:x\:\mathrm{for}\:8x-33=7[/tex]

[tex]8x=40[/tex]

[tex]\mathrm{Divide\:both\:sides\:by\:}8[/tex]

[tex]\frac{8x}{8}=\frac{40}{8}[/tex]

[tex]x=5[/tex]

[tex]y= 2x-11 \\ \mathrm{Substitute\:}x=5[/tex]

[tex]y=2\cdot \:5-11[/tex]

[tex]y=-1[/tex]

Answer:

(5, - 1 )

Step-by-step explanation:

2x + 3y = 7 → (1)

y = 2x - 11 → (2)

substitute y = 2x - 11 into (1)

2x + 3(2x - 11) = 7

2x + 6x - 33 = 7

8x - 33 = 7 ( add 33 to both sides )

8x = 40 ( divide both sides by 8 )

x = 5

substitute x = 5 into (2)

y = 2(5) - 11 = 10 - 11 = - 1

solution is (5, - 1 )

The side lengths and the angle measures is PQ = 2, PR ≈ 3.61, QR = 3

m∠P = 90°, m∠Q = 90°, m∠R ≈ 33.7°

Tο find the side lengths, we need tο use the distance fοrmula:

PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]

PR = √[(x₃ - x₁)² + (y₃ - y₁)²]

QR = √[(x₃ - x₂)² + (y₃ - y₂)²]

PQ = √[(-4 - (-2))² + (4 - 4)²] = √[2²] = 2

PR = √[(-4 - (-2))² + (1 - 4)²] = √[2² + 3²] = √13 ≈ 3.61

QR = √[(-4 - (-4))² + (1 - 4)²] = √[3²] = 3

Tο find the angle measures, we can use the Law οf Cοsines:

cοs(θ) = (b² + c² - a²) / 2bc

where a, b, and c are the side lengths οppοsite tο the cοrrespοnding angles. Fοr example, the angle measure at vertex P is οppοsite tο side PQ, sο we can use the lengths PR and QR tο find its angle measure.

cοs(m∠P) = (PR² + QR² - PQ²) / 2PRQR

cοs(m∠P) = (13 + 9 - 4) / (2 * √13 * 3)

cοs(m∠P) = 18 / (2√39)

m∠P = cοs⁻¹(18 / (2√39)) ≈ 90°

cοs(m∠Q) = (PQ² + QR² - PR²) / 2PQQR

cοs(m∠Q) = (4 + 9 - 13) / (2 * 2 * 3)

cοs(m∠Q) = 0

m∠Q = cοs⁻¹(0) = 90°

cοs(m∠R) = (PQ² + PR² - QR²) / 2PQPR

cοs(m∠R) = (4 + 16 - 13) / (2 * 2 * √13)

cοs(m∠R) = 7 / (4√13)

m∠R = cοs⁻¹(7 / (4√13)) ≈ 33.7°

Therefοre, the answer is:

PQ = 2, PR ≈ 3.61, QR = 3

m∠P = 90°, m∠Q = 90°, m∠R ≈ 33.7°

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Answer:

height = 15 cm

Step-by-step explanation:

Base area of a prism B = hw where h = height, w = width

So B = lw = 200 cm²

Volume V = lwh = 3000 cm³

V/B = lwh/lw = h

h = 3000cm³/200cm = 15 cm

Answer:

Since the number of bacteria doubles every 3 hours, after 3 hours there will be 400 x 2 = 800 bacteria. After 6 hours, there will be 800 x 2 = 1600 bacteria. After 9 hours, there will be 1600 x 2 = 3200 bacteria. After 12 hours, there will be 3200 x 2 = 6400 bacteria. Finally, after 15 hours, there will be 6400 x 2 = 12800 bacteria. Therefore, there will be 12800 bacteria at the end of 15 hours.

Step-by-step explanation:

6 pounds of chocolate worth $1.20 per pound must be mixed with 10 pounds of chocolate worth 90 cents per pound in order to create a mixture worth $1.00 per pound.

To produce a mixture worth $1.00 a pound, 6 pounds of chocolate worth $1.20 per pound must be mixed with 10 pounds of chocolate worth 90 cents per pound.

Let us proceed as follows:

x is the amount of chocolate worth $1.20 per pound that must be mixed with 10 pounds of chocolate worth 90 cents per pound in order to create a mixture worth $1.00 per pound.

Using the formula:

Price of chocolate (x) + Price of chocolate ($0.90) = Price of mixture ($1.00)

The following equation can be created:

1.20x + 0.9(10) = 1(10 + x)

Simplify: 1.20x + 9 = 10 + x1.20x - x = 10 - 91.20x - x = 1

Divide both sides by 0.2 to simplify the equation:

6x = 5x = 6

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By secant line formula, the slopes corresponding to the lines are listed below:

In this problem we have nine representations of lines, whose slopes must be determined. A line is represented by equations of the form:

y = m · x + b

Where:

According to analytic geometry, slope can be found by knowning the location of two points (initial, final) set on Cartesian plane and secant line formula:

m = (y₂ - y₁) / (x₂ - x₁)

Where:

Case 1: (x₁, y₁) = (0, - 3), (x₂, y₂) = (2, 1)

m = (2 - 0) / [1 - (- 3)]

m = 2 / 4

m = 1 / 2

Case 2: (x₁, y₁) = (- 3, 2), (x₂, y₂) = (1, - 3)

m = (- 3 - 2) / [1 - (- 3)]

m = - 5 / 4

Case 3: (x₁, y₁) = (- 3, - 4), (x₂, y₂) = (2, 3)

m = [3 - (- 4)] / [2 - (- 3)]

m = 7 / 5

Case 4: (x₁, y₁) = (0, 3), (x₂, y₂) = (3, - 3)

m = (- 3 - 3) / (3 - 0)

m = - 6 / 3

m = - 2

Case 5: (x₁, y₁) = (- 2, 1), (x₂, y₂) = (3, 2)

m = (2 - 1) / [3 - (- 2)]

m = 1 / 5

Case 6: (x₁, y₁) = (- 4, 3), (x₂, y₂) = (4, 1)

m = (1 - 3) / [4 - (- 4)]

m = - 2 / 8

m = - 1 / 4

Case 7: (x₁, y₁) = (2, - 4), (x₂, y₂) = (4, 2)

m = [2 - (- 4)] / (4 - 2)

m = 6 / 2

m = 3

Case 8: (x₁, y₁) = (- 4, - 1), (x₂, y₂) = (0, - 3)

m = [- 3 - (- 1)] / [0 - (- 4)]

m = - 2 / 4

m = - 1 / 2

Case 9: (x₁, y₁) = (- 2, 0), (x₂, y₂) = (4, 1)

m = (1 - 0) / [4 - (- 2)]

m = 1 / 6

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It will take approximately 4.4 years to earn enough money to go on the trip if we invest our 2200 in a savings account with a 1.55% interest rate compounded annually.

First, we need to calculate the amount of money that we need to save in order to cover the cost of the trip. This can be done by subtracting the amount we have already saved from the total cost of the trip:

Total cost of trip = 2500

Amount already saved = 2200

Amount to save = 2500 - 2200 = 300

Next, we can use the compound interest formula to calculate how long it will take to earn 300 with an interest rate of 1.55% compounded annually. We can set up the formula as follows:

A = P(1 + i)n

where:

A = accumulated amount = 300 + 2200 = 2500 (the total cost of the trip)

P = principal = 2200

i = interest rate per year = 1.55%

n = number of years we need to save for

We can now solve for n:

2500 = 2200(1 + 0.0155)n

Divide both sides by 2200:

1.13636 = 1.0155n

Take the natural logarithm of both sides:

ln(1.13636) = n ln(1.0155)

Divide both sides by ln(1.0155):

n = ln(1.13636)/ln(1.0155) ≈ 4.4 years

Therefore, it will take approximately 4.4 years to earn enough money to go on the trip if we invest our 2200 in a savings account with a 1.55% interest rate compounded annually.that we rounded our final answer to the nearest tenth as instructed.

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The calling plan was a logical decision that was made. For calls with a shorter estimated time (10 minutes), the first plan is selected; for calls with an extended expected duration, the second plan is favoured (15 minutes).

Let the costs for the first and second plans, respectively, be denoted by h₁(x) and h₂(x), while the call lasts for x minutes.

The expressions for these two cost functions are as follows, according to the information available:

h₁(x) = 10x

h₂(x) = 99; if x ≤ 20 and 99 + 10(x - 20); if x > 20

Let X represent how long the call was. The exponential distribution of X has a parameter value of λ = 1/10 if the anticipated call time is 10 minutes.

That is, X ~ exp(λ = 1/10)

The exponential distribution's density function with parameter λ = 1/10 is,

f(x) = 1/10[tex]e^{-x/10}[/tex]; if x > 0 and 0; if otherwise

Calculate the anticipated cost of the initial plan E[h₁(x)].

E[h₁(x)] = [tex]\int^{\infty}_{-\infty}h_{1}(x)\cdot f(x)dx[/tex]

E[h₁(x)] = [tex]\int_{-\infty}^{0}10x\cdot 0dx+\int^{\infty}_{0}10x\cdot \frac{1}{10}e^{-x/10}dx[/tex]

E[h₁(x)] = 0 + [tex]\int^{\infty}_{0}x\cdot e^{-x/10}dx[/tex]

E[h₁(x)] = [tex][-10xe^{-x/10}-100e^{-x/10}]^{\infty}_{0}[/tex]

E[h₁(x)] = 100

Calculate the anticipated cost of the initial plan E[h₂(x)].

E[h₂(x)] = [tex]\int^{\infty}_{-\infty}h_{2}(x)\cdot f(x)dx[/tex]

E[h₂(x)] = [tex]\int_{-\infty}^{0}99dx+\int^{20}_{0}99\cdot \frac{1}{10}e^{-x/10}dx + \int^{\infty}_{20}(99+10(x-20))\cdot \frac{1}{10}e^{-x/10}dx[/tex]

E[h₂(x)] = 0 + [tex]\frac{99}{10}\int^{20}_{0}e^{-x/10}dx - \frac{101}{10}\int^{\infty}_{20}e^{-x/10}dx+\int^{\infty}_{20}xe^{-x/10}dx[/tex]

E[h₂(x)] = [tex]\frac{99}{10}[-10e^{-x/10}]^{20}_{0} - \frac{101}{10}[-10e^{-x/10}]^{\infty}_{20} + [-10xe^{-x/10}-100e^{-x/10}]^{\infty}_{20}[/tex]

E[h₂(x)] = -99e⁻² + 99 - 101e⁻² + 200e⁻² + 100e⁻²

E[h₂(x)] = 99 + 100e⁻²

E[h₂(x)] = 112.5335

E[h₂(x)] > E[h₁(x)]  (112.53 > 100) is what has been seen. Hence, the first strategy is chosen when a 10-minute call is anticipated.

The exponential distribution's density function with parameter λ = 1/15 is,

f(x) = 1/15[tex]e^{-x/15}[/tex]; if x > 0 and 0; if otherwise

Calculate the anticipated cost of the initial plan E[h₁(x)].

E[h₁(x)] = [tex]\int^{\infty}_{-\infty}h_{1}(x)\cdot f(x)dx[/tex]

E[h₁(x)] = [tex]\int_{-\infty}^{0}10x\cdot 0dx+\int^{\infty}_{0}10x\cdot \frac{1}{15}e^{-x/15}dx[/tex]

E[h₁(x)] = 0 + [tex]\frac{2}{3}\int^{\infty}_{0}x\cdot e^{-x/15}dx[/tex]

E[h₁(x)] = [tex]\frac{2}{3}[-15xe^{-x/15}-225e^{-x/15}]^{\infty}_{0}[/tex]

E[h₁(x)] = 150

Calculate the anticipated cost of the initial plan E[h₂(x)].

E[h₂(x)] = [tex]\int^{\infty}_{-\infty}h_{2}(x)\cdot f(x)dx[/tex]

E[h₂(x)] = [tex]\int^{20}_{0}99\cdot \frac{1}{15}e^{-x/15}dx + \int^{\infty}_{20}(99+10(x-20))\cdot \frac{1}{15}e^{-x/15}dx[/tex]

E[h₂(x)] =  [tex]\frac{99}{15}\int^{20}_{0}e^{-x/15}dx - \frac{101}{15}\int^{\infty}_{20}e^{-x/15}dx+\frac{2}{3}\int^{\infty}_{20}xe^{-x/15}dx[/tex]

E[h₂(x)] = [tex]\frac{99}{15}[-15e^{-x/15}]^{20}_{0} - \frac{101}{15}[-15e^{-x/15}]^{\infty}_{20} + [-15xe^{-x/15}-225e^{-x/15}]^{\infty}_{20}[/tex]

E[h₂(x)] = -99[tex]e^{-4/3}[/tex] + 99 - 101[tex]e^{-4/3}[/tex] + 200[tex]e^{-4/3}[/tex] + 150[tex]e^{-4/3}[/tex]

E[h₂(x)] = 99 + 150[tex]e^{-4/3}[/tex]

E[h₂(x)] = 138.54

E[h₂(x)] < E[h₁(x)]  (138.54 < 150) is what has been seen. Hence, the first strategy is chosen when a 15-minute call is anticipated.

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The first statement is correct,  that the value of x is 55°.

The second statement is incorrect, the correct statement is the measure of the smaller angle is 35°.

Complementary angles are those whose combined angle is exactly 90°

Part A :

The figure shows right-angled or complementary angle which means the sum of angles is 90°

Therefore we can say that,

x + x-20 = 90°

2x - 20 =90°

2x = 90° + 20

2x = 110°

x = 55°

Therefore the value of the x = 55°, the first statement correct.

The measure of the smaller angle is

= x - 20°

= 55° - 20°

= 35°

The second statement is incorrect.

The correct statement is a measure of the smaller angle is 35°.

Part B:

The figure shows the supplementary angle and the sum of the supplementary angle is 180°.

3x + 6x = 180°

9x = 180°

x = 20°

The first statement is incorrect, the correct statement is the value of x = 20.

The angle measure is 3x = 3 * 20 = 60° and 6x = 6 * 20 = 120°

This means that the second statement is correct.

Part C:

Given that the angle is complementary. the first angle is 2x and the second angle is (3x-5).

A complementary angle is the sum of the angles equal to 90°

2x + 3x-5 = 90

5x = 90 - 5

5x = 85

x = 17°

The first statement is incorrect. the correct statement is the value of x = 17.

The measure of the angle,

2x = 2 * 17 = 34

3x - 5 = 3 * 17 - 5 = 51 - 5 = 46

From the above result, second statement is also incorrect, the correct statement is the measure of the larger angle is 46°.

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a) The equation that represents the total amount that the twins earned is; j + r = 96000

b) An amount that represents what Ron earned in terms of what Jon earned is; r = 3j + 8000

Algebraic word problems are defined as questions that require translating sentences to equations, then solving those equations. The equations we need to write will only involve basic arithmetic operations and a single variable. Usually, the variable represents an unknown quantity in a real-life scenario.

We are told that;

j represents the amount that Jon earned

r represents the amount that Ron earned

Thus, if together they earned $96000, then we have;

j + r = 96000

Ron earned $8000 more than three times what Jon earned. Thus, we have;

r = 3j + 8000

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