Consider a world with 10 countries. Each country must select its level of abatement of pollution zi, which is a continuously defined variable. The cost of abatement C(zi) for any country i depends upon the abatement effort thatcountry exerts and is given by C(zi) = 50 z? where i E {1, ..., 10}. The benefit that i gets on pollution abatement depends upon the total level of abatement by all countries and is given by B;(Z) = 10 Z – 0.005Z2 where 2 = k zk and i E {1,...,10}. The payoff for each country i is given by Ti = = B;(Z) – C(zi) Based on the above information, you are required to complete the following tasks a) What is the resulting level of abatement zi for each country if all act in their self- interest? b) Now assume that the countries cooperate to maximize the following collective pay-off 10 10 II = B;(Z) - C(zi). ) i=1 i=1 where B; and C have the same expressions as given previously. What are the new levels of abatement? c) Does cooperation result in a Pareto improvement in thiscase? Calculate the magnitude of efficiency gain obtained from full cooperation.

The probability that at least one WST111 student is in time for the 7h30 lecture on a Tuesday morning is 0.9961.

1. First, let's find the probability that a randomly selected student is on time for the lecture. Since the probability that a student is late is 0.25, the probability that a student is on time is 1 - 0.25 = 0.75.

2. Now, we need to calculate the probability that all five randomly selected students are late for the lecture. Since punctuality is independent, we can simply multiply each student's probability of being late: 0.25×0.25×0.25×0.25× 0.25 = 0.0009765625.

3. Finally, we want to find the probability that at least one student is on time. To do this, we'll subtract the probability that all students are late from 1:

1 - 0.0009765625 = 0.9961.

So, the probability that at least one WST111 student is in time for the 7h30 lecture on a Tuesday morning is 0.9961.

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The value of the unknown angle is 68°

Trigonometric Ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin(tetha) = opp/hyp

cos(tetha) = adj/hyp

tan(tetha) = opp/adj

In the triangle, 51 is the opposite and 55 is the hypotenuse.

therefore;

sin(tetha) = 51/55

sin(tetha) = 0.927

tetha = sin^-1( 0.927)

tetha = 67.97

approximately to 68°

therefore the value of the unknown angle is 68°

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The midpoint of the two points of P (-3,-7) and Q (3,-5) is (0, -6).

When you have the vertices of two points, you can find the midpoint by the formula :

= ( ( x 1 + x 2 ) / 2 , ( y 1 + y 2 ) / 2 )

Solving for the midpoint therefore gives:

= ( ( - 3 + 3 ) / 2 , ( - 7 + ( - 5 ) ) / 2 )

=  ( 0 / 2 , ( - 12 ) / 2 )

= (0, -6)

In conclusion, the midpoint is (0, -6).

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The number of distinct solutions of the given congruences and find the solutions.

a) 7x = 9 (mod 14) has no solution

b) 8x = 9 mod (mod 11) [tex]x\equiv 8 \hspace{0.1cm}(mod \hspace{0.1cm}11)[/tex]

c) 16x = 20 (mod 36) [tex]8, 17, 26, 35 \hspace{0.2cm}mod(36)[/tex]

(a) 7x = 9mod(14) 20

Here, gcd(7,14) =7 , and we know that 7 does not divide 9.

Thus, from Theorem 1, we can say that it has no solution.

(b)8x =  9 mod(11)

Here, gcd(8,11) = 1, so using theorem 2, we can say that it has a unique solution.

For that we need to find [tex]\phi (11)[/tex],  Since 11 is an prime number, therefore the gcd of 11 with any positive integer smaller than 11 will be 1. So,

[tex]\phi (11)[/tex] = 10  = |{1,2,3,..., 10}| ,

So, the solution for the congruence is given by using theorem 2:

[tex]x\equiv a^{\phi (m)-1}b \hspace{0.1cm}(mod \hspace{0.1cm}m)[/tex]

x = 810-19 (mod 11) (

x = 88*9*8 (mod 11)

[tex]x\equiv 64^{4}*72 \hspace{0.1cm}(mod \hspace{0.1cm}11)x\equiv 9^{4}*6 \hspace{0.1cm}(mod \hspace{0.1cm}11)x\equiv 81^{2}*6 \hspace{0.1cm}(mod \hspace{0.1cm}11)[/tex]

x = 16 * 6 (mod 11)

2 = 5*6 (mod 11

[tex]x\equiv 8 \hspace{0.1cm}(mod \hspace{0.1cm}11)[/tex]

which is the final solution.

(c) [tex]16x\equiv 20 \hspace{0.1cm}(mod \hspace{0.1cm}36)[/tex]

Here, d=gcd(16,36) =4 and 4 divides 20, so it has 4 unique solutions.

So, we will use theorem 3.

Divide by 4 whole congruence:

[tex]16x/4\equiv 20/4 \hspace{0.1cm}(mod \hspace{0.1cm}36/4)[/tex]

[tex]4x\equiv 5 \hspace{0.1cm}(mod \hspace{0.1cm}9)[/tex]

[tex]So, \phi (9)=\left | \left \{ 1,2,4,5,7,8 \right \} \right |=6[/tex]

[tex]So, x\equiv 4^{\phi (9)-1}*5 \hspace{0.1cm}(mod \hspace{0.1cm}9)[/tex]

[tex]x\equiv 4^{5}*5 \hspace{0.1cm}(mod \hspace{0.1cm}9)[/tex]

[tex]x\equiv 4^{4}*20 \hspace{0.1cm}(mod \hspace{0.1cm}9)[/tex]

[tex]x\equiv 16^{2}*20 \hspace{0.1cm}(mod \hspace{0.1cm}9)[/tex]

x = 72 * 2 (mod 9)

[tex]x\equiv 8 \hspace{0.1cm}(mod \hspace{0.1cm}9)[/tex]

Thus, the 5 unique solutions using theorem3 are given as follows:

[tex]t,t+\frac{m}{d}, t+\frac{2m}{d},. . ., t+\frac{(d-1)m}{d} \hspace{0.2cm} mod(m)[/tex]

[tex]8, 17, 26, 35 \hspace{0.2cm}mod(36)[/tex].

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Applying the Friedman test, we conclude that there is evidence that at least two population locations differ at a significance level of 10%, since our calculated [tex]$\chi^2$[/tex] value (979.5) is greater than the critical value (7.81).

To apply the Friedman test, we need to first rank the data within each block (column) and calculate the average ranks for each treatment (row). The ranks are calculated by assigning a rank of 1 to the smallest value, 2 to the second-smallest value, and so on. Ties are given the average rank of the tied values.

Treatment Block 1 Block 2 Block 3 Block 4 Ranks

1 2 1.5 3 3.5 10

2 1 3 5 5 14

3 3 2.5 4 2 11.5

4 2 4 2 1 9

5 1 4.5 1 4.5 11

The Friedman test statistic is calculated as:

[tex]$ \chi^2 = \frac{12}{n(k-1)} \left[ \sum_{j=1}^k \left( \sum_{i=1}^n R_{ij}^2 - \frac{n(n+1)^2}{4} \right) \right] $[/tex]

where [tex]$n$[/tex] is the number of blocks, [tex]$k$[/tex] is the number of treatments, and [tex]$R_{ij}$[/tex] is the rank of the [tex]$j^t^h[/tex] treatment in the [tex]$i^t^h[/tex] block.

In this case, [tex]$n=4$[/tex] and [tex]$k=5$[/tex], so:

[tex]$ \chi^2 = \frac{12}{4(5-1)} \left[ \sum_{j=1}^5 \left( \sum_{i=1}^4 R_{ij}^2 - \frac{4(4+1)^2}{4} \right) \right] $[/tex]

[tex]$ \chi^2 = \frac{3}{2} \left[ (10^2 + 14^2 + 11.5^2 + 9^2 + 11^2) - \frac{4(5^2)}{4} \right] $[/tex]

[tex]$ \chi^2 = \frac{3}{2} \left[ 727 - 50 \right] = 979.5 $[/tex]

The critical value for the Friedman test with [tex]$k=5$[/tex] treatments and [tex]$n=4$[/tex]blocks, at a significance level of [tex]\alpha = 0.1$,[/tex] is:

[tex]$ \chi_{0.1}^2 = 7.81 $[/tex]

Since our calculated [tex]$\chi^2$[/tex] value (979.5) is greater than the critical value (7.81), we reject the null hypothesis that there is no difference between the population locations, and conclude that there is evidence that at least two population locations differ at a significance level of 10%.

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The standard deviation of the total sales per day for the shoe store is approximately $1743.28.

To find the standard deviation of the total sales per day for the shoe store, we need to first find the mean and variance of the total sales.

The number of customers visiting the store follows a Poisson distribution with a rate of 70 per day. Let X be the number of customers per day, then X ~ Poisson(70).

The proportion of customers that make a purchase is 45%. Let Y be the number of customers that make a purchase, then Y ~ Binomial(X, 0.45).

The amount spent by a paying customer follows an exponential distribution with a mean of $120. Let Z be the amount spent by a paying customer, then Z ~ Exp(1/120).

Now, let's find the mean and variance of the total sales per day.

E[XY] = E[E[XY|X]] = E[X * 0.45] = 70 * 0.45 = 31.5

E[Z] = 120

E[XYZ] = E[E[XYZ|XY]] = E[XY * 120] = 31.5 * 120 = 3780

Var(XY) = E[Var(XY|X)] + Var(E[XY|X]) = E[X * 0.45 * 0.55] + Var(X * 0.45) = 70 * 0.45 * 0.55 + 70 * 0.45 * 0.55 = 17.325

Var(Z) = 120^2

Var(XYZ) = E[Var(XYZ|XY)] + Var(E[XYZ|XY]) = E[XY * 120^2] + Var(XY * 120) = 31.5 * 120^2 + 17.325 * 120^2 = 3035250

Therefore, the variance of the total sales per day is Var(XYZ) = 3035250.

The standard deviation of the total sales per day is the square root of the variance:

SD(XYZ) = sqrt(3035250) = 1743.28

Therefore, the standard deviation of the total sales per day for the shoe store is approximately $1743.28.

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When conducting a hypothesis test, a(an) **b. t-statistic** is more appropriate than a z-score when you don't know the population variance or the population standard deviation. The t-statistic takes into account the sample size and is better suited for situations where population parameters are unknown.

When conducting a hypothesis test, a t-statistic is more appropriate than a 2-score when you don't know the population variance or the population standard deviation. The t-statistic is used to test hypotheses about population means when the sample size is small or when the population standard deviation is unknown.

The t-statistic is calculated by dividing the difference between the sample mean and the hypothesized population mean by the standard error of the mean, which takes into account the variability of the sample.

The t-statistic is compared to a critical value from a t-distribution with n-1 degrees of freedom, where n is the sample size. The level of significance, or alpha value, is also used to determine the critical value. Sample variance and Cohen's d are other statistical measures used in hypothesis testing but are not specifically related to the use of t-statistics.

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Answer: You can convert it by dividing it by 12 which would give you 3.5 and since half of 12 is 6 the answer is 3ft and 6inches

Step-by-step explanation:

Answer: 10x - 1 and 20x - 12

Step-by-step explanation:

perimeter is defined as the distance around a figure:

as such, the perimeter of this triangle is: (2x -3) + (3x + 5) + (5x - 3)

which simplifies to 10x - 1

Area is a little bit tougher. the area of a triangle is defined as 0.5 base x height.

both are given, so we simply plug in: 0.5 x (5x-3) x 8 = 4(5x-3) = 20x - 12

And thats it!

the Perimeter is 10x - 1 units.

the Area is 20x - 12 square units.

Answer:

The mean weight of six boys as per the incorrect data is 59.1667 kilograms.

The actual mean weight of the boy's group is 58.5 kilograms.

To find the mean weight of the six boys as per the incorrect data, we add up all the weights and divide by 6:

(55 + 58 + 57 + 60 + 59 + 65)/6 = 354/6 = 59.1667 kilograms

To find the actual mean weight of the boy's group, we add up the weights of the first five boys and the corrected weight of the sixth boy, and divide by 6:

(55 + 58 + 57 + 60 + 59 + 56)/6 = 345/6 = 58.5 kilograms

Step-by-step explanation:

Answer: red hot= $4

gummies=$9

Step-by-step explanation:

3r+1g=21

-(3r+3g=39)

3r+1g=21

-3r-3g=-39

-2g=-18

g=9

3r+1(9)=21

3r=12

r=4

Answer:

The correct answer is the option 3

The best measure of variability for the data and the player which was more consistent include the following: B. Player B is the most consistent, with a range of 9.

In Mathematics and Statistics, interquartile range (IQR) of a data set and it is typically calculated as the difference between the first quartile (Q₁) and third quartile (Q₃):

Interquartile range (IQR) of Player A = Q₃ - Q₁

Interquartile range (IQR) of Player A = 4 - 1.5

Interquartile range (IQR) of Player A = 2.5.

Range of Player A = Highest number - Lowest number

Range of Player A = 8 - 1

Range of Player A = 7

Interquartile range (IQR) of Player B = Q₃ - Q₁

Interquartile range (IQR) of Player B = 5 - 1.5

Interquartile range (IQR) of Player B = 4.5.

Range of Player B = Highest number - Lowest number

Range of Player B = 10 - 1

Range of Player B = 9

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Part A - The pH of the buffer solution is 6.37.

Net ionic equation is [tex]H_2CO_3[/tex] + [tex]OH^-[/tex] → [tex]HCO^{3-}[/tex] + [tex]H_2O[/tex]

Part B - The pH of the buffer solution is 6.38.

Net ionic equation is [tex]H_2CO_3[/tex] + [tex]I^-[/tex] → [tex]HCO^{3-}[/tex] + [tex]H_3O^+[/tex]

Part A: To find the pH of the buffer solution, we first need to calculate the pKa of the weak acid. The pKa is -log(Ka1), so pKa1 = -log(4.20 x [tex]10^{-7}[/tex]) = 6.38.

Next, we can use the Henderson-Hasselbalch equation to find the pH: pH = pKa1 + log([[tex]A^-[/tex]]/[HA]).

Plugging in the values for the buffer solution, we get pH = 6.38 + log(0.304/0.304) = 6.38. Therefore, the pH of the buffer solution is 6.38.

The net ionic equation for the reaction when 0.088 mol KOH is added to 1.00 L of the buffer solution is:

[tex]H^+[/tex] + [tex]OH^-[/tex] → [tex]H_2O[/tex]

Part B: Similar to Part A, we first need to calculate the pKa of the weak acid. pKa1 = -log(4.20 x [tex]10^{-7}[/tex]) = 6.38.

Then, we can use the Henderson-Hasselbalch equation to find the pH: pH = pKa1 + log([A-]/[HA]).

Plugging in the values for the buffer solution, we get pH = 6.38 + log(0.311/0.311) = 6.38. Therefore, the pH of the buffer solution is 6.38.

The net ionic equation for the reaction when 0.089 mol HI is added to 1.00 L of the buffer solution is:

[tex]H_3O^+[/tex] + [tex]I^-[/tex] → HI + H2O

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Answer:

Part A:

Sunday newspaper: Start at $36 and add $4 for each paper delivered.

Saturday newspaper: Start at $12 and add $2.50 for each paper delivered.

Part B:

To calculate Stevic's total earnings after delivering 15 papers on each day:

Earnings from Sunday papers = $36 + ($4 x 15) = $96

Earnings from Saturday papers = $12 + ($2.50 x 15) = $49.50

Total earnings = Earnings from Sunday papers + Earnings from Saturday papers

Total earnings = $96 + $49.50

Total earnings = $145.50

Therefore, Stevic's total earnings after delivering 15 papers on each day is $145.50.

Step-by-step explanation:

The number of textbooks of each type sold is found by solving the system of equations and got as,

Number of chemistry textbooks = 210

Number of history textbooks = 158

Given that,

A textbook store sold a combined total of 368 chemistry and history textbooks in a week.

let c be the number of chemistry textbooks sold and h be the number of history textbooks sold.

c + h = 368

The number of history textbooks sold was 52 less than the number of chemistry textbooks sold.

h = c - 52

Substituting the second equation in first,

c + (c - 52) = 368

2c = 420

c = 210

h = 210 - 52 = 158

Hence the number of each textbooks is 210 and 158.

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Option C, "A study shows that the average student leaves a four-year college with a student loan debt of more than $30,000", involves descriptive statistics.

The area of statistics known as descriptive statistics deals with the gathering, organizing, organizing, analyzing, interpreting, and presenting of data. It summarizes and describes the main features of a dataset, including measures of central tendency (such as mean, median, and mode) and measures of variability (such as range, standard deviation, and variance). Option C presents a descriptive statistic (the average student loan debt) that summarizes a larger dataset, making it the correct answer.

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Answer:

1414 tickets, in explanation

Hope this helps!

Step-by-step explanation:

1 ticket = $9.50

? tickets = $13,433

13,433 ÷ 9.50 = 1414

9.50 × 1414 = 13,433

1 ticket × 1414 = ? tickets

? tickets = 1414 tickets

Using the considered ordered pairs:

The domain is {1,3,5}

The range is {2,4,5,6}

It is not a function

We have,

To determine the domain and range of a function, we need to know the set of possible input values (domain) and the set of possible output values (range).

To determine if a relation is a function, we need to check if every input has a unique output.

In other words, if there are no two distinct ordered pairs with the same first element.

This means,

A function can be one-to-one or onto.

For example,

Let's consider the relation given by the set of ordered pairs:

{(1,2), (3,4), (1,5), (5,6)}

To determine if this is a function, we first need to check if there are any two distinct ordered pairs with the same first element.

In this case, we see that both (1,2) and (1,5) have a first element of 1, so this relation is not a function.

The domain of this relation is the set of all first elements of the ordered pairs, which is {1,3,5}.

The range of this relation is the set of all second elements of the ordered pairs, which is {2,4,5,6}.

Thus,

Domain: {1,3,5}

Range: {2,4,5,6}

Not a function

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The partial sum for the power series which represents the function 1/(1-x³) consisting of the first 5 nonzero terms is: 1 + x³ + x⁶ + x⁹ + x¹² and the radius of convergence is 1.

The formula for the partial sum of a power series is given by:

Sₙ(x) = a₀ + a₁x + a₂x² + ... + aₙxⁿ

where a₀, a₁, a₂, ..., aₙ are the coefficients of the power series.

In this case, we can use the formula for the geometric series to find the coefficients:

1/(1-x³) = 1 + x³ + x⁶ + x⁹ + x¹² + ...

a₀ = 1

a₁ = 1

a₂ = 1

a₃ = 0

a₄ = 0

and so on.

Therefore, the first 5 nonzero terms of the power series are 1, x³, x⁶, x⁹, and x¹².

The radius of convergence for this power series can be found using the ratio test:

lim┬(n → ∞)⁡|aₙ₊₁/aₙ| = lim┬(n → ∞)⁡|x³/(1-x³)| = 1

Since the limit equals 1, the radius of convergence is 1.

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The number of hours that it will take Lindsey to rake the lawn alone will also be 3 hours just like Camilla.

The total number of hours it takes two people to rake the lawn = 2 hours.

The more people the less number of hours it will take to take the lawn.

That is;

If 2 people = 2 hours

Camilla = 3 hours

1 person (Lindsey) = 3 hours.

Therefore, for either Lindsey or Camilla, they will rake separately for 2 hours when working alone.

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The condition that is one of the assumptions of a valid multiple linear regression model is: the relationship between the dependent and independent variables is linear.

Condition that is one of the assumptions of a valid multiple linear regression model is that the relationship between the dependent and independent variables is linear. This means that the change in the dependent variable is proportional to the change in each independent variable, and there is no curved or nonlinear relationship between them. The assumption of linear independence of the independent variables is also important, meaning that they are not highly correlated with each other.

The assumption of constant residuals means that the errors in the model are consistent across all values of the independent variables. The assumption of a quadratic relationship between the dependent and independent variables is not appropriate for a multiple linear regression model.

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The answer of the compound inequality 3 < 3x + 9 < 24 is -2 < x < 5.

To solve the compound inequality 3 < 3x + 9 < 24, we need to isolate the variable x.

First, we will subtract 9 from all parts of the inequality:

3 - 9 < 3x + 9 - 9 < 24 - 9

-6 < 3x < 15

Next, we will divide all parts of the inequality by 3 (remembering to flip the direction of the inequality if we divide by a negative number):

-6/3 < 3x/3 < 15/3

-2 < x < 5

Therefore, the solution to the compound inequality 3 < 3x + 9 < 24 is -2 < x < 5.

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Since we know that the heptagon is convex, all of its interior angles are less than 180 degrees. Let's call the smallest angle x degrees.

According to the problem, the other six angles are each 1 degree larger than the angle just smaller than it. This means the second angle is x+1, the third angle is x+2, and so on, until we get to the seventh angle which is x+6.

We know that the sum of the interior angles of a heptagon is (7-2) * 180 = 900 degrees. So we can set up an equation:

x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) = 900

Simplifying this equation, we get:

7x + 21 = 900

Subtracting 21 from both sides:

7x = 879

Dividing both sides by 7:

x = 125.57

So the smallest angle is approximately 125.57 degrees.

To find the fourth largest angle, we need to find the value of x+3.

x+3 = 125.57 + 3 = 128.57

So the fourth largest angle is approximately 128.57 degrees.

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In an English literature course, the professor asks students to read three books by selecting one memoire, one book of poetry, and one novel to read. The students can select these books from a list of 8 memoires, 9 poetry books, and 4 novels.

To determine how many different ways a student can select their reading assignment of three books, we will use the multiplication principle.

1. Choose one memoire:  There are 8 memoires to choose from, so there are 8 ways to make this choice.
2. Choose one book of poetry:  There are 9 poetry books to choose from, so there are 9 ways to make this choice.
3. Choose one novel:  There are 4 novels to choose from, so there are 4 ways to make this choice.

Now, multiply the number of choices for each step together to find the total number of ways to select the reading assignment:

8 (memoires) x 9 (poetry books) x 4 (novels) = 288 different ways to select the reading assignment of three books.

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Given a ray passing through a line at an angle of 29 degrees, the angle opposite to it (angle R) can be found by subtracting 29 degrees from 180 degrees. Therefore, the value of angle R is 151 degrees.

We are given that a ray passes through a line, making an angle of 29 degrees with the line. Let us represent this situation as follows

The angle R represents the angle opposite to the angle of 29 degrees. Since the ray and the line form a straight line, their angles add up to 180 degrees. Therefore, we can write

angle R + 29 degrees = 180 degrees

To solve for angle R, we can subtract 29 degrees from both sides of the equation

angle R = 180 degrees - 29 degrees

Simplifying the expression, we get

angle R = 151 degrees

Therefore, the value of angle R is 151 degrees.

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The value of the iterated integral  ∫85∫√x12ye−xdydx is

-[tex]4e^(-5) + 7e^(-8)[/tex] where the inner integral is first integrated with respect to y.

We are inquiring to assess the iterated integral:

[tex]∫85∫√x12ye−xdydx[/tex]

We are able to coordinate the internal integral, to begin with regard to y:

[tex]∫√x12ye−xdy = (-1/2)e^(-x) y√x1/2 | from y = to y = √x^1/2[/tex]

[tex]= (-1/2)e^(-x) (√x^1/2)^2 - (-1/2)e^(-x) (0)[/tex]

[tex]= (-1/2)x e^(-x)[/tex]

Substituting this into the first necessity, we get:

[tex]∫85∫√x12ye−xdydx = ∫85(-1/2)x e^(-x)dx[/tex]

To assess this necessarily, we utilize integration by parts with u = x and [tex]dv = e^(-x) dx, so that du/dx = 1 and v = -e^(-x):[/tex]

[tex]∫85(-1/2)x e^(-x)dx = (-1/2)xe^(-x) + ∫85(1/2)e^(-x)dx[/tex]

[tex]= (-1/2)xe^(-x) - (1/2)e^(-x) | from x = 8 to x = 5[/tex]

[tex]= (-1/2)(8e^(-8) - 5e^(-5)) - (1/2)(e^(-8) - e^(-5))[/tex]

[tex]= -4e^(-5) + 7e^(-8)[/tex]

therefore, the value of the iterated integral is [tex]-4e^(-5) + 7e^(-8).[/tex]

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The degree of the polynomial 2 + 2w⁶ + 15y²w + 64u² - uy⁶ is found to be 7 as the term with highest power is 7.

A degree of the polynomial is the highest power to which any of its term is expressed as. For finding the degree we have to find the term with the highest degree in the polynomial. The given polynomial is,

2 + 2w⁶ + 15y²w + 64u² - uy⁶,

The term with the highest degree is uy⁶, which has a degree of 7 (the sum of the exponents of u and y ). Therefore, the degree of the polynomial is 7.

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Complete question - Give the degree of the polynomial. 2 + 2w⁶ + 15y²w + 64u² - uy⁶.

From the graph, the complex number with the greatest modulus is z1

From the question, we have the following parameters that can be used in our computation:

The complex numbers z1, z2, z3 and z4

The general rule of modulus of complex numbers is that

The complex number that has the greatest modulus is the complex number that is at the farthest distance from the origin

Using the above as a guide, we have the following:

From the graph, the complex number that is at the farthest distance from the origin is the complex number z1

Hence, the complex number with the greatest modulus is z1

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