11. Dave is building a doghouse that is a scale model of his shed in the farm yard, One of
the windows on the shed is 117 cm wide and 130 cm high. On the doghouse the window
is 9 cm wide,
a) What is the scale factor Dave is using to build the doghouse?
a.
2 mar
b) How high is the window in the scale model?

Answer:

C. 28

Step-by-step explanation:

40% + 30% = 70%

70% of 40 students

0.70 x 40 = 28

Let me know if this helps!

Around 4 times, maybe 3

Step-by-step explanation:

there's 2 red 3s and there's 52 cards in a deck. so u have a 2/52 chance to get a red 3 so its about a 4/104 chance to get a red 3 if u draw 1 card 100 times

Same I also need help ;-;

In a Young's double-slit experiment, the center of a bright fringe occurs wherever waves from the slits differ in phase by a multiple of λ/2.

In Young's double-slit experiment, a coherent light source, such as a laser, is passed through two narrow slits, creating two sources of waves that interfere with each other. When the waves from the two slits meet, they create an interference pattern of bright and dark fringes on a screen placed behind the slits.

The bright fringes occur when the waves from the two slits reinforce each other constructively, resulting in a bright spot. The central bright fringe is the brightest and occurs at the center of the pattern. This is because at the center, the waves from both slits travel the same distance to the screen.

For the waves to interfere constructively at the center of the pattern, they must be in phase. In other words, the waves from the two slits must have a phase difference of an integer multiple of the wavelength (λ) of the light. Mathematically, this phase difference can be expressed as an integer multiple of λ/2.

Therefore, the correct answer is D) λ/2, where λ represents the wavelength of the light used in the experiment.

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Answer:

10 days

Step-by-step explanation:

Mogaka and Ondiso working together can do a piece of job in 6 days. Mogaka working alone takes 5 days longer than Ondiso. How many days does it take Ondiso to do the work alone?

Let us represent :

The number of days

Mogaka worked alone = x

Ondiso worked alone = y

Total days worked together = T

Mogaka and Ondiso working together can do a piece of job in 6 days.

Hence,

1/x + 1/y = 1/T

Mogaka working alone takes 5 days longer than Ondiso.

x = y + 5

Therefore:

1/y + 5 + 1/y = 1/6

Multiply all through by (y+5)(y)

= y + y + 5 = (y+5)(y)/6

= 2y + 5/1 = (y+5)(y)/6

Cross Multiply

6( 2y + 5) = (y+5)(y)

12y + 30 = y² + 5y

= y² + 5y - 12y - 30 = 0

= y² - 7y - 30 = 0

Factorise

= y² +3y - 10y - 30 = 0

y(y + 3) - 10(y + 3) = 0

(y + 3)(y -10)= 0

y + 3 = 0, y = -3

y - 10 = 0

y = 10 days

Note that

The number of days Ondiso worked alone = y

Hence, it takes 10 days for Ondiso to work alone

9514 1404 393

Answer:

  x = 2

Step-by-step explanation:

To find g(x), we can start with the inverse of f(x).

  f(y) = x . . . . . . . solve this to find f^-1(x)

  3y -2 = x

  3y = x +2 . . . . add 2

  y = (x +2)/3 = f^-1(x) . . . . divide by 3

__

Now, we can find g(x):

  f^-1(f(g(x)) = g(x)

  f^-1(6x -2) = ((6x -2) +2)/3 = g(x)

  6x/3 = g(x) = 2x

__

Now, we want g(f(x)) = 8

  g(3x -2) = 8

  2(3x -2) = 8

  6x -4 = 8

  6x = 12

  x = 2 . . . . makes g(f(x)) = 8

The work shown is not correct. The correct result of (11 + 2i) - (3 - 10i) is 8 + 12i.

To evaluate the given expression, we need to subtract the real parts and the imaginary parts separately.

Real part:

11 - 3 = 8

Imaginary part:

2i - (-10i) = 2i + 10i = 12i

Combining the real and imaginary parts, we get the correct result:

8 + 12i

So, the correct answer is 8 + 12i, not 8 - 8i as shown in the work.

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The average salary for assistant professors is $80,776.0. The 95% confidence interval for B1 cannot be determined. The interpretation of ß1, ß2, and average salaries of associate and full professors cannot be provided due to missing coefficients.

(a) The average salary for assistant professors can be determined by the coefficient B1, which is reported as $80,776.0 in the regression model.

(b) To calculate a 95% confidence interval for B1, we need the standard error associated with B1. Unfortunately, the standard error is not provided in the given information, so we cannot determine the confidence interval or assess its statistical significance.

(c) The coefficient B1 represents the average difference in salary between associate professors and assistant professors. However, since the given information does not provide the coefficient for "Asst Prof," we cannot estimate the average salary of associate professors based on the given model.

(d) The coefficient B2 represents the average difference in salary between full professors and assistant professors. However, since the given information does not provide the coefficient for "Prof," we cannot estimate the average salary of full professors based on the given model.

(e) We cannot estimate the model yi = Bo + B11(AssocProf) + B21(Prof) + B31(Asst Prof) + u because the variable "Asst Prof" is collinear with the other two binary indicator variables (AssocProf and Prof). Collinearity occurs when predictor variables are highly correlated, leading to unreliable estimates of their coefficients.

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Answer:

The percentage discount is 37.5

Answer:

The significance of the sample is that at it is statistically significant at 0.01 level ( i.e. of observing the sample < 0.01 )

Step-by-step explanation:

The sample observation which is 121 mothers that take vitamin supplements have male babies with mean weight of : 3.67 kg

while the National average birth weight = 3.39 kg

probability of observing such a sample = 0.0015

Hence the significance of the sample is that at it is statistically significant at 0.01 level ( i.e. of observing the sample < 0.01 )

Answer:

3 years approx

Step-by-step explanation:

Given data

Principal=£100

Amount= £109.27

Rate= 3%

The expression for the compound interest is

A=P(1+r)^t

Make t subject of formula we have

t= ln(A/P) / r

t= ln(109.27/100)/ 3

t= ln(1.0927)/0.03

t= 0.088/0.03

t= 2.93

Hence the time is 3 years approx

Answer:

113.04cm

............

The calculated z-score of -47.76 falls outside the critical range of -1.96 to 1.96 indicating a statistically significant difference.

In order to determine whether this difference is significant, we will perform a one-sample z-test, as we know the population standard deviation.

The null hypothesis (H0) is that there is no difference between the national per capita monthly fuel oil cost and the average cost in Southeastern cities.

The alternative hypothesis (H1) is that there is a difference.

Sample mean (x): $78

Population mean (μ): $110

Sample standard deviation as an estimate: $4

Sample size (n): 36

z = (x - μ) / (σ/√n)

Substituting numbers:

z = ($78 - $110) / ($4/√36)

z = -32 / (4/6)

z = -48

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Answer:

D- A translation 2 units down followed by a 270-degree counterclockwise rotation about the origin.

Step-by-step explanation:

I'm sorry that it's late, I still posted it tho so you can give other person branliest.

Hope this helps for other readers :)

Explanation:

If you focus on one point, I'm doing R

if you first translate it down you will go from (3,4) to (3,2)

Then we have R' at (2, -3) which means we need (y, -x)

This can be found with 90 degree clockwise OR 270 degree counterclockwise.

Answer:

65,760

Step-by-step explanation:

Answer:

x = 7.2

Step-by-step explanation:

I am pretty sure this is right, but I apologize if I am wrong.

ii)  f[2,4,5] = -6.

iii)f[24, 25, 26) = -2.

iv) f[27] = 56.

v) f[24, 25, 26] = 56.

vi) The values of a, b, c, d, e, f, and g for the Hermite interpolation polynomial are: a = 0, b = -8, c = 40, d = 72, e = 40, f = -8, g = 72

The Hermite divided difference table for the given data:

i x f(x) f'(x) f"(x)

0 -1 2 -8 56

1 0 1 0 0

2 1 2 8 56

(ii)To find the value of f[2,4,5], we need to determine the divided difference for the corresponding values of x.

Using the divided difference formula, we have:

f[2,4,5] = (f[4,5] - f[2,4]) / (4-2) = (f[5] - f[4]) / (5-4)

f[2,4,5] = (2 - 8) / (5-4) = -6

Therefore, f[2,4,5] = -6.

(iii) To find the value of f[24, 25, 26), we need to determine the divided difference for the corresponding values of x.

f[24, 25, 26) = (f[25, 26) - f[24, 25]) / (25-24)

= (f[26) - f[25]) / (26-25)

f[24, 25, 26) = (0 - 2) / (26-25) = -2

Therefore, f[24, 25, 26) = -2.

(iv) To find the value of f[27], we can directly extract it from the divided difference table.

From the table, we can see that f[27] = 56.

(v) To find the value of f[24, 25, 26], we can directly extract it from the divided difference table.

From the table, we can see that f[24, 25, 26] = 56.

(vi) Given that the Hermite interpolation polynomial is:

Pₙ(X) = 2 - a (x+1) + b (x+1)² + c (x+1)³+ d x (x+1)³+ e x² (x+1)³ + f x³ (x+1)³ + g x³ (x+1)²

Comparing the Hermite interpolation polynomial:

Pₙ(X) = 2 - a (x+1) + b (x+1)² + c (x+1)³ + d x (x+1)³ + e x² (x+1)³ + f x³ (x+1)³ + g x³ (x+1)²

a + b = f[0,1]

a - 2b + c = f[0,1,2]

b + 3c - 3d = f'[0,1,2]

b - 4c + 6d - e = f''[0,1,2]

c - 5d + 10e - f = f[1,2]

d - 6e + 15f - g = f'[1,2]

e - 7f + 21g = f''[1,2]

we substitute the corresponding values from the divided difference table:

f[0] = 2, f[0,1] = -8, f[0,1,2] = 56, f'[0,1,2] = 0, f''[0,1,2] = 0, f[1,2] = 8, f'[1,2] = 56

f''[1,2] = 56

2 - a = 2 -> a = 0

-a + b = -8 -> b = -8

a - 2b + c = 56 -> c = 40

-b + 3c - 3d = 0 -> d = 72

b - 4c + 6d - e = 0 -> e = 40

c - 5d + 10e - f = 8 -> f = -8

d - 6e + 15f - g = 56 -> g = 72

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Answer:

752.884 cubic feet

Step-by-step explanation:

Brainliest?

Applying the formula, it is found that the flow rate is of 753 cubic feet per second.

The flow rate is modeled by:

[tex]Q = 3.367lh^{\frac{3}{2}}[/tex]

[tex]Q = 3.367l\sqrt{h^3}[/tex]

In which the parameters are:

In this problem:

Then:

[tex]Q = 3.367l\sqrt{h^3}[/tex]

[tex]Q = 3.367(20)\sqrt{5^3}[/tex]

[tex]Q = 753[/etx]

The rate is of 753 cubic feet per second.

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Answer:

-3.2

Step-by-step explanation:

The vector of length four that points in the same direction as u = (1, 2, 2) is v = (4/3, 8/3, 8/3).

To determine the equation of the sphere with a radius of five units, we need the coordinates of its center.

From the given information, we know that the sphere intersects the zy-plane along the circle with the equation [tex](x - 1)^2 + (y + 4)^2 = 9[/tex].

The center of this circle can be found by setting x = 1 and y = -4 in the equation since the circle intersects the zy-plane.

Thus, the center of the sphere is (1, -4, 0).

Now, we can write the equation of the sphere using the center and the radius.

The equation of a sphere in 3D space is given by:

[tex](x - h)^2 + (y - k)^2 + (z - l)^2 = r^ 2[/tex]

where (h, k, l) represents the center coordinates and r represents the radius.

Substituting the values, we have:

[tex](x - 1)^2 + (y + 4)^2 + (z - 0)^2 = 5^2[/tex]

Simplifying the equation, we get:

[tex](x - 1)^2 + (y + 4)^2 + z^2 = 25[/tex]

Therefore, the equation of the sphere with a radius of five units and a center at a positive number is:

[tex](x - 1)^2 + (y + 4)^2 + z^2 = 25[/tex]

Now, let's determine a vector of length four that points in the same direction as u = (1, 2, 2).

To find a vector with the same direction, we can normalize vector u to have a length of 1 and then scale it by a factor of 4.

The normalization of a vector u is given by:

[tex]u_{normalized}[/tex] = u / ||u||

where ||u|| represents the magnitude or length of vector u.

Calculating the magnitude of vector u:

||u|| = [tex]\sqrt{(1^2 + 2^2 + 2^2)} = \sqrt{(1 + 4 + 4)} = \sqrt{9} = 3[/tex]

Now, we can normalize vector u:

[tex]u_{normalized}[/tex] = (1/3, 2/3, 2/3)

To get a vector of length four pointing in the same direction as u, we can scale the normalized vector by 4:

vector v = 4 *[tex]u_{normalized}[/tex]= (4/3, 8/3, 8/3)

Therefore, the vector of length four that points in the same direction as u = (1, 2, 2) is v = (4/3, 8/3, 8/3).

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Answer:

9

Step-by-step explanation:

the prism has 9 edges because you said "a prism with 9 edges"

Hope this helps!!! have a great day!!

Step-by-step explanation:

b=books

3b[tex]\leq[/tex]23

[tex]b\leq 23/3[/tex]

[tex]b\leq 7.67[/tex]

L=lesson

25L[tex]\leq[/tex]150

L[tex]\leq[/tex]6

Hope that helps :)

Answer: a)7, B)8

Step-by-step explanation:

Problem a) If Kai wants to buy just one poster that costs 2 dollars, he has 25-2=23 dollars left. If each book is 3 dollars, then he  can buy 23/3 books. The inequality that results is that if b=#ofbooks, then b< or = (25-2)/3, because you cant buy a book for 2 and a half dollars, hence the less than. 23/3 is 7 r2. We get that b< or = to 7 2/3. However, he can't buy 2/3 of a book, so 7. The final inequality we get is that 2+3b<=25

Problem 2) She spends 50 initially, so 200-50=150 dollars left. Thus the number of lessons, or n, cover the rest of the money. Using the same thory as above, the final inequality is 50+25n<=200. n=8.

The probability of a student passing the module is less than 0.75. Therefore, the lecturer should reconsider his method of teaching.

Question analysis A statistics module has been running for many years, and the module is given to n students each year. It has been discovered that in each year, the number of students passing the exam is distributed Bi(n, 0.75). In the current year, 150 students took the module for the first time, and 105 students passed the exam.

Using the 0.05 level of significance, we will conduct a hypothesis test to decide if the module's pass rate this year is less than 0.75.AssumptionsIf the number of trials is huge, the distribution of successes will be nearly normal. The number of trials n is greater than 30 in this situation. The probability of success in each trial is the same, namely p = 0.75. This condition is also satisfied. Therefore, we may use a normal distribution to approximate the binomial distribution.

What is the conclusion?

Null hypothesis: H₀: P = 0.75

Alternative hypothesis: H₁: P < 0.75The level of significance is 0.05, which implies that the rejection area will be in the left tail because the alternative hypothesis is one-tailed. Since the distribution of successes is approximately normal with a mean of np and a variance of np(1−p), we may find the p-value using this formula:
[The probability that X ≤ 105]
= [Z = (X − µ)/σ]
= [Z = (105 − (150 × 0.75))/sqrt(150 × 0.75 × (1 − 0.75))]
= [Z = (105 − 112.5)/3.2958]
= -2.2782
The p-value is [P(Z < -2.2782)] = 0.011. Because the p-value is less than 0.05, we reject the null hypothesis and accept the alternative hypothesis.

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Given : A statistics module has been running for many years and, in the past, it has been found that each year the number of students passing the exam has distribution Bi(n. 0.75), where n are the number of students taking the module that year. A lecturer is teaching the module for the first time and 105 out of 150 students pass the exam. The conclusion is that we fail to reject the null hypothesis.

The null and alternative hypotheses are given as follows:

Null hypothesis: The probability of a student passing the module is 0.75.

Alternative hypothesis: The probability of a student passing the module is less than 0.75.

We need to perform a hypothesis test at the 0.05-significance level.

The given probability distribution Bi(n,0.75) can be approximated to the normal distribution N(np,npq) under the following conditions:

The sample size n is large enough.

np≥5 and nq≥5, where q=1-p.

Here, n=150 and

p = 0.75

q = 1−p

= 1−0.75

= 0.25

Since np and nq are both greater than 5, the distribution Bi(150,0.75) can be approximated by the normal distribution N(150×0.75,150×0.75×0.25) = N(112.5,28.125).

Let X be the number of students that passed the module.

Under the null hypothesis, X follows the binomial distribution Bi(150,0.75).

Let μ be the mean of X under the null hypothesis.

μ = np

= 150×0.75

= 112.5

Since the alternative hypothesis is the probability of passing the module is less than 0.75, we need to perform a one-tailed test in the left tail at the 0.05-significance level.

The test statistic is given by,

Z=(X−μ)/σ

Z=(105−112.5)/√28.125/150

Z ≈ −1.5

This is a left-tailed test, so the critical value for a 0.05-significance level is z=−1.645.

Since the test statistic z=-1.5 > critical value z=-1.645, we fail to reject the null hypothesis.

Hence, there is not enough statistical evidence to conclude that the probability of a student passing the module is less than 0.75.

Therefore, the conclusion is that we fail to reject the null hypothesis.

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The probability distribution of the random variable Y = X²+1 is;

P(Y=1) = 81/256,

P(Y=2) = 243/1024,

P(Y=5) = 27/256,

P(Y=10) = 1/64,

P(Y=17) = 1/256

The probability distribution of the random variable Y = X² + 1 can be obtained as follows;

Explanation:

We know that the binomial probability distribution function is given by;

P(X=k) = (nCk)pk(1−p)n−k

Here, X is a binomial random variable with parameters;

n = 4 and p = 1/4

For X = 0;

P(X=0) = (4C0)(1/4)0(3/4)4−0

=81/256

For X = 1;

P(X=1) = (4C1)(1/4)1(3/4)4−1

=243/1024For X = 2;

P(X=2) = (4C2)(1/4)2(3/4)4−2

=27/256

For X = 3;

P(X=3) = (4C3)(1/4)3(3/4)4−3

=1/64

For X = 4;

P(X=4) = (4C4)(1/4)4(3/4)4−4

=1/256

Now we find the distribution function of Y;

P(Y=y) = P(X²+1=y)

Using X=0;

Y = X²+1

= 0+1

= 1;

P(Y=1) = P(X²+1=1)

= P(X=0)

= 81/256

Using X=1;

Y = X²+1

= 1+1

= 2;

P(Y=2) = P(X²+1=2)

= P(X=0)

= 243/1024

Using X=2;

Y = X²+1

= 4+1

= 5;

P(Y=5) = P(X²+1=5)

= P(X=2)

= 27/256

Using X=3;

Y = X²+1

= 9+1

= 10;

P(Y=10) = P(X²+1=10)

= P(X=3)

= 1/64

Using X=4;

Y = X²+1

= 16+1

= 17;

P(Y=17) = P(X²+1=17)

= P(X=4)

= 1/256

Therefore, the probability distribution of the random variable

Y = X²+1 is;

P(Y=1) = 81/256,

P(Y=2) = 243/1024,

P(Y=5) = 27/256,

P(Y=10) = 1/64,

P(Y=17) = 1/256

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The indefinite integral of √[tex]x^{11[/tex] × sin(3[tex]x^{(13/2)[/tex]) dx is -2/39 × cos(3[tex]x^{(13/2)[/tex]) + C, where C represents the constant of integration.

To evaluate the indefinite integral of √[tex]x^{11[/tex] × sin(3[tex]x^{(13/2)[/tex]) dx, we can use substitution. Let's substitute u = [tex]x^{(13/2)[/tex]:

Step 1: Find du/dx:

Differentiating both sides with respect to x:

du/dx = (13/2) × [tex]x^{(11/2)[/tex]

Step 2: Solve for dx:

Rearrange the equation to solve for dx:

dx = (2/13) × du / [tex]x^{(11/2)[/tex]

Step 3: Substitute the values in the integral:

∫ √[tex]x^{11[/tex] × sin(3[tex]x^{(13/2)[/tex]) dx = ∫ √[tex]x^{11[/tex] × sin(3u) × (2/13) × du / [tex]x^{(11/2)[/tex]

Step 4: Simplify the integral:

∫ √[tex]x^{11[/tex] × sin(3[tex]x^{(13/2)[/tex]) dx = (2/13) × ∫ sin(3u) du

Step 5: Integrate with respect to u:

∫ sin(3u) du = - (1/3) × cos(3u) + C,

where C is the constant of integration.

Step 6: Substitute back the value of u:

∫ √[tex]x^{11[/tex] × sin(3[tex]x^{(13/2)[/tex]) dx = (2/13) × (-1/3) × cos(3u) + C

= -2/39 × cos(3[tex]x^{(13/2)[/tex]) + C.

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Answer:

Our radius is already given, 8.

8 x 3.14 x 2 = 50.24cm <-------- perimeter/circumference

8^2 x 3.14

64 x 3.14 = 200.96cm2 <------- area

Formula of a circumference:

2πr

2 x pie x radius (pie can be used as 22/7 or 3/4 most likely 3.14)

Formula of an area of a Circle:

πr^2

pie x radius squared (pie can be used as 22/7 or 3.14)

The density of the soda is 3.6 g/cm³

Density is the measurement of how tightly a material is packed together. It is defined as the mass per unit volume.

Given that, a cylindrical glass of soda has a mass of 700g. The glass itself has a mass of 80g, the glass has a radius of 4 cm and a height of 8 cm,

We are asked to find the density of the soda,

Density = mass / volume

Volume = 2π×radius×height

Therefore,

Density = 700/2π×4×8 [we will not add the mass of glass because we need to find the density of soda only]

Density = 700/194.68

= 3.59

= 3.6

Hence, the density of the soda is 3.6 g/cm³

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Sample size is a key component of any scientific experiment or study. The sample size is a vital factor to consider when conducting a research study as it allows you to identify how many participants should be involved in the study, and it can also assist you in interpreting and analyzing the results.

When collecting data for a study, the sample size is an important consideration. Some reasons to consider sample size include:

Representativeness: A larger sample size allows for a more representative sample of the population under study. It helps to reduce sampling bias and increase the likelihood that the sample accurately represents the characteristics of the larger population.

Precision and Accuracy: A larger sample size generally leads to more precise and accurate estimates. With a larger sample, statistical measures such as means, proportions, or regression coefficients tend to have smaller margins of error, providing more reliable and precise results.

Statistical Power: Sample size affects the statistical power of a study, which is the ability to detect true effects or relationships. A larger sample size increases the power of statistical tests, allowing researchers to more confidently detect significant effects or relationships.

Generalizability: A larger sample size enhances the generalizability of study findings. With a larger sample, the results are more likely to be applicable to the broader population from which the sample was drawn.

Subgroup Analysis: A larger sample size allows for more robust subgroup analyses. It enables researchers to examine and analyze smaller subgroups within the sample, potentially identifying important differences or patterns that may be overlooked with smaller sample sizes.

Therefore, the reasons to consider sample size include representativeness, precision and accuracy, statistical power, generalizability, and the ability to conduct subgroup analysis.

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In the system of linear equations -y = 2 kx - y = k,

(a) The augmented matrix can be reduced to row-echelon form by performing row operations.

(b) The system has (a) no solutions: None, (b) exactly one solution: x = 2k, y = k (in terms of k).

(c) infinitely many solutions: x = t, y = 0 (in terms of t) when k = 0.

(a) To reduce the augmented matrix for the system to row-echelon form, we can perform row operations.

Starting with the augmented matrix:

[ -1 | 2k ]

[ -1 | k ]

We can perform the following row operations to obtain row-echelon form:

Replace R2 with R2 + R1:

[ -1 | 2k ]

[ -2 | 3k ]

Now, the augmented matrix is in row-echelon form.

(b)To find the values of k for different cases, we can observe the row-echelon form:

[ -1 | 2k ]

[ 0 | k ]

From the row-echelon form, we can conclude the following:

(i) If k ≠ 0, then the system has a unique solution. The solution is x = 2k and y = k.

(ii) If k = 0, then the system has infinitely many solutions. The solution can be expressed as x = t and y = 0, where t is a parameter.

(iii) There are no values of k for which the system has no solutions.

Therefore, the system has (a) no solutions: None, (b) exactly one solution: x = 2k, y = k (in terms of k), and (c) infinitely many solutions: x = t, y = 0 (in terms of t) when k = 0.

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