Given f(x) and g(x) = f(x) + k, use the graph to determine the value of k. Graph of f of x and g of x. f of x equals 1 over 3 x minus 2 and g of x equals 1 over 3 x plus 3. 2 3 4 5

Ans .: The minimum score required for admission to the university is 559.

To find the minimum score required for admission to the university, we need to find the score that corresponds to the top 30% of the distribution.

First, we need to find the z-score that corresponds to the top 30% of the distribution. We can use a standard normal distribution table or a calculator to find this value. The area to the left of the z-score corresponding to the top 30% is 1 - 0.30 = 0.70. Looking this up on a standard normal distribution table or using a calculator, we find that the z-score is approximately 0.5244.

Next, we can use the formula z = (x - mu) / sigma to find the corresponding score x. We know that mu (the mean) is 501 and sigma (the standard deviation) is 110. Plugging in these values and solving for x, we get:

0.5244 = (x - 501) / 110

Multiplying both sides by 110, we get:

57.68 = x - 501

Adding 501 to both sides, we get:

x = 558.68

Rounding this to the nearest whole number, we get:

x = 559

Therefore, the minimum score required for admission to the university is 559.

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True. After testing H0: p = 0.33 versus HA: p < 0.33 at α = 0.05, with a sample proportion of 0.20 and a sample size of n = 100, we do not reject H0.

True. After conducting a hypothesis test with the given parameters, if the p-value is greater than the significance level (α = 0.05), we do not reject the null hypothesis (H0: p = 0.33). This means that there is not enough evidence to support the alternative hypothesis (HA: p < 0.33) and we conclude that the proportion is not significantly less than 0.33 based on the sample data.

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Given the information provided, we can construct a table of values for the function f(x) that satisfies the given conditions.

Since f'(3) = 2, we know that the slope of the tangent line to the graph of f at x = 3 is 2. This suggests that f is increasing around x = 3. Additionally, since f"(3) < 0, we know that the concavity of f changes from upward to downward at x = 3.

Based on this information, we can create a possible table of values for f(x):

x f(x)

2.0 a

2.5 b

3.0 c

3.5 d

4.0 e

Here, a, b, c, d, and e represent the values of f(x) at the corresponding x-values. Since f is continuous on the closed interval [2, 4], the function must take on all values between f(2) and f(4). Therefore, we have flexibility in choosing the specific values of a, b, c, d, and e, as long as they satisfy the given conditions.

To reflect that f'(3) = 2 and f"(3) < 0, we can choose values such that f is increasing but with a decreasing rate of change. For example, we can set f(2) = 0, f(2.5) = 1, f(3) = 2, f(3.5) = 3, and f(4) = 4. This table of values satisfies the given conditions and demonstrates an increasing function with decreasing rate of change at x = 3.

x f(x)

2.0 0

2.5 1

3.0 2

3.5 3

4.0 4

Note that there can be infinitely many possible tables of values for f(x) that satisfy the given conditions, as long as the function is continuous on the closed interval [2, 4], twice differentiable on the open interval (2, 4), and the specified conditions for f'(3) = 2 and f"(3) < 0 are met.

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The hexadecimal instruction 2888006416 is associated with the MIPS32 command "d. sub $v0, $t9, $t8".

In MIPS32 assembly language, "sub" is a command used for subtraction. In this particular instruction, the command is subtracting the value stored in register $t8 from the value stored in register $t9 and storing the result in register $v0.

It's important to note that hexadecimal instructions are machine code instructions that are represented in hexadecimal format for ease of reading. They are not typically used by programmers directly. Instead, programmers write code in assembly language and then use an assembler to translate it into machine code.

In summary, the MIPS32 command associated with the hexadecimal instruction 2888006416 is "sub $v0, $t9, $t8", which subtracts the value stored in register $t8 from the value stored in register $t9 and stores the result in register $v0.

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The number of outcomes possible when a spinner has three equally sized sections labeled from 1 to 3 and the spinner is spun three times is 27

The total number of outcomes refers to the possible events that can occur if an event takes place. These are helpful in calculating probability. For example, when a coin is tossed, the outcomes possible are heads and tails.

In the given question, the possible outcomes when a spinner has three equally sized sections labeled from 1 to 3 and the spinner is spun three times are calculated by the number of outcomes raised to the power the number of times the event occurs.

Possible outcome possible in  event = 3

Number of times the event occurs =  3

Thus the number of outcomes = [tex]3^3[/tex] = 27

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The solution to the differential equation y'''+3y''-4y=e^(-2x) using the method of variation of parameters is y(x) = c1 + c2e^(-4x) + c3e^x + [(1/4)e^(2x) - (1/2)x^2 + C1x + C2]e^(-2x) where c1, c2, c3, C1, and C2 are constants.

To find the particular solution y_p(x) using the method of variation of parameters, we first need to find the complementary solution y_c(x) by solving the characteristic equation: r^3 + 3r^2 - 4r = 0. Factoring out an r gives us r(r+4)(r-1) = 0, so the roots are r=0, r=-4, and r=1. Therefore, the complementary solution is y_c(x) = c1 + c2e^(-4x) + c3e^x.

Next, we assume that the particular solution has the form y_p(x) = u1(x)e^(-2x). Taking the derivatives of this form, we get y_p'(x) = u1'(x)e^(-2x) - 2u1(x)e^(-2x) and y_p''(x) = u1''(x)e^(-2x) - 4u1'(x)e^(-2x) + 4u1(x)e^(-2x). Substituting these into the differential equation and simplifying, we get:

u1''(x) = e^(2x)

To solve this equation for u1(x), we integrate twice: u1(x) = (1/4)e^(2x) - (1/2)x^2 + C1x + C2, where C1 and C2 are constants of integration.

Therefore, the particular solution is y_p(x) = [(1/4)e^(2x) - (1/2)x^2 + C1x + C2]e^(-2x).

Combining the complementary and particular solutions gives the general solution: y(x) = y_c(x) + y_p(x) = c1 + c2e^(-4x) + c3e^x + [(1/4)e^(2x) - (1/2)x^2 + C1x + C2]e^(-2x).

Thus, the solution to the differential equation y'''+3y''-4y=e^(-2x) using the method of variation of parameters is y(x) = c1 + c2e^(-4x) + c3e^x + [(1/4)e^(2x) - (1/2)x^2 + C1x + C2]e^(-2x) where c1, c2, c3, C1, and C2 are constants.

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Method of sampling applies to populations that are divided into natural subsets and allocates the appropriate proportion of samples to each subset: Stratified sampling. The correct answer is D.

Stratified sampling is a technique used in research where a population is divided into natural subsets or strata based on specific characteristics or attributes. Each stratum is then proportionally represented in the sample to ensure accurate representation of the overall population.

This method helps to improve the accuracy and precision of the results obtained, as it takes into consideration the variability within the different subgroups of the population. In contrast, continuous process sampling, systematic sampling, and cluster sampling are other types of sampling methods that do not specifically allocate proportional samples to each subset in a divided population. The correct answer is D.

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Complete question:

Which method of sampling applies to populations that are divided into natural subsets and allocates the appropriate proportion of samples to each subset?

a. Continuous process sampling

b. Systematic sampling

c. Cluster sampling

d. Stratified sampling

The approximate number of selected concerts will have decibel level less than 112 is equal to 9 .

Average number of decibel at full concert = 120

Standard deviation = 6

Sample size 'n' = 100

Distribution of decibel levels at a full concert is approximately normal with a mean of 120 .

Let X be the random variable representing the decibel level at a full concert.

Probability that a randomly selected concert will have a decibel level less than 112.

Probability using the standard normal distribution.

Standardize the random variable X to the standard normal distribution Z ~ N(0,1) using the formula,

Z = (X - μ) / σ

where μ is the mean of the distribution 120 and σ is the standard deviation of the distribution 6.

Z = (112 - 120) / 6

  = -1.33

Probability of a standard normal variable being less than -1.33 using a standard normal distribution table.

Attached table.

The probability is approximately 0.0918.

Probability of a randomly selected concert having a decibel level less than 112 is approximately 0.0918.

Expected number of concerts out of 100 that will have a decibel level less than 112, we multiply the probability by 100.

Expected number of concerts = 0.0918 × 100

                                                  = 9.18

Therefore, approximately 9 concerts out of 100 will have a decibel level less than 112.

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The indefinite integral is: (2x + 5)/2 - (5/2) * ln|2x + 5| + C

To evaluate the indefinite integral, ∫(2x)/(2x+5) dx, we can use a change of variables, also known as substitution. Let's set:

u = 2x + 5

Now, differentiate u with respect to x:

du/dx = 2

So, dx = du/2

Substitute u and dx in the original integral:

∫(2x)/(u) * (du/2) = ∫(u - 5)/(u) * (du/2)

Now, split the fraction:

∫(u/u - 5/u) * (du/2) = ∫(1 - 5/u) * (du/2)

Now, integrate with respect to u:

(1/2) * ∫(1 - 5/u) du = (1/2) * (u - 5 * ln|u|) + C

Now, substitute back the original variable, x:

(1/2) * ((2x + 5) - 5 * ln|2x + 5|) + C

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9.9.9.9 is the expression  which is equivalent to 3 the power of 8

The given expression is 3⁸

We have to find the equivalent expression of  3⁸

Equivalent expressions are expressions that work the same even though they look different.

=[tex]3^2^\times^4[/tex]

=(3²)⁴

=3²×3²×3²×3²

=9×9×9×9

=9.9.9.9

Hence, 9.9.9.9 is the expression  equivalent to 3 the power of 8

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Answer:

We can use the formula for continuous compounding:

A = Pe^(rt)

where A is the amount of money in the account after t years, P is the principal amount (initial deposit), e is the constant 2.71828 (from natural logarithms), r is the interest rate as a decimal, and t is the time in years.

We want to solve for t when the amount in the account is $30,000:

30,000 = 8,000e^(0.056t)

Divide both sides by 8,000:

3.75 = e^(0.056t)

Take the natural logarithm of both sides:

ln(3.75) = 0.056t

Solve for t by dividing both sides by 0.056:

t = ln(3.75) / 0.056 ≈ 20.1 years

Therefore, it will take Mark approximately 20.1 years for his account to reach $30,000.

The rectangular base has a length of 70 meters and a width of 31 meters.

An area refers to the amount of space occupied by a two dimensional object or figure. The area (A) of a rectangle is: A = length * width

Let w represent the width, hence:

l = 2w + 8

Area = (2w + 8)w

2170 = 2w² + 8w

2w² + 8w - 2170 = 0

w = 31 m

Substituting the value in "l = 2w + 8"

l = 2(31) + 8

i = 70 m

Therefore, the base has a length of 70 meters and a width of 31 meters.

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The region bounded by the two curves has an area of approximately 80.357 square units.

The total area of the region bounded by the two curves is approximately  0.843 square units.

We can see that the region we are interested in lies between the two curves and extends from θ = 0 to θ = π. To compute the area of this region, we can integrate the difference in the areas enclosed by the two curves over the interval [0,π]. That is,

Area = ∫(1/2)(15cos(θ))² dθ - ∫(1/2)(7+cos(θ))² dθ

Simplifying the integrals and evaluating them over the given interval, we obtain the area of the region to be approximately 80.357 square units.

The second problem involves finding the area of the region that lies inside both curves, which are given in polar coordinates as r = √3 cos(θ) and r = sin(θ). To visualize the region of interest, we can again sketch the two curves as shown below:

To compute these areas, we can integrate the corresponding expressions over the appropriate intervals.

The area of the region inside the circle and outside the cardioid is given by:

Area1 = ∫(1/2)(√3cos(θ))² dθ - ∫(1/2)(sin(θ))² dθ

Simplifying the integrals and evaluating them over the intervals [π/6,π/2] and [π/2,π], we obtain the area of this region to be approximately 0.798 square units.

The area of the region inside both curves is given by:

Area2 = ∫(1/2)(sin(θ))² dθ - ∫(1/2)(√3cos(θ))² dθ

Simplifying the integrals and evaluating them over the interval [0,π/6], we obtain the area of this region to be approximately 0.045 square units.

Therefore, the total area of the region bounded by the two curves is approximately 0.798 + 0.045 = 0.843 square units.

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To find the area of the ellipse cut from the plane z=9x by the cylinder x2 y2=4, we need to first visualize the situation. The cylinder x2 y2=4 is a circular cylinder with a radius of 2 in the xy-plane. The plane z=9x is a tilted plane passing through the origin. The intersection of these two surfaces will be an ellipse.

To find the equation of the ellipse, we need to substitute z=9x into the equation of the cylinder:

x2 y2 = 4 becomes 9x2 y2 = 36

Dividing both sides by 36, we get:

x2/4 + y2/4 = 1

This is the equation of an ellipse centered at the origin with a semi-major axis of length 2 and a semi-minor axis of length 2. The area of an ellipse is given by the formula A = πab, where a and b are the lengths of the semi-major and semi-minor axes, respectively.

In this case, a = 2 and b = 2, so the area of the ellipse is:

A = π(2)(2) = 4π

Therefore, the area of the ellipse cut from the plane z=9x by the cylinder x2 y2=4 is 4π. To find the area of the ellipse cut from the plane z=9x by the cylinder x^2 + y^2 = 4, follow these steps:

1. Express the equation of the cylinder in terms of x: x^2 = 4 - y^2.
2. Substitute this expression into the equation of the plane: z = 9(4 - y^2).
3. Find the range of y-values within the cylinder: -2 ≤ y ≤ 2.
4. Use the formula for the area of an ellipse: A = πab, where a and b are the semi-major and semi-minor axes, respectively.
5. Determine the lengths of the semi-major and semi-minor axes using the equation for z (found in step 2) at the endpoints of the range of y-values (found in step 3). For y = -2, a = 9(4 - 4) = 0, and for y = 2, b = 9(4 - 0) = 36.

So, the area of the ellipse cut from the plane z=9x by the cylinder x^2 + y^2 = 4 is A = π(0)(36) = 0.

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The equation of the tangent line to the curve y = f(x) at the point p = (a, f(a)) can be determined using the point-slope form, which is y - f(a) = f'(a)(x - a).

The equation of the tangent line to a curve at a specific point can be found using calculus. The point-slope form of a line is y - y₁ = m(x - x₁), where (x₁, y₁) represents the coordinates of a point on the line and m is the slope of the line.

In this case, the point on the tangent line is p = (a, f(a)), where f(a) represents the y-coordinate of the point on the curve. The slope of the tangent line at point p is given by f(a), which represents the derivative of the function f(x) evaluated at x = a.

Therefore, the equation of the tangent line becomes y - f(a) = f'(a)(x - a). This equation describes the line that touches the curve y = f(x) at point p = (a, f(a)) and has the same slope as the curve at that point.

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There are 420 rabbits in the park.

Given that, on Monday 20 rabbits were tagged on Tuesday 12 out of 42 rabbits were found tagged,

So,

Tuesday = 12 / 42 = 2/7

2/7 of the total rabbits on the farm = 20 tagged on Monday

Let r represent the total rabbits on the farm

then (2/7)r = 20

r = 120(7/2)

r ≈ 420

Hence, there are 420 rabbits in the park.

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The standard equation of the circle is (x - 5)^2 + (y - 8)^2 = 18.

The midpoint formula is:

((x1 + x2) / 2, (y1 + y2) / 2)

Applying the midpoint formula for the given endpoints:

((2 + 8) / 2, (5 + 11) / 2) = (10 / 2, 16 / 2) = (5, 8)

So, the center (h, k) of the circle is (5, 8).

distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Using the center (5, 8) and the endpoint (2, 5):

radius = sqrt((5 - 2)^2 + (8 - 5)^2) = sqrt(3^2 + 3^2) = sqrt(18)

So, the radius r of the circle is sqrt(18).

x - h)^2 + (y - k)^2 = r^2

Substituting the center (h, k) and radius r:

(x - 5)^2 + (y - 8)^2 = (sqrt(18))^2

Simplifying the equation:

(x - 5)^2 + (y - 8)^2 = 18

The standard equation of the circle is (x - 5)^2 + (y - 8)^2 = 18.

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The solution to the given differential equation is y = C₁ * (x + 8)⁻² + C₂ * (x + 8)⁻⁷ where C₁ and C₂ are constants determined by the initial or boundary conditions.

To solve the given differential equation, (x + 8)²y" - B(x + 8)y' + 14y = 0, using Y = (X - X₀)m, follow these steps:

1. Substitute Y = (X - X₀)m into the differential equation: (X - X₀ + 8)^2m" - B(X - X₀ + 8)m' + 14m = 0.
2. Solve for m: m" - (B/((X - X₀) + 8))m' + (14/((X - X₀) + 8)²)m = 0.
3. Find the general solution for m: m = C₁[tex]e^(r1X)[/tex] + C₂[tex]e^(r2X)[/tex], where r₁ and r₂ are the roots of the characteristic equation, and C₁ and C₂ are constants.
4. Determine the roots of the characteristic equation: r₁ and r₂.
5. Substitute the roots into the general solution for m.
6. Finally, substitute m back into the original substitution, Y = (X - X₀)m.

y(x), will be a function involving the roots, r₁ and r₂, and constants C₁ and C₂. The explanation involves substituting Y = (X - X₀)m into the differential equation, solving for m, finding the general solution for m, determining the roots of the characteristic equation, and substituting the roots back into the original substitution to find y(x).

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The probability that one of the dice has the number 5 or less facing up given that the other has at least the number 5 facing up is 2/4 or 1/2, which is equivalent to 0.5 or 50%.

There are 36 equally likely outcomes when rolling a pair of dice. To find the probability that one of the dice has the number 5 or less facing up given that the other has at least the number 5 facing up, we need to consider the outcomes where one die has a number 5 or less and the other has a number greater than or equal to 5.

Out of the 36 possible outcomes, there are 6 outcomes where both dice have a number greater than or equal to 5, namely (5,5), (5,6), (5,6), (6,5), (6,6), and (6,5).

Out of these, there are 2 outcomes where one of the dice has a number 5 or less and the other has a number greater than or equal to 5, namely (5,6) and (6,5).

Similarly, there are 25 outcomes where one of the dice has a number 5 or less, namely (1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (1,4), (2,4), (3,4), (4,1), (4,2), (4,3), (1,3), (2,3), (3,1), (3,2), (1,2), and (2,1).

Out of these, there are 2 outcomes where the other die also has a number greater than or equal to 5, namely (5,1) and (1,5).

Therefore, the probability that one of the dice has the number 5 or less facing up given that the other has at least the number 5 facing up is 2/4 or 1/2, which is equivalent to 0.5 or 50%.

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The mean mass of the women given the conditions are 52.5 kg and 54 kg

(a) a woman of mass 60kg leaves the group

Given that

Women = 15

Mean mass = 53 kg

So, we have

Total mass = 15 * 53 kg

Total mass = 795 kg

When a mass of 60 kg leaves, we have

Mean mass = (795 - 60)/(15 - 1)

Mean mass = 52.5 kg

(b) a woman of mass 69kg joins the original group ​

Given that

Women = 15

Mean mass = 53 kg

So, we have

Total mass = 15 * 53 kg

Total mass = 795 kg

When a mass of 69 kg joins, we have

Mean mass = (795 + 69)/(15 + 1)

Mean mass = 54 kg

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The measurement that is closest to the distance between point M and point M in units is 12 units

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

The distance between the points is the sqauare root of the sum of the squares of the translated uinits

So, we have

Distance = √(8^2 + 9^2)

Evaluate

Distance = 12 units (approx)

Hence, the distance is 12 units

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This is the way to find the approximate carrying capacity for the population.

To determine the approximate carrying capacity for the sheep population in Tasmania using the logistic growth equation, please follow these steps:

1. Recall the graph of sheep population size over time for Tasmania.
2. Identify the logistic growth equation, which is P(t) = K / (1 + (K - P0) / P0 * e^(-r * t)), where P(t) is the population at time t, K is the carrying capacity, P0 is the initial population, r is the growth rate, and t is the time.
3. Observe the graph and find the point where the population growth starts to level off, which is the carrying capacity (K).
4. Estimate the value of K from the graph, which represents the approximate carrying capacity for the sheep population in Tasmania.

Please note that without the actual graph, I cannot provide an exact value for the carrying capacity. However, you can follow the steps above to find the approximate carrying capacity using the logistic growth equation and the given graph.

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Answer:

[tex]m = \frac{ - 6 - ( - 4)}{ - 2 - ( - 5)} = - \frac{2}{3} [/tex]

First, let's check why V1, which is the vector [..)-(-3) 1:1, spans R2 but does not form a basis. We can see that V1 has two linearly independent components, which means it can span R2. However, V1 is not a basis because it is not linearly independent.

To express 15 -3 as a linear combination of V1, V2, V3, we need to solve the equation aV1 + bV2 + cV3 = 15 -3, where a, b, and c are coefficients. We can rewrite this equation as a system of linear equations:

a + 2b - 3c = 15
-3b + c = -3

Solving this system of linear equations, we get:

a = -1
b = -6
c = -15

Therefore, we can express 15 -3 as a linear combination of V1, V2, V3 as:

-1V1 - 6V2 - 15V3 = 15 -3

Now, let's find another way to express 15 -3 as a linear combination of V1, V2, V3 when the coefficient of V3 is 0. This means we need to solve the equation aV1 + bV2 = 15 -3, where a and b are coefficients. We can rewrite this equation as a system of linear equations:

a + 2b = 15
-3b = -3

Solving this system of linear equations, we get:

a = 3
b = 1

Therefore, we can express 15 -3 as a linear combination of V1, V2, V3 when the coefficient of V3 is 0 as:

3V1 + V2 + 0V3 = 15 -3

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The coordinates of the vertices of triangle A'B'C' are:

A'(-2, 1), B'(0, -2), and C'(1, 2)

From inspection of the given diagram, the coordinates of the vertices of triangle ABC are:

A = (-2, 1)

B = (2, -5)

C = (4, 3)

If the figure is translated left 2 units and up 1 unit, then the mapping rule of the translation is:

(x,y) ---> (x-2, y +1)

If a figure is dilated by scale factor k with the origin as the center of dilation, the mapping rule is:

(x,y) ---> (kx, ky)

Therefore, given the scale factor is 0.5, the final mapping rule that translates and dilates triangle ABC is:

(x,y) ---> (0.5 (x-2), 0.5 (y +1 ))

To find the coordinates of the vertices of triangle A'B'C', substitute the coordinates of the vertices of triangle ABC into the final mapping rule:

A'=(-2,1)

B'= (0, -2)

C'= (1,2)

Therefore, the coordinates of the vertices of triangle A'B'C' are:

A'(-2, 1), B'(0, -2), and C'(1, 2)

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Answer:

Step-by-step explanation:

No, Noah's arithmetic is not correct.

Lin is correct that when you add or multiply two complex numbers, the result can always be written in the form a + bi.

In the first example, (7+2)+(3-2), we can simplify by adding the real and imaginary parts separately: (7+3)+(2-2) = 10 + 0i, which can be written in the form a + bi.

In the second example, (2+2)(2+2), we can expand using FOIL: 2(2) + 2(2i) + 2i(2) + 2i(2i) = 4 + 4i + 4i - 4 = 8i, which can also be written in the form a + bi.

Therefore, Noah's exceptions are not valid, and the statement made by Lin is true.

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The probability that a salesperson selected at random has above-average sales ability and has excellent potential for advancement is 0.27.

We are given that;

Number of samples salespeople=500

Now,

The probability of a salesperson having above-average sales ability is given by:

P(A)=50093+72+135​=0.6

The probability of a salesperson having excellent potential for advancement given that they have above-average sales ability is given by:

P(B∣A)=93+72+135135​=0.45

Using the formula for joint probability, we get:

P(A∩B)=P(A)×P(B∣A)=0.6×0.45=0.27

Therefore, by the probability the answer will be 0.27.

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The student(s) that successfully outlined a series of steps to complete the construction is both Elias and Greyson

Both Elias and Greyson have presented techniques for generating a line that passes through point C in a perpendicular manner to line AB.

Although their processes differ slightly, they both necessitate the utilization of intersecting arcs prior to forming a connection between a point of intersection and point C by laying down a line.

In conclusion, choosing either Elias or Greyson as the correct answer would be accurate as shows that they both delivered relevant methods that could work.

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