Solve the system of equations by the substitution method
{x=4/9y-1
{18x-7y=-15

The absolute maximum value of f on the interval [0, 5] is 15.

The absolute minimum value of f on the interval [0, 5] is 5.

To find the absolute maximum and absolute minimum values of f on the given interval:

We need to evaluate f(x) at the endpoints of the interval and at any critical points within the interval.
First, we find the derivative of f(x):
f'(x) = 15 - 2x
Then, we set f'(x) = 0 and solve for x:
15 - 2x = 0
x = 7.5
However, 7.5 is not within the interval [0, 5], so we do not have any critical points within the interval.
Next, we evaluate f(x) at the endpoints of the interval:
f(0) = 15
f(5) = 5
Therefore, The absolute maximum value of f on the interval [0, 5] is 15 and the absolute minimum value of f on the interval [0, 5] is 5.

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To find the distance Add the distance between each point and the y-axis, the correct option is D.

We are given that;

Two points  A(-8,-4) and B(2,-4)

Now,

To find the distance between two points on a coordinate plane, you can use the distance formula: d = √((x2 - x1)² + (y2 - y1)²), where (x1, y1) and (x2, y2) are the coordinates of the points. In this case, the points are A(-8, -4) and B(2, -4). Plugging in the values, we get:

d = √((2 - (-8))² + (-4 - (-4))²) d = √(10² + 0²) d = √(100) d = 10

The distance between the points is 10 units.

Therefore, by the distance answer will be Add the distance between each point and the y-axis.

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Answer:

Interest Earned: $360

Total value of the Account: $1260

Step-by-step explanation:

To find the interest earned, we can use the formula:

Interest = Principal x Rate x Time

where the principal is the amount deposited, the rate is the interest rate expressed as a decimal, and the time is the number of years the money is left in the account.

In this case, the principal is $900, the rate is 0.04 (since 4% is equivalent to 0.04 as a decimal), and the time is 25 years. Plugging in these values, we get:

Interest = $900 x 0.04 x 25 = $900 x 1 = $360

So the interest earned over 25 years is $360.

To find the total value of the account, we can add the interest earned to the original deposit:

Total Value = Principal + Interest

Total Value = $900 + $360 = $1260

So the total value of the account after 25 years is $1260.

Hope this helps!

el aumento en el salario mensual base de Verónica será de 1,650.00.

para ayudar a Verónica a determinar cuánto dinero le van a aumentar en su salario mensual base con un aumento del

15%, debemos seguir estos pasos:

Identificar el salario mensual base de Verónica, que es de 11,000.00.

Identificar el porcentaje de aumento, que es del 15%.

Convertir el porcentaje de aumento a decimal dividiendo por 100 (15 ÷ 100 = 0.15).

Multiplicar el salario mensual base por el porcentaje de aumento en decimal (11,000.00 × 0.15 = 1,650.00).

Por lo tanto, el aumento en el salario mensual base de Verónica será de 1,650.00.

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(a) Approximately 30.85% of students have a GPA of at least 3.

(b) Approximately 56.12% of students have a GPA between 2 and 3.3.

(c) A student must have a GPA of approximately 3.902 to qualify.

(d) The chance that Kelly's GPA is higher than 3.3 is 15.87%.

(e) The chance that at least 3 out of 10 students have a GPA over 3 is approximately 87.61%.

(a) Calculate the z-score: (3 - 2.88) / 0.6 ≈ 0.2. Using a z-table, we find that 30.85% of students have a GPA of at least 3.
(b) Calculate the z-scores for 2 (z1 = -1.47) and 3.3 (z2 = 0.7). The proportion between these z-scores is 56.12%.
(c) Find the z-score for the top 1.5% (z ≈ 1.96). Then, calculate the GPA: 2.88 + (1.96 * 0.6) ≈ 3.902.
(d) Calculate the z-score for 3.3: (3.3 - 2.88) / 0.6 ≈ 0.7. From the z-table, 15.87% of students with a GPA of 3 or higher have a GPA > 3.3.
(e) Use the binomial probability formula with n=10, p=0.3085, and at least 3 successes. Calculate the probability and sum the probabilities for 3 to 10 successes, resulting in approximately 87.61%.

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By using the method of Lagrange multipliers, minimum and maximum values of function are  -3√17 and 3√17, respectively.

We can use the method of Lagrange multipliers to find the minimum and maximum values of the function subject to the given constraint. The Lagrange function is

L(x, y, z, λ) = 3x + 2y + 4z - λ(x^2 + 2y^2 + 6z^2 - 17)

Taking partial derivatives with respect to x, y, z, and λ, we get

∂L/∂x = 3 - 2λx = 0

∂L/∂y = 2 - 4λy = 0

∂L/∂z = 4 - 12λz = 0

∂L/∂λ = x^2 + 2y^2 + 6z^2 - 17 = 0

Solving these equations, we get:

x = ±√(17/2), y = ±√(17/8), z = ±√(17/24), and λ = 1/17

We evaluate the function at all eight possible combinations of these values and find that the minimum value is -3√17 and the maximum value is 3√17.

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Inequality comparing 13 is;

13/4-1/13¹/² >= 13/2

We need to show that 13/4-1/13¹/²ˣ² is greater than or equal to 13/2.

First, let's simplify 13/4-1/13¹/² by finding a common denominator:

13/4 - 1/13¹/² = 13/4 - 113¹/²/13 = (1313¹/² - 4)/13¹/²²) = (13¹/²)^2/13¹/²² - 4/13¹/²ˣ²

Simplifying further, we get:

13/4 - 1/13¹/² = (13/13) - 4/13¹/²ˣ²) = 13/13 - 4/169 = 159/169

So, we need to show that 159/169 is greater than or equal to 13/2:

159/169 >= 13/2

Multiplying both sides by 169/2, we get:

159*169/338 >= 169/2 * 13/2

Simplifying, we get:

159/2 >= 169/4

Which is true, so we can conclude that:

13/4-1/13¹/² >= 13/2

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The function f(x)=3x^2+3x+(4) has maximum value 64 at x=4 and minimum value of the function is 4 at x=0

Explanation: -

For the function f(x)=3x^2+3x+(4), we need to find the maximum and minimum values on the interval [0, 4], To find the absolute maximum and minimum values, we can use the first derivative test and the second derivative test.

To find the absolute maximum and minimum values, we can use the first derivative test and the second derivative test.

STEP1:-First, we take the derivative of f(x) and equate f'(x) to zero and solve for x to get the critical points.

Thus,

f'(x) = 6x + 3

Setting f'(x) = 0 and solving for x, we get:
6x + 3 = 0
x = -1/2

This critical point is not in the interval [0, 4], so we don't need to consider it.

STEP2:- If the critical point is not belonged to the provided interval, we check the endpoints of the interval,

Thus,

f(0) = 4
f(4) = 64

So, the absolute minimum value of f(x) on the interval [0, 4] is 4, which occurs at x = 0. The absolute maximum value of f(x) on the interval [0, 4] is 64, which occurs at x = 4.

Therefore, the absolute maximum and minimum values on the interval [0, 4] for the function f(x)=3x^2+3x+(4) are:

Absolute maximum = 64 at x = 4
Absolute minimum = 4 at x = 0

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Answer:

C. x-intercept = (-6, 0)

y-intercept = (0, -48/5)

Step-by-step explanation:

as you can see, x-intercept means y = 0.

and y-intercept means x = 0.

so, for x = 0 we have

5y = -48

y = -48/5

for y = 0 we have

8x = -48

x = -48/8 = -6

Each side of the square room is 9 feet long.

A square is a two-dimensional geometric shape with four equal sides and four equal angles of 90 degrees each. It is a type of rectangle, but it has the additional property that all its sides are of equal length.

Since the room is square and the floor tiles are also square, the number of tiles required to cover the floor is equal to the area of the room divided by the area of each tile.

Let's assume that each side of the square room is "x" feet long. Then, the area of the room can be expressed as x² square feet. If each floor tile measures "y" feet on each side, then the area of each tile can be expressed as y² square feet.

Given that 81 tiles are required to cover the floor, we can set up the following equation:

x² / y² = 81

To solve for "x", we need to first determine the value of "y". Since each floor tile is a square, we can assume that y is the length of one side of a tile. Let's suppose that each tile measures 1 foot on each side. Then, the area of each tile is y² = 1² = 1 square foot.

Substituting y = 1 in the above equation, we get:

x² / 1² = 81

x² = 81

x = √(81)

x = 9

Therefore, each side of the square room is 9 feet long.

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The domain of the function [tex]H(x) = -3x^2 + 12[/tex] is all real numbers.

To find the domain of the function [tex]H(x) = -3x^2 + 12[/tex], we need to determine the set of all possible x-values for which the function is defined.

Since this is a quadratic function, it is defined for all real numbers.

Here's a brief explanation:
Identify the function type:

In this case,[tex]H(x) = -3x^2 + 12[/tex] is a quadratic function, as it has the form [tex]f(x) = ax^2 + bx + c[/tex] , where a, b, and c are constants.

Determine the domain for the function type:

Quadratic functions are defined for all real numbers, meaning there are no restrictions on the x-values.

This is because you can input any real number for x, and the function will still output a real number.
State the domain:

Based on the above information, the domain of[tex]H(x) = -3x^2 + 12[/tex] is all real numbers.

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For every n ∈ Z(integer),  n² + 2  is not divisible by 4

Here the proposition is:  for every n ∈ Z, n² + 2 is not divisible by 4

We need to prove this proposition by contradiction.

Proof by contradiction:

Let us assume that  n² + 2 is divisible by 4

So,  n² + 2 must be multiple of 4

n² + 2 = 4k; where k is an integer

n² + 2 - 4k = 0

Now we write above equation for n.

⇒ n² = 4k - 2

⇒ n = √(4k - 2)

For n = 1,

⇒ (1)² = 4k - 2

⇒ 1 = 4k - 2

⇒ 4k = 1 + 2

⇒ 4k = 3

⇒ k = 3/4

The value of k is in the form p/q (p, q ∈ Z; q ≠ 0) and which is contradtion to our assumption k ∈ Z

Therefore, for every n ∈ Z, then n^2 + 2 is not divisible by 4

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The approximate area under the curve y=x^3 from x=1 to x=4 using a right endpoint approximation with 6 subdivisions is 49.0125 square units.

To approximate the area under the curve y=x^3 from x=1 to x=4 using a right endpoint approximation with 6 subdivisions, we can use the following steps:

1. Determine the width of each subdivision by dividing the total width (4-1) by the number of subdivisions (6):

width = (4-1)/6 = 0.5

2. Determine the right endpoint of each subdivision by adding the width to the left endpoint:

x1 = 1, x2 = 1.5, x3 = 2, x4 = 2.5, x5 = 3, x6 = 3.5

3. Evaluate the function at each right endpoint:

f(x1) = f(1) = 1
f(x2) = f(1.5) = 3.375
f(x3) = f(2) = 8
f(x4) = f(2.5) = 15.625
f(x5) = f(3) = 27
f(x6) = f(3.5) = 42.875

4. Calculate the area of each rectangle by multiplying the width by the function value at the right endpoint:

A1 = 0.5*1 = 0.5
A2 = 0.5*3.375 = 1.6875
A3 = 0.5*8 = 4
A4 = 0.5*15.625 = 7.8125
A5 = 0.5*27 = 13.5
A6 = 0.5*42.875 = 21.4375

5. Add up the areas of all the rectangles to get an approximate area under the curve:

A ≈ A1 + A2 + A3 + A4 + A5 + A6 = 49.0125

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(a) The maximum likelihood estimator for α, assuming β is known, is found by differentiating the likelihood function with respect to α, setting it equal to zero, and solving for α. This leads to the equation α-cap= n/∑(y_i^β), where n is the sample size and y_i is the i-th observed failure time.

(b) When both α and β are unknown, the likelihood function must be maximized with respect to both parameters simultaneously.

This involves taking partial derivatives of the likelihood function with respect to both α and β, setting them equal to zero, and solving the resulting equations.

The solutions for α-cap and β-cap will depend on the specific data observed, but they can be found using numerical optimization methods or by solving the equations iteratively. The resulting estimators will provide the best fit of the Weibull distribution to the observed failure times.

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The value of the given integral is [tex]\int_{1-c}^{2-c} f(u) du= \int_1^2 f(x-c) dx= 5[/tex]

Calculating an integral is called integration. Mathematicians utilise integrals to determine a variety of useful quantities, including areas, volumes, displacement, etc. When we discuss integrals, we typically refer to definite integrals. For antiderivatives, indefinite integrals are utilised. Aside with differentiation, which quantifies the rate at which any function changes in relation to its variables, integration is one of the two main calculus topics in mathematics.

We can use the substitution u = x - c for the first integral to get:

[tex]\int_{1}^{2} f(x-c) dx[/tex] = ∫ f(u) du from 1-c to 2-c

Since the integral is from 1 to 2, the limits of integration in terms of u become (1-c)-c = 1-2c and (2-c)-c = 2-3c. Thus:

[tex]\int_{1-c}^{2-c} f(u) du= \int_1^2 f(x-c) dx= 5[/tex]

Therefore, ∫ f(x) dx from 1-c to 2-c = ∫ f(u) du from 1-c to 2-c = ∫ f(x-c) dx from 1 to 2 = 5.

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Answer:

4/(x-2)

Step-by-step explanation:

(x²+2x-4)/(x-2)

x + 4/(x-2)

4/(x-2)

The answer are :

x = 5

y = 20

z = 17.3

Sorry for bad handwriting

if i was helpful Brainliests my answer ^_^

The confidence interval is given by: [F(b) - F(a)] - z_alpha/2 * sqrt{ [F(b)(1 - F(b)) + F(a)(1 - F(a)) - 2F(a)F(b) + 2F(a)^2] / n } <= theta <= [F(b) - F(a)] + z_alpha/2 * sqrt{ [F(b)(1 - F(b)) + F(a)(1 - F(a)) - 2F(a)F(b) + 2F(a)^2] / n }

The estimated standard error of theta can be found using the following formula:

SE(theta) = sqrt{ [F(b)(1 - F(b)) / n] + [F(a)(1 - F(a)) / n] }

where n is the sample size.

To find an approximate 1 - alpha confidence interval for theta, we first need to find the standard error of the estimator. Let X1, X2, ..., Xn be the random sample. Then, the estimator T(Fn) is given by:

T(Fn) = Fn(b) - Fn(a)

The variance of T(Fn) can be estimated as:

Var(T(Fn)) = Var(Fn(b) - Fn(a)) = Var(Fn(b)) + Var(Fn(a)) - 2Cov(Fn(b), Fn(a))

Using the fact that Fn is a step function with jumps of size 1/n at each observation, we can calculate the variances and covariance as:

Var(Fn(x)) = Fn(x)(1 - Fn(x)) / n

Cov(Fn(b), Fn(a)) = - Fn(a)(F(b) - F(a)) / n

Substituting these into the expression for Var(T(Fn)), we get:

Var(T(Fn)) = [F(b)(1 - F(b)) + F(a)(1 - F(a)) - 2F(a)(F(b) - F(a))] / n

Simplifying this expression, we get:

Var(T(Fn)) = [F(b)(1 - F(b)) + F(a)(1 - F(a)) - 2F(a)F(b) + 2F(a)^2] / n

Now, the standard error of T(Fn) can be calculated as the square root of the variance:

SE(T(Fn)) = sqrt{ [F(b)(1 - F(b)) + F(a)(1 - F(a)) - 2F(a)F(b) + 2F(a)^2] / n }

To construct an approximate 1 - alpha confidence interval for theta, we use the following formula:

T(Fn) +/- z_alpha/2 * SE(T(Fn))

where z_alpha/2 is the (1 - alpha/2)th quantile of the standard normal distribution. Therefore, the confidence interval is given by:

[F(b) - F(a)] - z_alpha/2 * sqrt{ [F(b)(1 - F(b)) + F(a)(1 - F(a)) - 2F(a)F(b) + 2F(a)^2] / n } <= theta <= [F(b) - F(a)] + z_alpha/2 * sqrt{ [F(b)(1 - F(b)) + F(a)(1 - F(a)) - 2F(a)F(b) + 2F(a)^2] / n }

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The value of t such that the area under the curve between −|t| and |t| equals 0.95, assuming 12 degrees of freedom, is approximately 1.782.

Using a t-distribution table or statistical software, we can find the t-value that corresponds to an area of 0.95 in the upper tail of the t-distribution with 12 degrees of freedom. From the t-distribution table, we find that the t-value with 0.95 area in the upper tail and 12 degrees of freedom is approximately 1.782.

Therefore, the value of t such that the area under the curve between −|t| and |t| equals 0.95, assuming 12 degrees of freedom, is approximately 1.782.

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The solution for the given expression is given below:

a. vertical asymptote. b. the graph of f will pass through the point (0, 15). c. to keep f(x) within 0.01 units of 7, we need to keep x between 2.9986 and 3.0014.

What is expression?

In general, an expression is a combination of symbols, numbers, and/or operators that can be evaluated to produce a value. In programming, an expression typically refers to a sequence of one or more operands and operators that can be evaluated to produce a single value.

a. When x = 3, the denominator of f(x) becomes zero, and therefore f(3) is undefined (or does not exist). This is called a vertical asymptote.

b. To plot the graph of f, we can factor the numerator as follows:

f(x) = (4x-3)(x-5)/(x-3)

The graph of f will have a vertical asymptote at x = 3, and the function will be undefined at that point. The factor (4x-3)(x-5) has zeros at x = 3/4 and x = 5, so the graph will cross the x-axis at those points. We can also find the y-intercept by setting x = 0:

f(0) = (4(0)-3)(0-5)/(0-3) = 15

Therefore, the graph of f will pass through the point (0, 15).

c. The limit of f(x) as x approaches 3 is given by:

lim[x→3] f(x) = lim[x→3] (4[tex]x^2[/tex]-17x+15)/(x-3) = 7

To find how close to 3 we need to keep x in order for f(x) to be within 0.01 units of 7, we can use the definition of a limit:

|f(x) - 7| < 0.01

This inequality can be rewritten as:

-0.01 < f(x) - 7 < 0.01

[tex]-0.01 < (4x^2-17x+15)/(x-3) - 7 < 0.01[/tex]

Solving for x using this inequality is difficult, but we can use a graphing calculator or a numerical method to find the values of x that satisfy it. For example, using a graphing calculator, we can graph the function (4x^2-17x+15)/(x-3) and the horizontal lines y = 7.01 and y = 6.99, and find the values of x where the graph intersects those lines. We get:

x ≈ 3.0014 and x ≈ 2.9986

Therefore, to keep f(x) within 0.01 units of 7, we need to keep x between 2.9986 and 3.0014.

Similarly, to find how close to 3 we need to keep x in order for f(x) to be within 0.0001 units of 7, we can use the inequality:

|f(x) - 7| < 0.0001

This inequality can be rewritten as:

-0.0001 < f(x) - 7 < 0.0001

[tex]-0.0001 < (4x^2-17x+15)/(x-3) - 7 < 0.0001[/tex]

Using a similar method as before, we can find that we need to keep x between approximately 2.99994 and 3.00006 to keep f(x) within 0.0001 units of 7.

To keep f(x) arbitrarily close to 7 just by keeping x close to 3 but not equal to 3, we can use the fact that the function approaches 7 as x approaches 3 from both sides. This means that we can make f(x) as close to 7 as we want by choosing a small enough positive or negative deviation from 3.

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a) The slope of the curve  y = √x at the given point P(4, 2) is 1/4

b)  An equation of the tangent line at P(4,2) is x - 4y + 4 = 0

a) Consider the equation of the curve y = √x

To find the slope of the curve at point P we find the derivative of y.

y'(x) = 1/2√x)

At point P(4, 2)

y' = 1/(2√4)

y' = 1/(2×2)

y' = 1/4

Therefore, the slope of the curve at the given point P is 1/4

b)

Now we need to find an equation of the tangent line at P

The equation of tangent line for the function f(x) at P(x₁, y₁) is:

(y - y₁) = m (x - x₁)

Here, slope m = 1/4

(x₁, y₁) = (4, 2)

(y - 2) = (1/4) (x - 4)

4y - 8 = x - 4

x - 4y + 4 = 0

This is a required equation.

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The answers to the trigonometric prompts are:

1)

1.1) -0.15

1.2) 0.87

2)

2.1) x = 60°

2.2) x = 70.54°

3) k = 1

1.1 To calculate the value of Tan^2 (316.4 degrees - 212.6 degrees), we first need to find the difference between the two angles:

316.4 degrees - 212.6 degrees = 103.8 degrees

Then, we can use the identity Tan^2 (A - B) = [Tan(A) - Tan(B)]/[1 + Tan(A) Tan(B)] to get:

Tan^2 (316.4 degrees - 212.6 degrees) = [Tan(316.4 degrees) - Tan(212.6 degrees)]/[1 + Tan(316.4 degrees) Tan(212.6 degrees)]

Using a calculator, we get:

Tan(316.4 degrees) ≈ -1.378

Tan(212.6 degrees) ≈ 1.378

So, substituting these values into the above equation, we get:

Tan^2 (316.4 degrees - 212.6 degrees) ≈ [(-1.378) - 1.378]/[1 + (-1.378)(1.378)] ≈ -0.19

Therefore, Tan^2 (316.4 degrees - 212.6 degrees) ≈ -0.19

1.2 To calculate the value of 2Sin(2x 103.4 degrees), we can use the double angle formula for sine:

2Sin(2x 103.4 degrees) = 2(2Sin(103.4 degrees)Cos(103.4 degrees))

Using a calculator, we get:

Sin(103.4 degrees) ≈ 0.974

Cos(103.4 degrees) ≈ -0.226

Substituting these values into the above equation, we get:

2Sin(2x 103.4 degrees) ≈ 2(2(0.974)(-0.226)) ≈ -0.88

Therefore, 2Sin(2x 103.4 degrees) ≈ -0.88

2.1 To solve the equation 2cos(x) = 1, we can first isolate cos(x) by dividing both sides by 2:

cos(x) = 1/2

To find the solutions in the given interval [0 degrees; 90 degrees], we can use the inverse cosine function (cos^-1) and find the principal value:

cos^-1(1/2) ≈ 60 degrees

Therefore, the solution to the equation 2cos(x) = 1 in the interval [0 degrees; 90 degrees] is x = 60 degrees.

To find the value of k in the equation k.sin 60 degrees = (2 cos 30 degrees)/tan 45 degrees, we can use the values of sin 60 degrees, cos 30 degrees, and tan 45 degrees:

sin 60 degrees = √3/2

cos 30 degrees = √3/2

tan 45 degrees = 1

Substituting these values into the given equation, we get:

k(√3/2) = (2 √3/2)/1

Simplifying, we get:

k = 2

Therefore, the value of k is 2.

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The average value of the function f(x,y) = 5 cos(x) cos(y) over the region R = {(x,y) : 0 ≤ x ≤ 5, 0 ≤ y ≤ 5} is (1/5) sin(5) sin(5).

The average value of a function f(x,y) over a region R is given by:

average value = (1/Area(R)) * double integral over R of f(x,y) dA

where dA represents the differential area element and Area(R) represents the area of the region R.

In this case, the region R is given by:

R = {(x,y) : 0 ≤ x ≤ 5, 0 ≤ y ≤ 5}

and the function f(x,y) is given by:

f(x,y) = 5 cos(x) cos(y)

So, we need to compute the double integral over R of f(x,y) dA and divide by the area of R.

To compute the double integral, we have:

∫∫R f(x,y) dA = ∫0^5 ∫0^5 5 cos(x) cos(y) dy dx

= 5 ∫0^5 cos(x) dx ∫0^5 cos(y) dy

= 5 sin(5) sin(5)

To find the area of R, we have:

Area(R) = ∫0^5 ∫0^5 1 dy dx = 25

So, the average value of f(x,y) over R is:

average value = (1/Area(R)) * double integral over R of f(x,y) dA

= (1/25) * 5 sin(5) sin(5)

= (1/5) sin(5) sin(5)

Therefore, the average value of the function f(x,y) = 5 cos(x) cos(y) over the region R = {(x,y) : 0 ≤ x ≤ 5, 0 ≤ y ≤ 5} is (1/5) sin(5) sin(5).

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By using proportion The option to the above question is The college will have about 1,440 students who prefer ice cream.

What is Proportion?

A proportion is an equation that states that two ratios are equal to each other. Ratios are a way of comparing two or more quantities or values, and a proportion is used to express the relationship between these ratios. Proportions are commonly used in various mathematical and real-world contexts to solve problems that involve comparing quantities or predicting values.

According to the given information:

According to the table:

Number of students who prefer Ice Cream = 81

To make a prediction about the number of students who prefer ice cream among the total student population of 4,000, we can set up a proportion:

Number of students who prefer Ice Cream / Total number of students surveyed = Number of students who prefer Ice Cream in the total student population / Total number of students in the total student population

Plugging in the values:

81 / 225 = x / 4000

Solving for x, the number of students who prefer ice cream in the total student population:

x = (81 / 225) * 4000 ≈ 1440

So, the best prediction about the scoops of ice cream the college will need is that they will have about 1440 students who prefer ice cream among the total student population of 4000. Therefore, the correct statement is "The college will have about 1,440 students who prefer ice cream."

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Option B is showing graph of given equations.

Rewrite the first equation in slope-intercept form, y = mx + b:

y = -3 + (1/2)x

This equation has a y-intercept of -3 and a slope of 1/2.

Rewrite the second equation in slope-intercept form:

3x + 2y = 2

2y = -3x + 2

y = (-3/2)x + 1

This equation has a y-intercept of 1 and a slope of -3/2.

Plot the y-capture of every situation on the y-hub:

The principal condition has a y-block of - 3, so plot the point (0,- 3).

The subsequent condition has a y-capture of 1, so plot the point (0,1).

Find additional points on each line by utilizing the slope of each equation:

Since the slope of the first equation is 1/2, the point (2,-2.5) can be obtained by moving up one unit and right two units from the y-intercept (0,-3). To get the point, move up one unit and right two units again.

Since the slope of the second equation is -3/2, move down three units and right two units from the y-intercept (0,1) to reach the point (2,-3.5). To get the point, move down three units and right two units once more.

Plot the points and draw lines through them:

Plot the points (0,-3), (2,-2.5), and (4,-2) for the first equation, and connect them with a straight line. Plot the points (0,1), (2,-3.5), and (4,-5) for the second equation, and connect them with a straight line.

The resulting graph should show two lines intersecting at a point and  the intersection point of the two lines is (2, -2). This point represents the solution to the system of equations.

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The sampling distribution for x Overbar Subscript Upper A Baseline minus x Overbar Subscript Upper C is 0.13.

The standard deviation of the sampling distribution for the difference in sample means can be calculated using the formula:

Standard deviation of the sampling distribution = √[(σ[tex]A^2[/tex]/nA) + (σ[tex]C^2[/tex]/nC)]

Where nA and nC are the sample sizes for Alex and Chris, respectively, and A and C are the standard deviations of the population for Alex and Chris, respectively.

Substituting the given values, we get:

Standard deviation of the sampling distribution = [tex]\sqrt{[(0.38^2/10) + (0.2^2/15)]}[/tex]

= [tex]\sqrt{0.01444 + 0.00222}[/tex]

= [tex]\sqrt{0.01666}[/tex]

= 0.129

Therefore, the answer is 0.13.

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