Let X be a random variable with pdf given by f(x) = 2x for 0 < x < 1 and f(x) = 0 otherwise. a. Find P(X > 1/2). b. Find P(X > 1/2 X > 1/4).

The capacitance of the given Si n-p junction is 2.52 x 10^-16 F by using the formula for the capacitance of a pn junction under reverse bias.

To calculate the capacitance of an n-p junction with donor doping of 8×10^15 cm^−3 on the n-side, we need to use the depletion approximation and the equation for the capacitance of a pn junction under reverse bias

C = sqrt(q * ε * N_a * N_d) / V_bi * [1 + (2 * V_bi / V_r)]

where

C is the capacitance per unit area of the junction

q is the elementary charge (1.602 x 10^-19 C)

ε is the permittivity of the semiconductor material (assumed to be 11.7 * ε0 for Si)

N_a and N_d are the acceptor and donor doping concentrations, respectively

V_bi is the built-in potential of the junction

V_r is the reverse bias voltage applied to the junction.

First, we need to find the built-in potential V_bi. For an n-p junction with doping concentrations N_a and N_d, the built-in potential is given by:

V_bi = (kT/q) * ln(N_a * N_d / ni^2)

where k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature (assumed to be room temperature, or 300 K), and ni is the intrinsic carrier concentration of the semiconductor material (for Si at room temperature, ni = 1.45 x 10^10 cm^-3).

Plugging in the values, we get

V_bi = (1.38 x 10^-23 J/K * 300 K / 1.602 x 10^-19 C) * ln(8 x 10^15 cm^-3 * 1.45 x 10^10 cm^-3 / (1.45 x 10^10 cm^-3)^2)

= 0.721 V

Next, we can calculate the capacitance per unit area of the junction

C = sqrt(q * ε * N_a * N_d) / V_bi * [1 + (2 * V_bi / V_r)]

= sqrt(1.602 x 10^-19 C * 11.7 * ε0 * 8 x 10^15 cm^-3 * 1 cm^-3) / 0.721 V * [1 + (2 * 0.721 V / 10 V)]

= 2.52 x 10^-8 F/cm^2

Multiplying by the cross-sectional area of the junction (1 μm^2 = 10^-8 cm^2), we get the capacitance of the junction

C_total = C * A = 2.52 x 10^-8 F/cm^2 * 10^-8 cm^2 = 2.52 x 10^-16 F

So the capacitance of the n-p junction under reverse bias of 10 V is approximately 2.52 x 10^-16 F.

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--The given question is incomplete, the complete question is given

" calculate the capacitance for the following si n -p junction  with donor doping of 8×10^15 cm ^−3 on the n-side with the cross sectional area of 1μm ^2 and a reverse bias of 10V. (Note: Include the built-in potential of this junction. To calculate the contact potential, assume that the p-side Fermi level is pinned at the valence band edge and the intrinsic Fermi level is exactly at mid-gap.)"--

the geometric mean of the given data is approximately 2.74.

to calculate the geometric mean of the given data, you need to multiply all the numbers together and then take the nth root, where n is the number of values. In this case, n = 12.

Geometric Mean = (3 × 2 × 1 × 2 × 1 × 5 × 9 × 1 × 2 × 3 × 3 × 10)[tex]^{1/12}[/tex]

After multiplying the numbers, we get:

Geometric Mean [tex]= (32,760)^{(1/12)}[/tex]

Now, take the 12th root:

Geometric Mean ≈ 2.74

So, the geometric mean of the given data is approximately 2.74.

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The amount that Bella will need to contribute every year, given earnings is $7,150 .

Bella's total expenses are $40,000 per year. From the given information, we can calculate the total amount of financial aid and contribution from family as follows:

Total financial aid = $9,750 (scholarships and grants) + $3,100 (work-study) = $12,850

Total contribution from family = $20,000

To find out how much Bella needs to contribute, we can subtract the total financial aid and contribution from family from the total expenses:

Bella's contribution = Total expenses - Total financial aid - Contribution from family

= $40,000 - $12,850 - $20,000

= $7,150

Therefore, Bella needs to contribute $7,150 each year to cover her expenses.

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First part of the question is:

Bella qualifies for $9,750 in scholarships and grants per year, and she will earn $3,100 through the work-study program.

Given;

f(x):3x5 -20x3 in the interval I-1,2]
The critical numbers for the function f(x) = 3x^5 - 20x^3 in the interval [-1, 2] are x = 0. The maximum value of f on the interval is 96, and the minimum value of f on the interval is -23.

To find the critical numbers for the function f(x) = 3x^5 - 20x^3 in the interval [-1, 2], follow these steps:

1. Find the derivative of the function:
f'(x) = 15x^4 - 60x^2

2. Set the derivative equal to zero and solve for x to find critical numbers:
15x^4 - 60x^2 = 0
x^2(15x^2 - 60) = 0
x^2(5x^2 - 20) = 0

Critical numbers are x = 0, x = ±2√2.

3. Check which critical numbers are within the given interval [-1, 2]:
Only x = 0 is within the interval.

4. Evaluate the function at the endpoints and critical numbers:
f(-1) = 3(-1)^5 - 20(-1)^3 = -23
f(0) = 0
f(2) = 3(2)^5 - 20(2)^3 = 96

5. Determine the maximum and minimum values on the interval:
The maximum value of f on the interval is 96, which occurs at x = 2.
The minimum value of f on the interval is -23, which occurs at x = -1.

The critical numbers for the function f(x) = 3x^5 - 20x^3 in the interval [-1, 2] are x = 0. The maximum value of f on the interval is 96, and the minimum value of f on the interval is -23.

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Answer:

84 inches

Step-by-step explanation:

7 x 12 = 84  There are 12 inches in a foot.

Helping in the name of Jesus.

Answer: 84 inches.

Step-by-step explanation:

Since 1 foot = 12 inches, we can multiply 7 feet by 12 inches to get 84 inches.

Sure, I can walk you through the steps to calculate the final price with tax on those shorts.

Here are the steps:

1. The original price of the shorts is $58.

2. BC charges Provincial Sales Tax (PST) at a rate of 7%. 7% of $58 is $4.06.

3. The PST amount is $4.06

4. The price after PST is $58 + $4.06 = $62.06

5. You also need to add Federal Goods and Services Tax (GST) of 5%. 5% of $62.06 is $3.10.

6. The final price with GST added is $62.06 + $3.10 = $65.16

So the final price of the $58 shorts with 7% PST and 5% GST in BC will be $65.16

Let me know if you have any other questions! I'm happy to help explain the steps.

Answer:

Step-by-step explanation:

The final price would be calculated as follows:

- First, calculate the total tax rate by adding the PST and GST: 7% + 5% = 12%

- Next, calculate the amount of tax to be paid on the shorts by multiplying the original price by the tax rate: $58.00 x 12% = $6.96

- Finally, add the tax amount to the original price to get the final price: $58.00 + $6.96 = $64.96

Therefore, the final price for a $58.00 pair of shorts sold in BC with 7% PST and 5% GST would be $64.96.

The value that represents the vertical translation from the graph of the parent function is 3.

What is translation?

A translation is a geometric transformation when each point in a figure, shape, or space is moved in a specific direction by the same amount. A translation can also be thought of as moving the origin of the coordinate system or as adding a constant vector to each point.

Here, we have

Given: function f(x) = x² , g(x)=(x+5)²+3

We have to find the value that the vertical translation from the graph of the parent function f(x) to the graph of the function g(x).

function

We apply the following function transformations:

Horizontal translations:

Suppose that h> 0

To graph y = f (x + h), move the graph of h units to the left:

For h = 5, we have:

f(x+5) = (x+5)²

Vertical translations:

Suppose that k> 0

To graph y = f (x) + k, move the graph of k units up.

For k = 3, we have:

g(x) = (x+5)²+3

Hence, The value that represents the vertical translation from the graph of the parent function is 3.

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Answer:

False

Step-by-step explanation:

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Answer:

False

Step-by-step explanation:

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Answer:

148.3° and 31.7°

Step-by-step explanation:

Supplementary angles add up to 180°

Let x represent the angle we are trying to find

Let y represent the supplementary angle of x

We have x + y = 180  [1]

We are given
x - y = 116.6  [2]

Add both equations. [1] + [2]

x + y + x - y = 180 + 116.6

2x = 296.6

x = 296.6 / 2 = 148.3

y = 180 - 148.3 = 31.7

Therefore the two angles measure 148.3° and 31.7°

Check:
148.3 - 31.7 = 116.6

The distance between the skew lines is √[30625t² - 244000ts + 12864000].

Skew lines are two lines in space that do not intersect and are not parallel. They are not planes, as they do not lie in a single plane.

Given,

x = 3t, y = 16t, z = 2t and x = 32s, y = 614s, z = -35s

We need to find the distance between the skew lines.

To solve this problem, we will use the formula for the distance between two skew lines.

Distance between two skew lines = √[(x₁ - x₂)² + (y₁ - y₂)² + (z₁ - z₂)²]

Substituting the given values in the above formula,

Distance between two skew lines = √[(3t - 32s)² + (16t - 614s)² + (2t - (-35s))²]

= √[(3t - 32s)² + (16t - 614s)² + (2t + 35s)²]

= √[9t² - 64ts + 1024s² + 256t² - 9696st + 38416s² + 4t² + 140ts + 1225s²]

= √[1225t² - 9760ts + 51456s²]

= √[(1225 x 25)t² - 9760ts + 51456 x 25]

= √[30625t² - 244000ts + 12864000]

Therefore, the distance between the skew lines is √[30625t² - 244000ts + 12864000].

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Thus, the area of the rectangular playground is found as 1500 sq. ft.

A parallelogram with four opposing, parallel, congruent sides is referred to as a rectangle. The rectangle's corners are at a right angle. The fact that a rectangle's sides are not all equal is the only distinction between it and a square.

A two-dimensional shape's area is the interior blank space. The quantity of space that a shape occupies is another way to define area. When calculating a rectangle's area, we multiply the length by the width of a rectangle.

Given that-

P = 2(l + w)

160 = 2 (50 + w)

80 = 50 +w

w = 80 - 50

w = 30 feet

area = length* width

area = 50*30

area = 1500 sq. ft

Thus, the area of the rectangular playground is found as 1500 sq. ft.

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Correct question-

The perimeter of the playground shown is 160 feet. Find the area.

Length is 50 ft.

The directional derivative of f at the point (3,2) in the direction of v = (-2,3) is 18/sqrt(13), which is approximately equal to 4.96.

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The maximum amount of revenue the bowling alley can generate is 1,440,000.

The revenue, R, at a bowling alley is given by the equation R = -1(x^2 - 2400x),

where x is the number of frames bowled. We want to find the maximum amount of revenue the bowling alley can

generate.

Recognize that the given equation is a quadratic function in the form of [tex]R = ax^2 + bx + c[/tex].

In this case, a = -1, b = 2400, and c = 0.

To find the maximum revenue, we need to find the vertex of the parabola represented by the quadratic function.

The x-coordinate of the vertex can be found using the formula x = -b / 2a.

Substitute the values of a and b into the formula:

x = -2400 / 2(-1) = 2400 / 2 = 1200.

Now that we have the x-coordinate of the vertex, plug it back into the equation to find the maximum revenue:

R = -1([tex]1200^2[/tex] - 2400 × 1200) = -1(-1440000) = 1,440,000.

The maximum amount of revenue the bowling alley can generate is 1,440,000.

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The shortest distance between the skew lines with parametric equations is |−74s/17 + 23/17|.

To find the distance between the skew lines, we need to find the shortest distance between any two points on the two lines. Let P be a point on the first line with coordinates (1t, 36t, 2t) and let Q be a point on the second line with coordinates (12s, 414s, −35s).

Let's call the vector connecting these two points as v:

v = PQ = <1−2s, 3−10s, 2+5s>

Now we need to find a vector that is orthogonal (perpendicular) to both lines. To do this, we can take the cross product of the direction vectors of the two lines.

The direction vector of the first line is <1, 6, 0> and the direction vector of the second line is <2, 14, −5>. So,

d = <1, 6, 0> × <2, 14, −5>

d = <−84, 5, 14>

We can normalize d to get a unit vector in the direction of d:

u = d / ||d|| = <−84/85, 5/85, 14/85>

Finally, we can find the distance between the two lines by projecting v onto u:

distance = |v · u| = |(1−2s)(−84/85) + (3−10s)(5/85) + (2+5s)(14/85)|

Simplifying this expression yields:

distance = |−74s/17 + 23/17|

Therefore, the distance between the two skew lines is |−74s/17 + 23/17|. Note that the distance is not constant and depends on the parameter s.

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The absolute minimum value of the function is -4390, which occurs at x = -3, and the absolute maximum value of the function is 15620, which occurs at x = 5.

To find the absolute maximum and minimum values of the function f(x)=5x^7−7x^5−5 on the interval [−3,5], we need to evaluate the function at the endpoints of the interval and at any critical points in between.

First, let's find the derivative of the function f(x):

f'(x) = 35x^6 - 35x^4

To find the critical points, we need to solve for f'(x) = 0:

35x^6 - 35x^4 = 0

35x^4(x^2 - 1) = 0

x = 0, ±1

Next, we evaluate f(x) at the endpoints and critical points:

f(-3) = -4390

f(0) = -5

f(1) = -6

f(5) = 15620

Therefore, the absolute minimum value of the function is -4390, which occurs at x = -3, and the absolute maximum value of the function is 15620, which occurs at x = 5.

In summary, the absolute minimum value of f(x) on the interval [-3,5] is -4390 and it occurs at x = -3. The absolute maximum value of f(x) on the interval [-3,5] is 15620 and it occurs at x = 5.

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To solve questions involving standard normal distribution. A distribution describes how frequently each possible outcome of an event occurs in a sample or population and can be represented by a graph, a formula, or a table of values.

a. To find P(Z < -2.51 or Z > 1.76), you need to calculate the individual probabilities and then add them together.

Step 1: Find P(Z < -2.51)
Using a standard normal distribution table or calculator, look for the probability associated with Z = -2.51. You will find P(Z < -2.51) ≈ 0.0062.

Step 2: Find P(Z > 1.76)
Since the normal distribution is symmetric, P(Z > 1.76) = P(Z < -1.76). Using the standard normal distribution table, look for the probability associated with Z = -1.76. You will find P(Z < -1.76) ≈ 0.0392.

Step 3: Add the probabilities together
P(Z < -2.51 or Z > 1.76) = P(Z < -2.51) + P(Z > 1.76) ≈ 0.0062 + 0.0392 = 0.0454.

b. To find P(Z < 1.76 or Z > -2.51), note that this covers the entire range of the distribution. Thus, the probability is equal to 1.

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A unique solution exists when the determinant of the coefficients is non-zero. In this case, the coefficients are 4 and -24s. So, we must ensure that the determinant is not equal to zero.
Determinant = 4 ≠ 0

Since 4 is always non-zero, the determinant is always non-zero, which means the system will have a unique solution for all values of 's'.

In this equation, "s" is a parameter or a variable that can take different values. To determine the values of "s" for which the system has a unique solution, we need to look at the coefficients of the variables x1 and x2.

The system of equations can be written as:

4x1 - 24sx2 = 5

To have a unique solution, the coefficients of x1 and x2 should not be proportional or multiples of each other. In other words, the determinant of the coefficient matrix should not be zero.

The coefficient matrix of the system is:

4 -24s
0  0

The determinant of this matrix is:

4(0) - (-24s)(0) = 0

Therefore, the system has a unique solution when the determinant is not zero, which is when s ≠ 0.

To describe the solution, we can solve for x1 and x2 in terms of s.

From the equation, 4x1 - 24sx2 = 5, we can isolate x1 by adding 24sx2 to both sides:

4x1 = 5 + 24sx2

Dividing both sides by 4, we get:

x1 = 5/4 + 6sx2

We can also isolate x2 by dividing both sides by -24s:

x2 = (4x1 - 5) / (24s)

Substituting x1 in terms of x2, we get:

x2 = (4(5/4 + 6sx2) - 5) / (24s)

Simplifying this equation, we get:

x2 = (5 - 24s^2) / (24s)

Therefore, when s ≠ 0, the solution to the system is:

x1 = 5/4 + 6sx2

x2 = (5 - 24s^2) / (24s)

This solution is unique for any value of s that is not equal to zero.

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We have come to find that the probability P(x < 299) = P(z < -0.505) = 0.3061 (rounded to 4 decimal places).

In statistics, standard deviation is a measure of the amount of variability or dispersion in a set of data. It measures how spread out the data is from the mean or average value.

To calculate the standard deviation, you first find the mean of the data set, then for each data point, you subtract the mean from the data point and square the result. Next, you take the average of all the squared differences, and finally, you take the square root of that average. This gives you the standard deviation of the data set.

To find P(x < 299) for a normally distributed random variable with mean (μ) of 350 and standard deviation (σ) of 101, we need to standardize the variable and use a standard normal distribution table or calculator.

z = (x - μ) / σ # Standardizing the variable

z = (299 - 350) / 101

z = -0.505

Using a standard normal distribution table or calculator, we can find the probability that z is less than -0.505. This probability is 0.3061 (rounded to 4 decimal places).

Therefore, P(x < 299) = P(z < -0.505) = 0.3061 (rounded to 4 decimal places).

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The double integral is equal to the volume of a rectangular prism which is 30.

The given double integral can be written as:

∬_R 3 dA

where R is the region in the xy-plane given by -1 ≤ x ≤ 1 and 3 ≤ y ≤ 8.

To identify this double integral as the volume of a solid, we can consider a solid with constant density 3 occupying the region R. The volume of this solid is then equal to the given double integral.

The solid in question can be visualized as a rectangular prism with a base that is a rectangle in the xy-plane and a height of 1 unit. The base of the prism corresponds to the region R in the xy-plane. The sides of the prism are perpendicular to the xy-plane and extend vertically from the base to a height of 1 unit.

Therefore, the volume of this solid is equal to the given double integral:

∬_R 3 dA

= 3 × (area of R)

= 3 × (2 × 5)

= 30.

Hence, the value of the double integral ∬_R 3 dA over the region R is equal to 30, which is the volume of the solid described above.

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We can observe that ET is the permutation matrix that reverses the permutation of columns performed by E.

Firstly, to generate the permutation matrix E defined in (2), we need to enter the commands provided in the example. This can be done in MATLAB by simply copying and pasting the commands into the command window.

Once we have the permutation matrix E, we can generate a 5 x 5 matrix A with integer entries using the command A = floor(10*rand(5)). This command generates a matrix A with random integers between 0 and 10.

Next, we need to compute the product EA and compare the answer with the matrix A. The product EA is computed in MATLAB by typing E*A. The resulting matrix is related to A by a permutation of its rows. Specifically, the rows of A are rearranged according to the permutation matrix E.

Left multiplication by the permutation matrix E has the effect of permuting the rows of the matrix A. Specifically, the ith row of A is replaced by the row of A corresponding to the ith row of E.

Similarly, we can compute the product AE and compare the answer with the matrix A. The product AE is computed in MATLAB by typing A*E. The resulting matrix is related to A by a permutation of its columns. Specifically, the columns of A are rearranged according to the permutation matrix E.

Right multiplication by the permutation matrix E has the effect of permuting the columns of the matrix A. Specifically, the ith column of A is replaced by the column of A corresponding to the ith column of E.

Moving on to part (b) of the question, we need to compute E-1 and ET. The inverse of the permutation matrix E can be computed in MATLAB using the command inv(E). The transpose of the permutation matrix E can be computed using the command E'.

Observing E-1 and ET, we can see that they are also permutation matrices. This is because the inverse of a permutation matrix is also a permutation matrix, and the transpose of a permutation matrix is also a permutation matrix.

Furthermore, we can observe that E-1 is the permutation matrix that reverses the permutation of rows performed by E. Specifically, the ith row of A is replaced by the row of A corresponding to the ith row of E-1.

Similarly, we can observe that ET is the permutation matrix that reverses the permutation of columns performed by E. Specifically, the ith column of A is replaced by the column of A corresponding to the ith column of ET.

Overall, we can conclude that permutation matrices are a powerful tool in linear algebra, allowing us to manipulate the rows and columns of a matrix in a precise and structured manner.

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The infinite geometric series 1 - 1/5 + 1/25 - 1/125 + ... is convergent and the sum of this infinite geometric series is 5/6.

To determine whether the infinite geometric series is convergent or divergent, and to find its sum if convergent, we'll consider the given series: 1 - 1/5 + 1/25 - 1/125 + ...

Step 1: Identify the common ratio (r).
In a geometric series, each term is a constant multiple of the previous term.

In this case, we can see that the common ratio is -1/5 because each term is obtained by multiplying the previous term by -1/5.

Step 2: Determine convergence or divergence.
An infinite geometric series converges if the absolute value of the common ratio (|r|) is less than 1, and diverges if |r| is greater than or equal to 1.

Since |-1/5| = 1/5 < 1, the series is convergent.

Step 3: Calculate the sum.
For a convergent geometric series, the sum can be found using the formula:
Sum = a / (1 - r)
where 'a' is the first term and 'r' is the common ratio.

In this case, a = 1 and r = -1/5, so:

Sum = 1 / (1 - (-1/5))
Sum = 1 / (1 + 1/5)
Sum = 1 / (6/5)
Sum = 5/6

Therefore, the sum of the infinite geometric series 1 - 1/5 + 1/25 - 1/125 + ... is 5/6.

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The means and variances of the given quantities are.

a. E(X+Y) = 4, Var(X+Y) = 6

b. E(X-Y) = -2, Var(X-Y) = 3

c. E(3X+2Y) = 9, Var(3X+2Y) = 29

d. E(5Y-2X) = 13, Var(5Y-2X) = 21

We can use the following properties of means and variances of linear combinations of random variables

If a and b are constants and X and Y are random variables, then E(aX+bY) = aE(X) + bE(Y).

If X and Y are independent random variables, then Var(X+Y) = Var(X) + Var(Y).

If X and Y are independent random variables and a and b are constants, then Var(aX+bY) = a^2Var(X) + b^2Var(Y).

Using these properties, we can find the means and variances of the given quantities:

a. X+Y

E(X+Y) = E(X) + E(Y) = 1 + 3 = 4

Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y) = 2^2 + 1^2 + 2(0.5)(2)(1) = 6

b. X-Y

E(X-Y) = E(X) - E(Y) = 1 - 3 = -2

Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y) = 2^2 + 1^2 - 2(0.5)(2)(1) = 3

c. 3X+2Y

E(3X+2Y) = 3E(X) + 2E(Y) = 3(1) + 2(3) = 9

Var(3X+2Y) = 3^2Var(X) + 2^2Var(Y) + 2(3)(2)(0.5) = 29

d. 5Y-2X

E(5Y-2X) = 5E(Y) - 2E(X) = 5(3) - 2(1) = 13

Var(5Y-2X) = 5^2Var(Y) + 2^2Var(X) - 2(5)(2)(0.5) = 21

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The amount Jack charges for mowing and raking leaves is $18 and $8 respectively.

Let

charge of mowing = x

charge of raking = y

4x + 10y = 152

6x + 8y = 172

Multiply (1) by 6 and (2) by 4

24x + 60y = 912

24x + 32y = 688

Subtract to eliminate x

60y - 32y = 912 - 688

28y = 224

divide both sides by 28

y = 224/28

y = 8

Substitute into (1)

4x + 10y = 152

4x + 10(8) = 152

4x + 80 = 152

4x = 152 - 80

4x = 72

divide both sides by 4

x = 72/4

x = 18

Therefore, $18 is charged for mowing and $8 is charged for raking.

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Answer:

The volume of Rectangular Prism A is greater than the volume of Rectangular Prism B

Step-by-step explanation:

Volume = area x height

Rectangle A:

Area = 12 x 8

Area = 96 in^2

Volume = 96 x 20

Volume = 1920 in^3

Rectangle B:

Area = 84 in^2

Volume = 84 x 20

Volume = 1680 in^3

Answer: Volume A is bigger than B

Step-by-step explanation:

V(A)= length x width x height =(20)(12)(8)=1920

V(B)=height x base =(20)(84)=1680

So Volume A is bigger than Volume B

To find the coordinate matrix of x in Rn relative to the standard basis, we need to express x as a linear combination of the standard basis vectors. In R2, the standard basis vectors are e1 = (1,0) and e2 = (0,1).

We can write x as:

x = 7(1,0) - 6(0,1)

This means that the coordinate matrix of x in R2 relative to the standard basis is:

[x] = [7 -6]

Note that the first column corresponds to the coordinate of x with respect to e1, and the second column corresponds to the coordinate of x with respect to e2.

To find the coordinate matrix of the vector x in R^n relative to the standard basis, you simply need to represent the vector x as a column matrix using its given components. In this case, x = (7, -6), so the coordinate matrix of x relative to the standard basis is:

[ 7 ]
[ -6 ]

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Therefore, a basis for the column space (image) of P is given by:
{[1; 0; 0], [0; 0; 1]}

The orthogonal projection onto the xz-plane in R3 can be represented by the transformation matrix

[tex]P = [1 0 0; 0 0 0; 0 0 1].[/tex]

To find the kernel and image of this transformation, we can solve for the null space and column space of P.

Null space (kernel) of P:
To find the null space of P, we need to solve the equation Px = 0. This is equivalent to the system of equations:
x1 = 0
x3 = 0
where [tex]x = [x_1; x_2; x_3][/tex]is a vector in R3. The solutions to this system form the kernel of P. We can see that any vector in the xz-plane will satisfy this system since x2 can take any value. Therefore, a basis for the kernel is given by:
{[0; 1; 0]}

Column space (image) of P:
To find the column space of P, we need to determine the span of its columns. Since the second column of P is zero, we only need to consider the first and third columns. These are the standard basis vectors for R3 in the xz-plane. Therefore, a basis for the column space (image) of P is given by:
{[1; 0; 0], [0; 0; 1]}

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Answer:  25% gain

Step-by-step explanation:

math :)