Solve for the series with center at 0 (Present up to 6th term): 1.1.e-2x 1.3. sin(2x – 1) 1 1.4. 1+3x2 1.2.cos(3x)

The value of F(3) -F(0). d) In which area is the probability of entering F (3) -F (0) is "0 ≤ X < 3".

a) The function of distribution by regions is:

For x < -4: F(x) = 0

For -4 ≤ x < -2: F(x) = 0.2

For -2 ≤ x < 0: F(x) = 0.2 + 0.1 = 0.3

For 0 ≤ x < 1: F(x) = 0.3 + 0.25 = 0.55

For 1 ≤ x < 3: F(x) = 0.55 + 0.2 = 0.75

For 3 ≤ x < 5: F(x) = 0.75 + 0.2 = 0.95

For x ≥ 5: F(x) = 1

b) Expressing the result obtained as  F(x) =:

F(x) = {0, x < -4

0.2, -4 ≤ x < -2

0.3, -2 ≤ x < 0

0.55, 0 ≤ x < 1

0.75, 1 ≤ x < 3

0.95, 3 ≤ x < 5

1, x ≥ 5

c) F(3) - F(0) = 0.95 - 0.55 = 0.4

d) The probability of entering F(3) - F(0) is the probability of the random variable falling between 0 and 3. This can be expressed as:

P(0 ≤ X < 3) = F(3) - F(0) = 0.4

Therefore, the answer is "0 ≤ X < 3".

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The measure of the central angle m ∠JIK is 160°.

Given that the circle I, in which IJ is the radius of 9 units, the area of shaded sector = 36π, we need to find the m ∠JIK, the central angle.

Area of the sector = central angle / 360° × π × radius²

∴ 36π = m ∠JIK / 360° × π × 9²

m ∠JIK = 360° × 4 / 9

m ∠JIK = 160°

Hence, the measure of the central angle m ∠JIK is 160°.

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To include two vectors u and v in component shape, you include the comparing components. The equation is: u + v = (u₁ + v₁, u₂ + v₂, u₃ + v₃)

where u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃) are the component vectors of u and v, separately.

So also, to subtract two vectors u and v in the component frame, you subtract the comparing components. The equation is:

u - v = (u₁ - v₁, u₂ - v₂, u₃ - v₃)

where u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃) are the component vectors of u and v, separately.

For illustration, on the off chance that u = (1, 2, -3) and v = (4, -2, 5), at that point u + v = (1 + 4, 2 - 2, -3 + 5) = (5, 0, 2) and u - v = (1 - 4, 2 + 2, -3 - 5) = (-3, 4, -8). 

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Complete Question: To add or subtract vectors in component form, you simply add or subtract the corresponding components, For example to add two vectors u and v, you can use the formula u v. Describe the component vectors of u and v.

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Answer: 720

Step-by-step explanation:

there are 6 letters in "choice"

and they must be permuted into 6 word letters:

6P6 = 720 distinct possibilities

The balance be 8 years from now will be :

A = $1,789.124

Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is different from simple interest, where interest is not added to the principal while calculating the interest during the next period.

In this problem we are going to apply the compound interest formula

[tex]A= P(1+r)^t[/tex]

A = final amount

P = initial principal balance

r = interest rate

t = number of time periods elapsed

P= $1,527.00

R= 2%= 2/100= 0.02

T= 8 years

[tex]A = 1,527.00(1+0.02)^8[/tex]

A = $1,789.124

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Using the Alternate angles theorem, the value of x in the given diagram is 11

From the question, we are to calculate the value of x that will make A || B

From the Alternate angles theorem which states that when two parallel lines are cut by a transversal, then the resulting alternate interior angles or alternate exterior angles are congruent.

In the given diagram,

The angle measures (4x + 41) and (6x + 19) are alternate interior angle measures

Thus.

For A to be parallel to B (A || B)

4x + 41 = 6x + 19

Solve for x

4x + 41 = 6x + 19

41 - 19 = 6x - 4x

22 = 2x

Divide both sides by 2

22 / 2 = x

11 = x

x = 11

Hence,

The value of x is 11

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The integral test states that if a series is a sum of terms that are positive and decreasing, and if the terms of the series can be expressed as the values of a continuous and decreasing function, then the series converges if and only if the corresponding improper integral converges.

Let's apply the integral test to the given series. We need to find a continuous, positive, and decreasing function f(x) such that the series is the sum of the values of f(x) for x ranging from 2 to infinity.

For the first series, we have:

∑n=2∞n(lnn)p

Let f(x) = x(lnx)p. Then f(x) is continuous, positive, and decreasing for x ≥ 2. Moreover, we have:

f'(x) = (lnx)p + px(lnx)p-1

f''(x) = (lnx)p-1 + p(lnx)p-2 + p(lnx)p-1

Since f''(x) is positive for x ≥ 2 and p > 0, f(x) is concave up and the trapezoidal approximation underestimates the integral. Therefore, we have:

∫2∞f(x)dx = ∫2∞x(lnx)pdx

Using integration by substitution, let u = lnx, then du = 1/x dx. Therefore:

∫2∞x(lnx)pdx = ∫ln2∞u^pe^udu

Since the exponential function grows faster than any power of u, the integral converges if and only if p < -1.

For the second series, we have:

∑n=2∞1/n(ln⁡n)²

Let f(x) = 1/(x(lnx)²). Then f(x) is continuous, positive, and decreasing for x ≥ 2. Moreover, we have:

f'(x) = -(lnx-2)/(x(lnx)³)

f''(x) = (lnx-2)²/(x²(lnx)⁴) - 3(lnx-2)/(x²(lnx)⁴)

Since f''(x) is negative for x ≥ 2, f(x) is concave down and the trapezoidal approximation overestimates the integral. Therefore, we have:

∫2∞f(x)dx ≤ ∑n=2∞f(n) ≤ f(2) + ∫2∞f(x)dx

where the inequality follows from the fact that the series is the sum of the values of f(x) for x ranging from 2 to infinity.

Using the comparison test, we have:

∫2∞f(x)dx = ∫ln2∞(1/u²)du = 1/ln2

Therefore, the series converges if and only if p > 1.

In summary, the series ∑n=2∞n(lnn)p converges if and only if p < -1, and the series ∑n=2∞1/n(ln⁡n)² converges if and only if p > 1. For values of p such that -1 ≤ p ≤ 1, the series diverges.
To find the values of p for which the series converges or diverges using the integral test, we will first write the series and then perform the integral test.

The given series is:

∑(n=2 to infinity) [1/n(ln(n))^p]

Now, let's consider the function f(x) = 1/x(ln(x))^p for x ≥ 2. The function is continuous, positive, and decreasing for x ≥ 2 when p > 0.

We will now perform the integral test:

∫(2 to infinity) [1/x(ln(x))^p] dx

To evaluate this integral, we will use the substitution method:

Let u = ln(x), so du = (1/x) dx.

When x = 2, u = ln(2).
When x approaches infinity, u approaches infinity.

Now the integral becomes:

∫(ln(2) to infinity) [1/u^p] du

This is now an integral of the form ∫(a to infinity) [1/u^p] du, which converges when p > 1 and diverges when p ≤ 1.

So, for the given series:

- It converges when p > 1.
- It diverges when p ≤ 1.

In conclusion, using the integral test, the series ∑(n=2 to infinity) [1/n(ln(n))^p] converges for values of p > 1 and diverges for values of p ≤ 1.

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If it is known that the cardinality of the set A X A is 16. Then the cardinality of A is: option d. 4

Cardinality refers to the number of elements or values in a set. It represents the size or count of a set. In other words, cardinality is a measure of the "how many" aspect of a set. We know that the cardinality of A X A is 16, which means that there are 16 ordered pairs in the set A X A. Each ordered pair in A X A consists of two elements, one from A and one from A. So, the total number of possible pairs of elements in A is the square root of 16, which is 4. Therefore, the cardinality of A is 4. So, the answer is d. 4.

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P1 is 51.9 bpm and P99 is 103.1 bpm.

The probability that a school's average female pulse rate is between 70 and 85 bpm is approximately 1.

a) The distribution of the average pulse rates is expected to follow a normal distribution.

b) To find the percentiles P1 and P99, we can use the z-score formula:

z = (x - μ) / σ

where x is the pulse rate, μ is the population mean (77.5 bpm), and σ is the population standard deviation (11.6 bpm).

For P1, we want to find the pulse rate such that only 1% of the population has a lower pulse rate. Using a standard normal distribution table or calculator, we find that the z-score corresponding to the 1st percentile is -2.33. Thus,

-2.33 = (x - 77.5) / 11.6

Solving for x, we get:

x = 51.9 bpm

For P99, we want to find the pulse rate such that only 1% of the population has a higher pulse rate. Using the same method as above, we find that the z-score corresponding to the 99th percentile is 2.33. Thus,

2.33 = (x - 77.5) / 11.6

Solving for x, we get:

x = 103.1 bpm

Therefore, P1 is 51.9 bpm and P99 is 103.1 bpm.

c) The mean of a sample of n = 36 female high school seniors is estimated to be the same as the population mean of 77.5 bpm. The standard deviation of the sample is estimated to be:

s = σ / sqrt(n) = 11.6 / sqrt(36) = 1.933 bpm

d) To find the probability that a school's average female pulse rate is between 70 and 85 bpm when the sample size is n = 25, we first need to calculate the standard error of the mean:

SE = s / sqrt(n) = 1.933 / sqrt(25) = 0.387 bpm

Next, we find the z-scores for 70 and 85 bpm:

z1 = (70 - 77.5) / 0.387 = -19.34

z2 = (85 - 77.5) / 0.387 = 19.34

Using a standard normal distribution table or calculator, we find that the area to the left of z1 is essentially 0 and the area to the left of z2 is essentially 1. Therefore, the probability that a school's average female pulse rate is between 70 and 85 bpm is approximately 1.

Shading the area of interest on a normal probability curve would show the entire curve as it represents the entire population of high school female seniors, but the area of interest would be shaded in the middle of the curve.

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3. Using  slope and/or the distance formula .The most precise name for the

figures A(-6, -7), B(-4,-2), C(2,-1), D(0) is: [A] rectangle.

4.  Using slope and/or the distance formula. The most precise name for the

figure: A(-3,-5), B(4, -2), C(7, -9), D(0,-12) is: [D] kite.

3. We must look at the sides and angles characteristics to identify the figure's name. We can plot the four points on a graph to see how the figure appears since we have four points.

When we plot the points on a graph we can see that BC and AD and AB and CD have the same length. In addition every angle is 90 degrees. The figure is a rectangle.

Therefore the correct option is A.

4. Once more we can graph the points and analyze the sides and angles characteristics.

Since AB, BC, and CD have different lengths when the points are plotted on a graph the figure is neither a rhombus nor a square. The figure is a kite since the diagonals AC and BD both connect at a straight angle.

Therefore the correct option is D.

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The critical region lies In the lower 5% of the null distribution.

Option E is the correct answer.

We have,
When our alternative hypothesis is mu < 1.2, it means we are testing if the population mean is less than 1.2.

The critical region is the area in the null distribution where we reject the null hypothesis.

Since our alternative hypothesis is a one-tailed test (less than), the critical region will be in the tail of the null distribution on the left side.

If alpha is .05, it means we want to reject the null hypothesis if the probability of observing our sample mean is less than 5% under the null distribution.

This corresponds to the lower 5% of the null distribution, which is our critical region.

Therefore, any sample mean that falls in the lower 5% of the null distribution will lead to rejection of the null hypothesis and acceptance of the alternative hypothesis that mu < 1.2.

Thus,

The critical region lies In the lower 5% of the null distribution.

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Answer: Y-intercept:

Axis of symmetry: X = - 1

Vertex: Y = 2(X + 1)^2-5

Maximum: -1

Minimum: - 5

Domain:

(−∞,∞),{x|x∈R}

Range:

[−5,∞),{y|y≥−5}

Step-by-step explanation:

So the perfect square factors of 792 are 4, 9, 36, 144, 484, and 1089.

The prime factorization of 792 by dividing it by its smallest prime factor repeatedly:

792 ÷ 2 = 396

396 ÷ 2 = 198

198 ÷ 2 = 99

99 ÷ 3 = 33

33 ÷ 3 = 11

So the prime factorization of 792 is [tex]2^3 * 3^2 * 11.[/tex]

To find the perfect square factors, we can look for pairs of factors where both factors have even exponents.

The factors of 792 are: 1, 2, 3, 4, 6, 8, 11, 12, 18, 22, 24, 33, 36, 44, 66, 72, 88, 132, 198, 264, 396, and 792.

The perfect square factors of 792 are:

[tex]2^2 = 4\\2^2 * 3^2 = 36\\2^2 * 11^2 = 484\\3^2 = 93^2 * 4^2 = 144\\3^2 * 11^2 = 1089[/tex]

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The series converges for all values of p < -1, and diverges for all values of p ≥ -1.

To apply the Integral Test, we need to verify that the terms of the series are positive and decreasing for all n greater than some fixed integer. For this series, note that the terms are positive since both the base and the natural logarithm are positive. To show that the terms are decreasing, we take the ratio of successive terms:

[tex]a(n+1)/a(n) = [(n+1)ln(n+1)]^p / [nln(n)]^p[/tex]

[tex]= [(n+1)/n]^p * [(1+1/n)ln(1+1/n)]^p[/tex]

Since (n+1)/n > 1 and ln(1+1/n) > 0 for all n, it follows that the ratio is greater than 1 and therefore the terms are decreasing.

To use the Integral Test, we need to find a function f(x) such that f(n) = a(n) for all n and f(x) is positive and decreasing for x ≥ 3. A natural choice is [tex]f(x) = x(ln(x))^p[/tex]. Note that f(n) = a(n) for all n and f(x) is positive and decreasing for x ≥ 3. Then we have:

integral from 3 to infinity of f(x) dx = integral from 3 to infinity of x(ln(x))^p dx

To evaluate this integral, we use integration by substitution with u = ln(x):

[tex]du/dx = 1/x, dx = x du[/tex]

So the integral becomes:

integral from ln(3) to infinity of [tex]u^p e^u du[/tex]

This integral converges for p < -1, by the Integral Test for Improper Integrals.

Therefore, the series converges for all values of p < -1, and diverges for all values of p ≥ -1.

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The value of S° in the given adjacent angles would be = 26.7°

Adjacent angles are those angles that are found on the same side of the plane and they share a common vertex.

The adjacent angles are different from the supplementary angles which are angles found in the same side but when measured together sums up to 180°.

The angles 41.6° and S° are two angles that share the same vertex with the sum of 68.3°

Therefore, S° which is the second part of the adjacent angles would be = 68.3+41.6 = 26.7°

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The value of j is equal to -9.

In Mathematics and Geometry, the slope of any straight line can be determined by using the following mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = rise/run

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

By substituting the given data points into the formula for the slope of a line, we have the following;

10 = (-5 - 5)/(-10 - j)

10(-10 - j) = -10

(-10 - j) = -1

j = -10 + 1

j = -9

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Complete Question:

A line with a slope of 10 passes through the points (j, 5) and (-10,-5). What is the value of j?

Answer:

No solution

Step-by-step explanation:

Simplifies to

15-12x=18-12x

15=18

No solution

The new dimensions of the shape that is being dilated by the scale factor of 3 would be = 180 by 270.

To calculate the new dimensions of a shape, the formula for a scale factor can be used.

Scale factor = Bigger dimensions/smaller dimensions

Scale factor = 3

Length of bigger dimension = 60

width = 90

Dilated length = 60×3 = 180

width of dilated shape = 90×3= 270

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Answer:

11

Step-by-step explanation:

Volume of right cone = (1/3) · π · r² · h

V = 968π

h = 24 units

Let's solve

968π = (1/3) · π · r² · 24

2904π =  π · r² · 24

121π = π · r²

121 = r²

r = 11

So, the radius is 11 units

The value of x in the triangle is 21, option B is correct.

The given triangle is right triangle

We know that the sine function is the ratio of opposite side and hypotenuse

Opposite side =19

Hypotenuse =x

We have to find the value of x

Sin 65 = 19/x

0.91 =19/x

x=19/0.91

x=20.8

x=21

Hence, the value of x in the triangle is 21, option B is correct.

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The probability that the sample mean would be less than 132.25 liters is 0.0564 (or 5.64% when expressed as a percentage), rounded to four decimal places.

We can use the central limit theorem to approximate the distribution of the sample mean as normal with a mean of 138 liters and a standard deviation of 28/sqrt(60) liters.

z = (132.25 - 138) / (28 / sqrt(60)) = -1.5811

Using a standard normal distribution table or a calculator, we can find the probability of getting a z-score less than -1.5811. The probability is approximately 0.0564.

Therefore, the probability that the sample mean would be less than 132.25 liters is 0.0564 (or 5.64% when expressed as a percentage), rounded to four decimal places.

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The summation notation [tex]\sum\limits^{2}_{n = 0} (-52 + n)[/tex] when evaluated has a value of -153

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

[tex]\sum\limits^{2}_{n = 0} (-52 + n)[/tex]

This means that we substitute 0 to 2 for n in the expression and add up the values

So, we have

[tex]\sum\limits^{2}_{n = 0} (-52 + n) = (-52 + 0) + (-52 + 1) + (-52 + 2)[/tex]

Evaluate the sum of the expressions

So, we have

[tex]\sum\limits^{2}_{n = 0} (-52 + n) = -153[/tex]

Hence, the solution is -153

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The null hypothesis is H₀: p ≥ 0.2 (the return rate is greater than or equal to 20%). The alternative hypothesis is H₁: p < 0.2 (the return rate is less than 20%).

We will first identify the null hypothesis (H₀) and alternative hypothesis (H₁). The null hypothesis represents the assumption that there is no significant difference or effect. In this case, the return rate is assumed to be equal to 20%. The alternative hypothesis represents the claim we want to test, which is that the return rate is less than 20%. So, the null hypothesis and alternative hypothesis are: H₀: p = 0.2 H₁: p < 0.2 Based on the provided options,

The correct answer is: OB. H₀: p = 0.2 OC. H₁: p < 0.2

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The transformations that move the pentagon FGHIJ onto the pentagon F'G'H'I'J' includes a rotation and a reflection. The correct option therefore is the option C.

C. Rotation, reflection

A rotation transformation is one in which a geometric figure is rotated about a fixed point or location, known as the center of rotation.

The coordinates of the image are; I(-5, 4), H(-2, 4), G(-1. 3), F(-2, 0), and J(-4, 1)

The coordinates of a point, (x, y), following a 90 degrees clockwise rotation are; (y, -x)

Therefore, the coordinates of the image of the after a 90 degrees rotation are; I'(4, 5), H'(4, 2), G'(3, 1), F'(0, 2), J'(1, 4)

The coordinates of the point on the preimage, (x, y), following a reflection over the y-axis is the point (x, -y), therefore;

The coordinates of the image of the figure F'G'H'I'J' after a reflection over the x-axis are;
I''(4, -5), H''(4, -2), G''(3, -1), F''(0, -2), and J''(1, -4)

The above points corresponds to the coordinates of the figure, F'G'H'I'J' in the diagram, therefore;

The series of transformations that maps the pentagon FGHIJ onto the pentagon F'G'H'I'J' is a 90 degrees clockwise rotation and a reflection over the y-axis. Option C is the correct option

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The local truncation error incurred in the approximation of y(2.01) using the next term in the corresponding Taylor series is O(0.0001)

We can use the Taylor series method of order three to estimate y(2.01) in one step. Let's first write the Taylor series expansion of y(x) about x=2 up to the third derivative:

[tex]y(x) = y(2) + (x-2)y'(2) + \frac{(x-2)^{2} }{2!} y''(2) + \frac{(x-2)^{3} }{2!} y'''(2) + O((x-2)^{4} )[/tex]

where  [tex]y'(x) = 3x^{2} y(x), y''(x) = 6xy(x) + 3x^{2} y'(x), y'''(x) = 9x^{2} y'(x) + 18xy'(x) + 6x^{2} y''(x).[/tex]

(a) To estimate y(2.01) in one step, we need to evaluate the above expression at x=2.01. Using y(2) = 3 and [tex]y'(2) = 3(2)^{2} (3) = 36[/tex], we get:

[tex]y(2.01) = y(2) + (2.01-2)y'(2) + \frac{(2.01-2)^{2} }{2!} y''(2)[/tex]

[tex]= 3 + 0.01(36) + \frac{(0.01)^{2} }{2!} (6(2)(3) + 3(2)^{2} (3)(3))[/tex]

=3.1089

Therefore, y(2.01) =3.1089.

(b) The local truncation error is given by the next term in the Taylor series expansion, which is O((x-2)⁴) in this case. Evaluating this term at x=2.01, we get:

O((2.01-2)⁴) = O(0.0001)

Therefore, the local truncation error incurred in the approximation of y(2.01) using the next term in the corresponding Taylor series is O(0.0001).

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The expected profit is $67.5K

To determine what ABC should do to maximize expected profits, we can use a decision tree to analyze the different possible outcomes and their probabilities.

First, let's define the events and their probabilities:

H: the show is a hit (P(H) = 0.25)

F: the show is a flop (P(F) = 0.75)

PH: the market research firm predicts a hit (P(PH|H) = 0.9, P(PH|F) = 0.2)

PF: the market research firm predicts a flop (P(PF|H) = 0.1, P(PF|F) = 0.8)

Using these probabilities, we can construct the following decision tree:

               / PH: P = 0.225 (0.25 * 0.9)

              /

             /

            /

    H: P = 0.25

           \

            \

             \ PF: P = 0.025 (0.25 * 0.1)

              \

               \

                \

                 \

                  \

                   \ PH: P = 0.15 (0.75 * 0.2)

                    \

                     \

                      F: P = 0.75

                     /

                    /

           PF: P = 0.6 (0.75 * 0.8)

starting from the top of the tree, we can calculate the expected profits for each decision:

If ABC launches the show without doing the market research, the expected profit is:

E1 = P(H) * $400K + P(F) * (-$100K) = $75K

If ABC does the market research and it predicts a hit, the expected profit is:

E2 = P(H and PH) * $400K - $40K + P(F and PH) * (-$40K) = $89K

If ABC does the market research and it predicts a flop, the expected profit is:

E3 = P(H and PF) * $400K - $40K + P(F and PF) * (-$100K - $40K) = -$52K

Therefore, the decision that maximizes expected profits is to do the market research and launch the show only if the market research predicts a hit.

The expected profit in this case is:

E = P(H and PH) * $400K - $40K + P(F and PH) * (-$40K) = 0.225 * $400K - $40K + 0.15 * (-$40K) = $67.5K

Therefore, the expected profit is $67.5K.

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