graph the following inequality

Simplify, [tex]3x \times 2y+5y[/tex]

Using order of operations, PEMDAS

[tex]P = > Parenthesis\\E= > Exponents \\M > Multiplication\\D= > Division\\A= > Addition\\S= > Subtraction[/tex]

[tex]\Longrightarrow 3x \times 2y+5y \Longrightarrow (3x \times 2y)+5y \Longrightarrow(6xy)+5y \Longrightarrow \boxed{6xy+5y}[/tex]

Thus, 6xy+5y is the correct simplified form.

The left and right Riemann sums are same which is equal to 35.

The given function is,

f(x) = 9 - (x - 3)²

Interval is [0, 6]

This can be divided in to 6 subintervals [0, 1], [1, 2], [2, 3], ....., [5, 6].

Δx = (6 - 0) / 6 = 1

[tex]x_i[/tex] = a + Δx (i - 1), where [a, b] is the interval.

x1 = 0 + (1 × (1 - 1) = 0

x2 = 1, x3 = 2, x4 = 3, x5 = 4 and x6 = 5.

Left Riemann sum = 1. f(0) + 1. f(1) + ..... + 1. f(5)

                                        = 0 + 5 + 8 + 9 + 8 + 5

                                        = 35

Similarly for right Riemann sum,

[tex]x_i[/tex] = a + Δx i, where [a, b] is the interval.

x1 = 1, x2 = 2, x3 = 3, x4 = 4, x5 = 5 and x6 = 6

Right Riemann sum = 1. f(1) + ..... + 1. f(5) + 1. f(6)

                                         = 5 + 8 + 9 + 8 + 5 + 0

                                         = 35

Hence both the sums are equal to 35.

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No, R is not transitive.
Explanation: A relation R on a set X is considered transitive if, for every pair of elements (x, y) and (y, z) in R, there exists a pair (x, z) in R. In this case, X = {a, b, c, d}, and R = {(a, b), (b, c), (c, a)}.

To show that R is not transitive, we need to find pairs that violate the transitivity condition. We have the pairs (a, b) and (b, c) in R, but we don't have the pair (a, c) in R. Therefore, R is not transitive.

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The expected ACV for the new anti-dandruff, 2-in-1 conditioning product is 1,266,500 households.

The expected trial volume for the new product is 312,900 households.

The expected repeat volume for the new product is 119,082 households.

The expected total volume for the new product in the simulated test market is 431,982 households.

What is the expected ACV (All Commodity Volume) for the new anti-dandruff, 2-in-1 conditioning product in the simulated test market?

The expected ACV% for the new product is 85%, which means that the product is expected to be available in 85% of the stores in the test market.

Assuming that the product will be equally available in all the households in the test market, the expected ACV can be calculated as follows:

Expected ACV = 85% of 1,490,000 = 1,266,500 households

The expected ACV for the new anti-dandruff, 2-in-1 conditioning product is 1,266,500 households.

What is the expected trial volume for the new product in the simulated test market?

The expected trial rate for the new product is 21%. Assuming that all households in the test market have an equal probability of trying the new product, the expected trial volume can be calculated as follows:

Expected trial volume = 21% of 1,490,000 = 312,900 households

The expected trial volume for the new product is 312,900 households.

What is the expected repeat volume for the new product in the simulated test market?

The expected repeat purchase rate for the new product is 38%. Assuming that all households that tried the new product have an equal probability of making a repeat purchase, the expected repeat volume can be calculated as follows:

Expected repeat volume = 38% of 312,900 = 119,082 households

The expected repeat volume for the new product is 119,082 households.

What is the expected total volume (trial + repeat) for the new product in the simulated test market?

The expected total volume for the new product can be calculated as the sum of the expected trial volume and the expected repeat volume:

Expected total volume = Expected trial volume + Expected repeat volume

= 312,900 + 119,082

= 431,982 households

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The value of the integral is:
∫[(36 – x2)^(5/2)]dx = -9/2 [(1 - x^2/36)^(5/2)] + C.

To evaluate the integral [(36 – x2)5/2dx, we can use the substitution method. Let u = 36 - x^2, then du/dx = -2x dx, and dx = -du/(2x).
Substituting, we have:
∫[(36 – x^2)^(5/2)]dx = ∫u^(5/2) (-du/(2x))
= (-1/2) ∫u^(5/2)/x du
Now, we need to express x in terms of u. From the original substitution, we have:
u = 36 - x^2
x^2 = 36 - u
x = ±(36 - u)^(1/2)
We will use the positive root for simplicity. Substituting, we have:
x = (36 - u)^(1/2)
Now, we can express x in terms of u and substitute back into the integral:
∫[(36 – x^2)^(5/2)]dx = (-1/2) ∫u^(5/2)/[(36 - u)^(1/2)] du
To evaluate this integral, we can use a trigonometric substitution. Let u = 36 sin^2 θ, then du/dθ = 72 sin θ cos θ dθ, and du = 36(2 sin θ cos θ) dθ = 18 sin 2θ dθ.
Substituting, we have:
∫u^(5/2)/[(36 - u)^(1/2)] du = ∫[(36 sin^2 θ)^(5/2)]/[(36 cos^2 θ)^(1/2)] (18 sin 2θ) dθ
= 18 ∫[sin^5 θ]/cos θ dθ
We can use the substitution v = cos θ, then dv/dθ = -sin θ, and sin θ = ±(1 - v^2)^(1/2). We will use the positive root for simplicity. Substituting, we have:
sin^5 θ = (1 - v^2)^(5/2)
dθ = -dv/(1 - v^2)^(1/2)
Substituting back into the integral, we have:
∫[sin^5 θ]/cos θ dθ = -∫(1 - v^2)^(3/2) dv
= (1/2) (1 - v^2)^(5/2) + C
Substituting back u = 36 sin^2 θ and v = cos θ, we have:
∫[(36 – x^2)^(5/2)]dx = (-1/2) 18 [(1/2) (1 - cos^2 θ)^(5/2)] + C
= -9/2 [(1 - cos^2 θ)^(5/2)] + C
= -9/2 [(1 - (1 - u/36))^(5/2)] + C
= -9/2 [(u/36)^(5/2)] + C
Finally, substituting u = 36 - x^2, we have:
∫[(36 – x^2)^(5/2)]dx = -9/2 [(36 - x^2)/36]^(5/2) + C
= -9/2 [(1 - x^2/36)^(5/2)] + C

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We have that the 80% of this type of seeds germinate, if we plant 90 seeds, the 80% is: 90 * 80/100 = 72

Then we know that 72 seeds will germinate.

a) The probability that fewer than 75 seeds germinate is 1 or 100%, having in count that at least 72 seeds will germinate.

Then the correct answer is 1 (100%)

b) The probability of 80 or more seeds germinating is 0, again, having in mind the percent of seeds that germinate. In other words, as just 72 of 90 seeds will germinate, it's impossible that 80 or more seeds will germinate.

Then the correct answer is 0 (0%).

c) To approximate the probability that the number of seeds germinated is between 67 and 75 is the average of the probability that 67 seeds have been germinated and the maximum probability because 72 are the seed that will germinate.

Then the correct answer is 0.965

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The area under the curve to the right of 64 is 0.2514.

To find the area under the curve to the right of 64 in a normal distribution with mean μ = 60 and standard deviation σ = 6, follow these steps:

1. Calculate the z-score: z = (x - μ) / σ, where x = 64.
  z = (64 - 60) / 6
  z = 4 / 6
  z ≈ 0.67

2. Use a z-table or calculator to find the area to the left of z = 0.67.
  The area to the left of z = 0.67 is approximately 0.7486.

3. Subtract the area to the left from 1 to find the area to the right of z = 0.67.
  Area to the right = 1 - 0.7486
  Area to the right ≈ 0.2514

So, the area under the curve to the right of 64 in this normal distribution is approximately 0.2514.

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The general solution to the differential equation is y = 1/(Cx + xln|x|)

One way to solve this equation is to use Bernoulli's equation, which is a technique used to solve a specific type of nonlinear differential equation. Bernoulli's equation has the form:

dy/dx + P(x)y = Q(x)y

where P(x) and Q(x) are functions of x and n is a constant. The given equation can be transformed into this form by dividing both sides by y and setting n = 1:

x(dy/dx)y⁻¹ - (x/y + y)y⁻¹ = 0

Letting v = y⁻¹, we can rewrite this equation as:

-x(dv/dx) + (1/x + v) = 0

This is now in the form of Bernoulli's equation with P(x) = 1/x and Q(x) = 1. To solve this equation, we first divide both sides by x:

-(dv/dx) + (1/x²)(1 + xv) = 0

Next, we make the substitution u = xv, which gives:

x(dv/dx) + u = 1

This is a linear differential equation, which can be solved using standard methods. We first find the integrating factor:

exp(integral of 1/x dx) = exp(ln|x|) = |x|

Multiplying both sides by |x|, we get:

|x|*(dv/dx) + |x|u = |x|

Next, we integrate both sides with respect to x:

|x|*v = |x|*ln|x| + C

Substituting back for v = 1/y, we get:

y = 1/(Cx + xln|x|)

where C is the constant of integration.

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There will be an average of approximately 43.38 grams of radium in the vault during the next 600 years.

We need to calculate the average amount of radium in the vault during the next 600 years. Given the half-life of radium is 1690 years and the initial amount is 50 grams, we can use the radioactive decay formula:
Final Amount = Initial Amount * (1/2)^(time elapsed/half-life)
After 600 years, the amount of radium left in the vault will be:
Final Amount = 50 * (1/2)^(600/1690)
Final Amount ≈ 50 * (1/2)^0.35503 ≈ 36.75979727 grams
Now, to find the average amount of radium during the 600 years, we can take the average of the initial amount and the final amount:
Average Amount = (Initial Amount + Final Amount) / 2
Average Amount = (50 + 36.75979727) / 2 ≈ 43.37989864 grams
Therefore, there will be an average of approximately 43.38 grams of radium in the vault during the next 600 years.

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To determine the area under the standard normal curve that lies to the right of a certain Z-value, we need to use a standard normal distribution table or calculator. This table or calculator gives us the area to the left of a Z-value. Therefore, to find the area to the right of a Z-value, we need to subtract the area to the left from 1.

(a) For Z=-1.89, the area to the left is 0.0301. Therefore, the area to the right is 1-0.0301=0.9699.
(b) For Z=-0.02, the area to the left is 0.4920. Therefore, the area to the right is 1-0.4920=0.5080.
(c) For Z=-0.93, the area to the left is 0.1762. Therefore, the area to the right is 1-0.1762=0.8238.
(d) For Z=1.46, the area to the left is 0.9265. Therefore, the area to the right is 1-0.9265=0.0735.

Therefore, the areas under the standard normal curve that lie to the right of Z=-1.89, Z=-0.02, Z=-0.93, and Z=1.46 are 0.9699, 0.5080, 0.8238, and 0.0735, respectively.

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Answer:

For a day VFR flight in a Piper Archer, you would need a fuel reserve of approximately 5 gallons.

Step-by-step explanation:

The amount of fuel reserve that must be carried on a day VFR (Visual Flight Rules) flight, the FAA requires a minimum fuel reserve of 30 minutes at cruising speed for day VFR flights. To calculate the gallons needed for a Piper Archer, follow these steps:

1. Determine the fuel consumption rate of the Piper Archer at cruising speed, which is typically around 10 gallons per hour (GPH).
2. Divide the required fuel reserve minutes (30) by the minutes in an hour (60): 30 / 60 = 0.5 hours.
3. Multiply the fuel consumption rate (10 GPH) by the required fuel reserve time in hours (0.5 hours): 10 * 0.5 = 5 gallons.

Therefore, for a day VFR flight in a Piper Archer, you would need a fuel reserve of approximately 5 gallons.

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The point estimate used to build this confidence interval is 7.

Given is a confidence interval: (4, 10), we need to find the point estimate used to build this confidence interval,

The two ends of the confidence interval are both at same distance away from.

Knowing this we can find the point that is directly in-between these points.

4 + 10 / 2 = 14 / 2 = 7

Hence, the point estimate used to build this confidence interval is 7.

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It would take them 9/10 of an hour to mow the lawn together.

To solve the problem, we can use the formula:
time = work / rate
Let's first find the rates of Fred and Melissa. Fred can mow the lawn in 90 minutes, so his rate is:
1 lawn / 90 minutes = 1/90 lawns per minute
Similarly, Melissa's rate is:
1 lawn / 30 minutes = 1/30 lawns per minute
When they work together, their rates add up:
rate together = rate of Fred + rate of Melissa
rate together = 1/90 + 1/30
rate together = 1/54 lawns per minute
Now we can use the formula to find the time it takes for them to mow the lawn together:
time = work / rate
time = 1 lawn / (1/54 lawns per minute)
time = 54 minutes
Therefore, it would take Fred and Melissa 54 minutes to mow the lawn if they worked together. This can be expressed as the reduced fraction 9/10.

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The function f(x) = 3√(2x) + 3 is concave down on the interval (0, +∞).

To determine the intervals where the function is concave up or concave down, we need to find the second derivative of the function and analyze its sign. The given function is f(x) = 3√(2x) + 3.

First, let's find the first derivative, f'(x):
[tex]f'(x) = \frac{d(3\sqrt{2x} + 3)}{dx} = 3 \cdot \frac{d(\sqrt{2x})}{dx} = 3 \cdot \frac{1}{2} \cdot (2x)^{-\frac{1}{2}} \cdot 2 = 3 \cdot \frac{1}{\sqrt{2x}}[/tex]

Now, let's find the second derivative, f''(x):
[tex]f''(x) = \frac{d}{dx}\left(3 \cdot \left(\frac{1}{\sqrt{2x}}\right)\right) = -3 \cdot \frac{1}{2} \cdot (2x)^{-\frac{3}{2}} \cdot 2 = -3 \cdot \frac{1}{2(2x)^{\frac{3}{2}}}[/tex]
Now, we need to analyze the sign of f''(x):

f''(x) is concave up when f''(x) > 0, which does not happen in this case.

f''(x) is concave down when f''(x) < 0, which occurs for all x > 0.

Thus, the function f(x) = 3√(2x) + 3 is concave down on the interval (0, +∞).

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When examining the potential linear association between two variables, there are several key things that you should look for. Firstly, you should examine the scatterplot of the two variables to see if there is a general trend or pattern in the data. A strong linear association will typically result in a clear linear pattern in the scatterplot, where the points follow a straight line.

You should also calculate the correlation coefficient between the two variables, which measures the strength and direction of the linear relationship. A correlation coefficient of +1 indicates a perfect positive linear relationship, while a correlation coefficient of -1 indicates a perfect negative linear relationship. A correlation coefficient of 0 indicates no linear relationship.

Additionally, you should consider the range of the data and the potential outliers, as extreme values can have a significant impact on the correlation coefficient and the strength of the linear relationship. It is also important to consider the context of the data and whether a linear model is appropriate for the relationship between the two variables, as some relationships may be better described by a nonlinear model.

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The two lines are neither parallel nor perpendicular. Therefore, the answer is C. Neither.

The ratio of the change in the y-coordinate to the change in the x-coordinate between any two points on the line is known as the slope, and it is a measure of an equation's steepness.

Compare the slopes of each equation;

The slope-intercept form of a line is y = mx + b, where m is the slope of the line and b is the y-intercept.

Let's rewrite the two given equations in slope-intercept form:

equation 1:   5x - y = 7

-y = -5x + 7

y = 5x - 7         (here Slope = 5)

equation 2:   2x - 10y = 20

-10y = -2x + 20

y = 0.2x - 2      (here Slope = 0.2)

A. If the slopes of two lines are equal it means the lines are parallel. Since 5 is not equal to 0.2, the two lines are not parallel.

B. Two lines are perpendicular if and only if the product of their slopes is -1.
Their product is 5 × 0.2 = 1, which is not equal to -1.
Therefore, the two lines are not perpendicular.

Therefore, the answer is C. Neither.

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It is $4 much cheaper to make cabinets from 151 feet of cedar than from 150 feet.

For 150 feet of cedar, the cost of making cabinets is $6.00 per foot, so the total cost would be:

150 feet x $6.00/foot = $900

For 151 feet of cedar or more, the cost of making cabinets is $4.00 per foot. So, the cost of making cabinets for 151 feet of cedar would be:

(151 feet - 150 feet) x $4.00/foot + $900 = $4 + $900 = $904

The difference in cost between making cabinets from 151 feet of cedar and 150 feet of cedar is:

$900 - $904 = -$4

Therefore, it is $4 more expensive to make cabinets from 151 feet of cedar than from 150 feet.

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The probability that a senior citizen takes either blood pressure-lowering or cholesterol-lowering medication is 0.96.

Let's start by defining some terms. The probability of an event is a number between 0 and 1 that represents the likelihood of that event occurring. An event with a probability of 0 is impossible, while an event with a probability of 1 is certain.

Now, let's apply these concepts to the problem at hand. We know that 65% of senior citizens take blood pressure-lowering medication, and 38% take cholesterol-lowering medication. We also know that 7% take both medications.

So, let A be the event of taking blood pressure-lowering medication, and B be the event of taking cholesterol-lowering medication. We want to find P(A or B), which can be written as P(A U B). Using the inclusion-exclusion principle, we have:

P(A U B) = P(A) + P(B) - P(A ∩ B)

Substituting the values we know, we get:

P(A U B) = 0.65 + 0.38 - 0.07

P(A U B) = 0.96

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The astronaut swings the hammer in one direction, an equal and opposite force acts on the astronaut in the opposite direction.    

The astronaut swings the hammer to hammer the loose rivet back in place outside the spaceship, they will experience an equal and opposite force known as "reaction force" as described by Newton's Third Law of Motion.

This means that for every action (force) in one direction, there is an equal and opposite reaction (force) in the opposite direction.

The astronaut swings the hammer, they will experience a small amount of recoil or pushback in the opposite direction.

The magnitude of the reaction force will be equal to the force exerted by the hammer on the rivet, but in the opposite direction.

The effect of this recoil on the astronaut will depend on the mass of the astronaut and the force exerted by the hammer.

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The value of x in the tangent intersection is 6.

The measure of the angle between the two tangents is also half the difference between the major and minor arcs between the two points of contact with the tangents.

Therefore, using the tangent intersection theorem, the value of x can be found as follows:

9x + 1 = 1 / 2(235 - (360 - 235))

9x + 1 = 1 / 2 (235 - 125)

9x + 1 = 1 / 2 (110)

9x + 1 = 55

Therefore,

9x = 55 - 1

9x = 54

divide both sides of the equation by 9

x = 54 / 9

x = 6

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The probability of event A, P(A), is 1/3, and the probability of event B, P(B), is also 1/3.

(a) To verify the probability of event A, P(A), using the Addition Rule, we need to add the probabilities of the outcomes in event A, which are 1 and 2. Since a die has six equally likely outcomes (1, 2, 3, 4, 5, 6), the probability of rolling a 1 or a 2 is 2 out of 6, or 1/3.

(b) Similarly, to verify the probability of event B, P(B), we need to add the probabilities of the outcomes in event B, which are 4 and 6. Again, since a die has six equally likely outcomes, the probability of rolling a 4 or a 6 is also 2 out of 6, or 1/3.

(c) Event D in Figure 3.2 is not explicitly mentioned in the prompt, so we cannot determine its outcomes.

(d) Events B and D are disjoint because they do not have any outcomes in common. Event B consists of outcomes 4 and 6, while event D is not mentioned in the prompt.

(e) Events A and D are also disjoint because event D is not mentioned in the prompt and event A consists of outcomes 1 and 2.

Therefore, the probability of event A, P(A), is 1/3 using the Addition Rule, and the probability of event B, P(B), is also 1/3. Events B and D are disjoint, and events A and D are also disjoint

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This means we can be 95% confident that the true population mean annual earnings of all students falls between $2908.29 and $3331.71.

To construct a confidence interval for the population mean, μ, we can use the formula:

CI = x ± z×(σ/√n)

where x is the sample mean, σ is the population standard deviation, n is the sample size, z is the critical value from the standard normal distribution for the desired confidence level (95% in this case), and CI is the confidence interval.

Plugging in the values given in the question, we get:

CI = 3120 ± 1.96×(677/√40)

Simplifying this expression, we get:

CI = 3120 ± 211.71

Therefore, the 95% confidence interval for the population mean, μ, is:

CI = (2908.29, 3331.71)

This means we can be 95% confident that the true population mean annual earnings of all students falls between $2908.29 and $3331.71.

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The rate of change of the soap's radius, when it has a height of 5 cm and a radius of 12 cm, is approximately -0.16 cm/h.

We are given that the soap puck is a perfect cylinder and is losing volume at a rate of 2 cm^3 per hour, and its height is decreasing at a rate of 0.5 cm per hour. We want to find the rate of change of the radius when the height is 5 cm and the radius is 12 cm.

The volume of the cylinder at any time t is given by:

V = πr^2h

Taking the derivative of both sides with respect to time t, we get:

dV/dt = 2πrh(dr/dt) + πr^2(dh/dt)

We are given that dV/dt = -2 cm^3/h, dh/dt = -0.5 cm/h, h = 5 cm, and r = 12 cm.

We want to find dr/dt when h = 5 cm and r = 12 cm.

Substituting the given values into the equation, we get:

-2 = 2π(12)(5/6)(dr/dt) + π(12)^2(-0.5)

Simplifying and solving for dr/dt, we get:

dr/dt = (-2 + 72π/5)/(40π) ≈ -0.16 cm/h

Therefore, the rate of change of the soap's radius when it has a height of 5 cm and a radius of 12 cm is approximately -0.16 cm/h.

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The solution to the given integral is 26 ln |√6 + in(x)| + C, where C is the constant of integration.

The given integral is ∫ 13 in (x)/ x √6 + in (x))² dx. To evaluate this integral, we need to use the substitution method. We substitute u = √6 + in(x) and obtain du/dx = 1/2x √6 + in(x), which implies dx = 2u/(√6 + in(x)) du.

Substituting these values in the integral, we get:

∫ (13/u²) (2u/(√6 + in(x))) du

Simplifying this expression, we get:

∫ (26/u) du

= 26 ln |u| + C

= 26 ln |√6 + in(x)| + C

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a) The probability P(X < 0.5) is 0.00050625.

b) The probability P(0.4 < X <0.7) is 0.00723502.

c) The expected value of X is and 0.9 the standard deviation of X is 0.2121.

d) The cumulative distribution function of X, F(x) = [tex]9x^{9}-10x^{10}+1[/tex]  for 0 ≤ x ≤ 2.

To show that f(x) is a probability density function, we need to show that it satisfies the following properties,

f(x) is non-negative for all x in the range of X.

The area under the curve of f(x) over the range of X is equal to 1.

a) To find P(X < 0.5), we need to integrate f(x) from 0 to 0.5,

P(X < 0.5) = ∫[0,0.5] f(x) dx

           = ∫[0,0.5] 90[tex]x^{8}[/tex] (1-x) dx

           = 0.00050625

Therefore, the probability that X is less than 0.5 is 0.00050625.

b) To find P(0.4 < X < 0.7), we need to integrate f(x) from 0.4 to 0.7,

P(0.4 < X < 0.7) = ∫[0.4,0.7] f(x) dx

                = ∫[0.4,0.7] 90[tex]x^{8}[/tex] (1-x) dx

                = 0.00723502

Therefore, the probability that X is between 0.4 and 0.7 is 0.00723502.

c) To find the expected value (mean) of X, we need to integrate x*f(x) over the range of X,

E(X) = ∫[0,2] x*f(x) dx

    = ∫[0,2] x*90[tex]x^{8}[/tex] (1-x) dx

    = 0.9

Therefore, the expected value of X is 0.9.

To find the standard deviation of X, we need to calculate the variance first,

Var(X) = E([tex]X^{2}[/tex]) - [tex][E(X)]^{2}[/tex]

We can calculate E([tex]X^{2}[/tex]) by integrating [tex]x^{2}[/tex]*f(x) over the range of X,

E(X^2) = ∫[0,2]  [tex]x^{2}[/tex]*f(x) dx

      = ∫[0,2]  [tex]x^{2}[/tex]*90[tex]x^{8}[/tex] (1-x) dx

      = 0.54

Therefore, the variance of X is,

Var(X) = 0.54 - [tex](0.9)^{2}[/tex]

      = 0.045

And the standard deviation of X is,

SD(X) = [tex]\sqrt{Var(X)}[/tex]

     = 0.2121

d) The cumulative distribution function of X, F(x), is given by,

F(x) = P(X ≤ x) = ∫[0,x] f(t) dt

We can calculate F(x) by integrating f(x) from 0 to x,

F(x) = ∫[0,x] f(t) dt

    = ∫[0,x] 90[tex]t^{8}[/tex] (1-t) dt

    = [tex]9x^{9}-10x^{10}+1[/tex]

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The derivative of the following is: 1. [tex]dy/dx = (1/2)x^(^-^1^/^2^)[/tex] ;                          2. [tex]dy/dx = (-1)(1 + tan(x))^(^-^2^) * (sec^2^(^x^))[/tex] ;                                                    3.  [tex]dy/dx = (-2)(1 + tan(x))^(^-^3^) * (sec^2^(^x^))[/tex]

To find the derivatives of the given functions.

1. For y = √x, we want to find dy/dx.
Step 1: Rewrite the function as [tex]y = x^(^1^/^2^)[/tex]
Step 2: Use the power rule [tex](dy/dx = nx^(^n^-^1^))[/tex] to find the derivative.
[tex]dy/dx = (1/2)x^(^-^1^/^2^)[/tex]

2. For y = 1/(1 + tan(x)), we want to find dy/dx.
Step 1: Rewrite the function as [tex]y = (1 + tan(x))^(^-^1^)[/tex]
Step 2: Apply the chain rule [tex](dy/dx = f'(g(x)) * g'(x)).[/tex]
[tex]dy/dx = (-1)(1 + tan(x))^(^-^2^) * (sec^2^(^x^))[/tex]

3. For [tex]y = 1/(1 + tan(x))^2[/tex], we want to find dy/dx.
Step 1: Rewrite the function as [tex]y = (1 + tan(x))^(^-^2^)[/tex]
Step 2: Apply the chain rule [tex](dy/dx = f'(g(x)) * g'(x))[/tex] to find the derivative.
[tex]dy/dx = (-2)(1 + tan(x))^(^-^3^) * (sec^2^(^x^))[/tex]

So, the derivatives are:
1. [tex]dy/dx = (1/2)x^(^-^1^/^2^)[/tex]
2. [tex]dy/dx = (-1)(1 + tan(x))^(^-^2^) * (sec^2^(^x^))[/tex]
3. [tex]dy/dx = (-2)(1 + tan(x))^(^-^3^) * (sec^2^(^x^))[/tex]

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The probability that x will be between 22 and 25 is approximately 0.6844 or 68.44%.

To find the probability that x will be between 22 and 25, we first need to standardize the values using the formula z = (x - mu) / (sigma/sqrt (n)),

where x is the given value, mu is the population mean, sigma is the population standard deviation, and n is the sample size.
So, for x = 22: z = (22 - 23) / (8 / ^(60)) = -1.50

And for x = 25: z = (25 - 23) / (8 / ^(60)) = 1.50

Next, we can use a standard normal distribution table or a calculator to find the probability of z being between -1.50 and 1.50. This is equivalent to finding the area under the standard normal curve between -1.50 and 1.50.
Using a calculator or table, we can find this probability to be approximately 0.6844 or 68.44%.
Therefore, the probability that x will be between 22 and 25 is approximately 0.6844 or 68.44%.

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The confidence interval for the given population is [7.68, 14.77], under the condition that standard deviation  of a random sample of 25 men who have a mean weight of 170.4 pounds

Constructing a 95% confidence interval for the given population
The standard deviation σ of a random sample of 25 men who have a mean weight of 170.4 pounds with a standard deviation of 10.3 pounds,
Then,
Confidence interval = [√ (n-1)s² /[tex]X^{2a/2}[/tex], √ (n-1)s² /[tex]X^{21-a/2}[/tex]]

here:
n= sample size
s²= sample variance
[tex]X^{2a/2}[/tex] = chi-square value with α/2 degrees of freedom
[tex]X^{21-a/2}[/tex] = chi-square value with 1-α/2 degrees of freedom

Now for a 95% confidence interval,
α = 0.05
n = 24 degrees of freedom.

Applying a chi-square distribution table, we can find that [tex]X^{20.025}[/tex] = 38.58 and [tex]X^{20.975}[/tex] = 11.07.

Staging  the values we have:

Confidence interval = [√ (24)(10.3)² /38.58, √ (24)(10.3)² /11.07]
= [7.68, 14.77]

Hence, we can say with 95% confidence that the population standard deviation σ lies between 7.68 and 14.77 pounds.

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