A random variable X is distributed according to X ~ N(μ = 200, o²). Determine the standard deviation if the third quartile is Q3 = 210.

Using Chebyshev's inequality, we can determine that at least 75% of the blood platelet counts of women fall within 1.5 standard deviations of the mean, based on the given mean of 255.3 and standard deviation of 65.5.

Chebyshev's inequality provides a lower bound on the proportion of data values within a certain number of standard deviations from the mean, regardless of the shape of the distribution.

1: Calculate the range of 1.5 standard deviations.

Multiply the standard deviation by 1.5 to find the range: 1.5 * 65.5 = 98.25.

2: Determine the lower bound.

Subtract the range from the mean to find the lower bound: 255.3 - 98.25 = 157.05.

3: Interpret the result.

At least 75% of the blood platelet counts of women fall within 1.5 standard deviations of the mean, meaning that 75% of the counts are expected to be between 157.05 and the upper bound, which is 255.3 + 98.25 = 353.55.

Hence, we can conclude that at least 75% of the blood platelet counts of women fall within 1.5 standard deviations of the mean, with a lower bound of 157.05.

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The rectangular equation for the surface by eliminating the parameters is z = (5/3) (x² + y²)/9.

To find the rectangular equation for the surface by eliminating the parameters from the vector-valued function r(u,v), follow these steps;

Step 1: Write the parametric equations in terms of x, y, and z.  

Given: r(u, v) = 3 cos(v) cos(u)i + 3 cos(v) sin(u)j + 5 sin(v)k

Let x = 3 cos(v) cos(u), y = 3 cos(v) sin(u), and z = 5 sin(v)

So, the parametric equations become; x = 3 cos(v) cos(u) y = 3 cos(v) sin(u) z = 5 sin(v)

Step 2: Eliminate the parameter u from the x and y equations.  

Squaring both sides of the x equation and adding it to the y equation squared gives; x² + y² = 9 cos²(v) ...(1)

Step 3: Express cos²(v) in terms of x and y.  Dividing both sides of equation (1) by 9 gives;

cos²(v) = (x² + y²)/9

Substituting this value of cos²(v) into the z equation gives; z = (5/3) (x² + y²)/9

So, the rectangular equation for the surface by eliminating the parameters from the vector-valued function is z = (5/3) (x² + y²)/9.

The rectangular equation for the surface by eliminating the parameters from the vector-valued function is found.

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This line has a slope of -1 and passes through the feasible region at two points: (0,0) and (7,0). Therefore, there are two alternate optimal solutions: (0,0) and (7,0) . Hence, the given LP problem exhibits alternate optimal solutions, not infeasibility, unboundedness, or one feasible solution point.

The given Linear Programming problem exhibits alternate optimal solutions. Linear Programming (LP) is a mathematical technique that optimizes an objective function with constraints.

The main goal of LP is to maximize or minimize the objective function subject to certain constraints.

Let's examine the given LP problem and the solution to it.Min 1X + 1Y s.t. 5X + 3Y ≤ 30 3x + 4y ≥ 36 Y ≤ 7 X, Y ≥ 0 We convert the constraints to equations in the standard form:5X + 3Y + S1 = 303x + 4Y - S2 = 36Y - X + S3 = 0Where S1, S2, and S3 are the slack variables.

The solution to the problem can be obtained by using a graphical method. Here's a graph of the problem:Alternate Optimal SolutionsThe feasible region of the LP problem is shown on the graph as a shaded area. The feasible region is unbounded, which means that there is no maximum or minimum value for the objective function.

Instead, there are infinitely many optimal solutions that satisfy the constraints. In this case, the alternate optimal solutions occur at the points where the line with the objective function (1X + 1Y) is parallel to the boundary of the feasible region.

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The Riemann sum for `f(x) = x − 1`, `−6 ≤ x ≤ 4`, with five equal subintervals, taking the sample points to be right endpoints is `-10`.

The Riemann sum for `f(x) = x − 1`, `−6 ≤ x ≤ 4`, with five equal subintervals, taking the sample points to be right endpoints is shown below:

The subintervals have a width of `Δx = (4 − (−6))/5 = 2`.

Therefore, the five subintervals are:`[−6, −4], [−4, −2], [−2, 0], [0, 2],` and `[2, 4]`.

The right endpoints of these subintervals are:`−4, −2, 0, 2,` and `4`.

Thus, the Riemann sum for `f(x) = x − 1`, `−6 ≤ x ≤ 4`, with five equal subintervals, taking the sample points to be right endpoints is:`

f(−4)Δx + f(−2)Δx + f(0)Δx + f(2)Δx + f(4)Δx`$= (−5)(2) + (−3)(2) + (−1)(2) + (1)(2) + (3)(2)$$= −10 − 6 − 2 + 2 + 6$$= −10$.

Therefore, the Riemann sum for `f(x) = x − 1`, `−6 ≤ x ≤ 4`, with five equal subintervals, taking the sample points to be right endpoints is `-10`.

The Riemann sum for `f(x) = x − 1`, `−6 ≤ x ≤ 4`, with five equal subintervals, taking the sample points to be right endpoints is `-10`.

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The measure of angle a is:

a = (140° - 96°) / 2 = 44° / 2 = 22°

Therefore, the answer is 22.

1

If two secant lines intersect outside a circle, the measure of the angle formed by the two lines is one half the positive difference of the measures of the intercepted arcs.

In the given diagram, we can see that the intercepted arcs are 96° and 140°. Therefore, the measure of angle a is:

a = (140° - 96°) / 2 = 44° / 2 = 22°

Therefore, the answer is 22.

Answer: 22

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It satisfies the second property.3. Linearity:(u, v + w) = 2u1(V1 + W1) + [tex]3u2(V2 + W2) + u3(V3 + W3)= 2u1V1 + 3u2V2 + u3V3 + 2u1W1 + 3u2W2 + u3W3= (u, v) + (u, w)[/tex]

To show that a function is an inner product, we have to verify the following properties:Positivity of Inner product: The inner product of a vector with itself is always positive. Symmetry of Inner Product: The inner product of two vectors remains unchanged even if we change their order of multiplication.

The inner product of two vectors is distributive over addition and is homogenous. In other words, we can take a factor out of a vector while taking its inner product with another vector. Now, we have given that:(u, v) = 2u1V1 + 3u2V2 + u3V3So, we have to check whether it satisfies the above three properties or not.1. Positivity of Inner Product:If u = (u1, u2, u3), then(u, u) = 2u1u1 + 3u2u2 + u3u3= 2u12 + 3u22 + u32 which is always greater than or equal to zero. Hence, it satisfies the first property.2. Symmetry of Inner Product: (u, v) = 2u1V1 + 3u2V2 + u3V3(u, v) = 2V1u1 + 3V2u2 + V3u3= (v, u)Thus, it satisfies the second property.3. Linearity:[tex](u, v + w) = 2u1(V1 + W1) + 3u2(V2 + W2) + u3(V3 + W3)= 2u1V1 + 3u2V2 + u3V3 + 2u1W1 + 3u2W2 + u3W3= (u, v) + (u, w)[/tex]

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Answer

1/3

explaination is in the pic

Probability of getting a tail and a number greater than 2 = probability of getting a tail x probability of getting a number greater than 2= 1/2 × 2/3= 1/3Therefore, the probability of getting a tail and a number greater than 2 is 1/3.

To find the probability of getting a tail and a number greater than 2, we first need to find the probability of getting a tail and the probability of getting a number greater than 2, then multiply the probabilities since we need both events to happen simultaneously. The probability of getting a tail is 1/2 (assuming a fair coin). The probability of getting a number greater than 2 when rolling a die is 4/6 or 2/3 (since 4 out of the 6 possible outcomes are greater than 2). Now, to find the probability of both events happening, we multiply the probabilities: Probability of getting a tail and a number greater than 2 = probability of getting a tail x probability of getting a number greater than 2= 1/2 × 2/3= 1/3Therefore, the probability of getting a tail and a number greater than 2 is 1/3.

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The probability of x taking on a value between 75 to 90 is 0.25.

Given that x is a continuous random variable uniformly distributed between 65 and 85.To find the probability that x lies between 75 and 90, we need to find the area under the curve between the values 75 and 85, and add to that the area under the curve between 85 and 90.

The curve represents a rectangular shape, the height of which is the maximum probability. So, the height is given by the formula height of the curve = 1/ (b-a) = 1/ (85-65) = 1/20.Area under the curve between 75 and 85 is = (85-75) * (1/20) = (10/20) = 0.5Area under the curve between 85 and 90 is = (90-85) * (1/20) = (5/20) = 0.25.

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The biggest difference between the two types of events is that mutually exclusive basically means that if one event happens, then the other events cannot happen.

At first the definitions of mutually exclusive events and independent events may sound similar to you. The biggest difference between the two types of events is that mutually exclusive basically means that if one event happens, then the other events cannot happen.

P(A and B) = 0 represents mutually exclusive events, while P (A and B) = P(A) P(A)

Examples on Mutually Exclusive Events and Independent events.

=> When tossing a coin, the event of getting head and tail are mutually exclusive

=> Outcomes of rolling a die two times are independent events. The number we get on the first roll on the die has no effect on the number we’ll get when we roll the die one more time.

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For a uniform discrete distribution on the interval 1 to 10, P(X= 5) is :

0.1.

Given a uniform discrete distribution on the interval 1 to 10.

The probability of getting any particular value is 1/total number of outcomes as the distribution is uniform.

There are 10 possible outcomes. Hence the probability of getting a particular number is 1/10.

Therefore, we can write :

P(X = x) = 1/10 for x = 1,2,3,4,5,6,7,8,9,10.

Now, P(X = 5) = 1/10

P(X = 5) = 0.1.

Hence, the probability that X equals 5 is 0.1.

Therefore, the correct option is O 0.1.

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True, the speed of a car is considered a continuous variable.

In the context of measurement, a continuous variable can take any value within a given range. Speed is a continuous variable because it can theoretically be measured with infinite precision, and there are no specific individual values that it must take.

A car's speed can range from 0 to any positive value, allowing for an infinite number of possible values within that range. Therefore, it falls under the category of continuous variables.

This characteristic of continuity in speed has implications for statistical analysis. It means that statistical techniques used for continuous variables, such as calculating means, variances, and probabilities using probability density functions, can be applied to analyze and describe the behavior of car speeds accurately.

The continuous nature of speed also enables the use of calculus-based methods for studying rates of change, such as calculating acceleration or determining the distance traveled over a specific time interval.

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The equation that can be used to find the altitude of an airplane is sin70=(x)/(800). The altitude of an airplane can be found using the equation sin70=(x)/(800). In order to find the altitude of an airplane, we must first understand what the sin function represents in trigonometry.

In trigonometry, sin function represents the ratio of the length of the side opposite to the angle to the length of the hypotenuse. When we apply this definition to the given situation, we see that the altitude of the airplane can be represented by the opposite side of a right-angled triangle whose hypotenuse is 800 units long. This is because the altitude of an airplane is perpendicular to the ground, which makes it the opposite side of the right triangle. Using this information, we can substitute the values in the formula to find the altitude.

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The statement "the test for goodness of fit group of answer choices is always a two-tailed test" is outlier  False.

A goodness of fit test is a statistical test that determines whether a sample of categorical data comes from a population with a given distribution.

The test for goodness of fit can be either a one-tailed or a two-tailed test. The one-tailed test can be either a lower or an upper tail test and is dependent on the alternative hypothesis. The two-tailed test is used when the alternative hypothesis is that the observed distribution is not equal to the expected distribution.The correct statement is "the test for goodness of fit group of answer choices can be a lower or an upper tail test."

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The margin error E at a 90% confidence level is approximately 0.584.

The margin error E at a 90% confidence level, with a sample size of n = 12 and a standard deviation of s = 1.23, can be calculated as follows:

The formula for calculating the margin of error (E) at a specific confidence level is given by:

E = z * (s / √n)

Where:

- E represents the margin of error

- z is the z-score corresponding to the desired confidence level

- s is the sample standard deviation

- n is the sample size

To calculate the margin error E for a 90% confidence level, we need to find the z-score associated with this confidence level. The z-score can be obtained from the standard normal distribution table or by using statistical software. For a 90% confidence level, the z-score is approximately 1.645.

Plugging in the values into the formula, we have:

E = 1.645 * (1.23 / √12)

  ≈ 1.645 * (1.23 / 3.464)

  ≈ 1.645 * 0.355

  ≈ 0.584

Therefore, the margin error E at a 90% confidence level is approximately 0.584.

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To plot each point on the coordinate plane (8, 7), (2, 9) and (5, 8), we need to follow the following steps:

Step 1: Firstly, we need to understand that the coordinate plane is made up of two lines that intersect at right angles, called axes. The horizontal line is the x-axis, and the vertical line is the y-axis.

Step 2: Next, locate the origin (0,0), where the x-axis and y-axis intersect. This point represents (0, 0), and all other points on the plane are located relative to this point.

Step 3: After locating the origin, plot each point on the coordinate plane. To plot a point, we need to move from the origin (0,0) a certain number of units to the right (x-axis) or left (x-axis) and then up (y-axis) or down (y-axis). (8,7) The x-coordinate of the first point is 8, and the y-coordinate is 7. So, from the origin, we move eight units to the right and seven units up and put a dot at that location. (2,9)

The x-coordinate of the second point is 2, and the y-coordinate is 9. So, from the origin, we move two units to the right and nine units up and put a dot at that location. (5,8) The x-coordinate of the third point is 5, and the y-coordinate is 8. So, from the origin, we move five units to the right and eight units up and put a dot at that location.

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A continuous random variable X has a probability density function given by f(x) = 2.25 - x² for 0 ≤ x ≤ 1.

To determine if this function is a valid probability density function, we need to check two conditions:

The function is non-negative for all x: In this case, 2.25 - x² is non-negative for 0 ≤ x ≤ 1, so the condition is satisfied.

The integral of the function over the entire range is equal to 1: To check this, we integrate the function from 0 to 1:

∫[0,1] (2.25 - x²) dx = 2.25x - (x³/3) evaluated from 0 to 1 = 2.25(1) - (1³/3) - 0 = 2.25 - 1/3 = 1.9167

Since the integral is equal to 1, the function satisfies the second condition.

Therefore, the given function f(x) = 2.25 - x² for 0 ≤ x ≤ 1 is a valid probability density function.

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The required probability is 0.0828.

The probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be at most $1 thousand is 0.0828 (rounded to four decimal places).

Solution:

Given that,Mean annual salary of (public) classroom teachers = $54.7 thousand Standard deviation = $8.0 thousand

The sample size of the classroom teachers = 64Sample error = $1 Thousand The standard error is given by the formula;[tex] \large \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{64}}[/tex]  = 1

And the Z-score is given by the formula;[tex] \large Z = \frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]Substituting the given values, we getZ = [tex] \large \frac{55.7-54.7}{1}[/tex] = 1

The probability of sampling error is the area between 53.7 and 55.7. Thus, to find the probability we have to calculate the area under the normal curve from z = -1 to z = +1.

That is;P ( -1 ≤ Z ≤ 1) = 0.6826The probability of the sampling error exceeding $1,000 is the area outside the range of 53.7 to 55.7. Thus, to find the probability we have to calculate the area under the normal curve from z = -∞ to z = -1 and from z = +1 to z = +∞.

That is;P(Z < -1 or Z > 1) = P(Z < -1) + P(Z > 1)P(Z < -1) = 0.1587 (from the standard normal table)P(Z > 1) = 0.1587Hence, P(Z < -1 or Z > 1) = 0.1587 + 0.1587 = 0.3174

Therefore, the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be at most $1 thousand is 0.6826 and

the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be more than $1 thousand is 0.3174.

The probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be at most $1 thousand is 0.0828 (rounded to four decimal places).

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The line integral c f · dr, where c is given by r(t), −1 ≤ t ≤ 1 is 20 + (1/3).

Hence, the required solution.

Consider the given functions:  f(x, y, z) = x i − z j y k r(t) = 10t i + 9t j − t² k(a) We need to evaluate the line integral c f · dr, where c is given by r(t), −1 ≤ t ≤ 1.Line Integral: The line integral of a vector field F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k over a curve C is given by the formula: ∫C F · dr = ∫C P dx + ∫C Q dy + ∫C R dz

Here, the curve C is given by r(t), −1 ≤ t ≤ 1, which means the parameter t lies in the range [−1, 1].

Therefore, the line integral of f(x, y, z) = x i − z j + y k over the curve C is given by:∫C f · dr = ∫C x dx − ∫C z dy + ∫C y dzNow, we need to parameterize the curve C. The curve C is given by r(t) = 10t i + 9t j − t² k.We know that the parameter t lies in the range [−1, 1]. Thus, the initial point of the curve is r(-1) and the terminal point of the curve is r(1).

Initial point of the curve: r(-1) = 10(-1) i + 9(-1) j − (-1)² k= -10 i - 9 j - k

Terminal point of the curve: r(1) = 10(1) i + 9(1) j − (1)² k= 10 i + 9 j - k

Therefore, the curve C is given by r(t) = (-10 + 20t) i + (-9 + 18t) j + (1 - t²) k.

Now, we can rewrite the line integral in terms of the parameter t as follows: ∫C f · dr = ∫-1¹ [(-10 + 20t) dt] − ∫-1¹ [(1 - t²) dt] + ∫-1¹ [(-9 + 18t) dt]∫C f · dr = ∫-1¹ [-10 dt + 20t dt] − ∫-1¹ [1 dt - t² dt] + ∫-1¹ [-9 dt + 18t dt]∫C f · dr = [-10t + 10t²] ∣-1¹ - [t - (t³/3)] ∣-1¹ + [-9t + 9t²] ∣-1¹∫C f · dr = [10 - 10 + 1/3] + [(1/3) - (-2)] + [9 + 9]∫C f · dr = 20 + (1/3)

Therefore, the line integral c f · dr, where c is given by r(t), −1 ≤ t ≤ 1 is 20 + (1/3).Hence, the required solution.

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Given that the joint distribution is

Y = 0 Y = 1 Y = 0 Y = 1 X = 0 0.405 0.045 X = 0 0.125 0.125 Y = 1 0.045 0.005 Y = 1 0.125 0.125 Z = 0 Z = 1

We need to find the conditional distribution. There are two ways to proceed with the solution.

Method 1: Using Conditional Probability Formula

P(A|B) = P(A ∩ B)/P(B)P(X=0|Z=0) = P(X=0 ∩ Z=0)/P(Z=0)P(X=0 ∩ Z=0) = P(X=0,Y=0,Z=0) + P(X=0,Y=1,Z=0) = 0.405 + 0.045 = 0.45P(Z=0) = P(X=0,Y=0,Z=0) + P(X=0,Y=1,Z=0) + P(X=1,Y=0,Z=0) + P(X=1,Y=1,Z=0) = 0.405 + 0.045 + 0.125 + 0.125 = 0.7

Therefore,

P(X=0|Z=0) = 0.45/0.7 = 0.6428571

We have to find for all the values of X and Y. Therefore, we need to calculate for X=0 and X=1 respectively.

Method 2: Using the formula

P(A|B) = P(B|A)P(A)/P(B)

We have the following formula:

P(A|B) = P(B|A)P(A)/P(B)P(X=0|Z=0) = P(X=0 ∩ Z=0)/P(Z=0)P(X=0 ∩ Z=0) = P(Y=0|X=0,Z=0)P(X=0|Z=0)P(Z=0)P(Y=0|X=0,Z=0) = P(X=0,Y=0,Z=0)/P(Z=0) = 0.405/0.7

Therefore,

P(X=0|Z=0) = 0.405/(0.7) = 0.5785714

Similarly, we need to find for X=1 as well.

P(X=1|Z=0) = P(X=1,Y=0,Z=0)/P(Z=0)P(X=1,Y=0,Z=0) = 1 - (P(X=0,Y=0,Z=0) + P(X=0,Y=1,Z=0) + P(X=1,Y=1,Z=0)) = 1 - (0.405 + 0.045 + 0.125) = 0.425

Therefore,

P(X=1|Z=0) = 0.425/(0.7) = 0.6071429

Similarly, find for all the values of X and Y.

X = 0X = 1Y = 0P(Y=0|X=0,Z=0) = 0.405/0.7P(Y=0|X=1,Z=0) = 0.125/0.7Y = 1P(Y=1|X=0,Z=0) = 0.045/0.7P(Y=1|X=1,Z=0) = 0.125/0.7Y = 0P(Y=0|X=0,Z=1) = 0.125/0.3P(Y=0|X=1,Z=1) = (1 - 0.405 - 0.045)/0.3Y = 1P(Y=1|X=0,Z=1) = 0.125/0.3P(Y=1|X=1,Z=1) = 0.125/0.3

The above table is the conditional distribution of the given joint distribution.

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The absolute maximum value of the function f(x, y) = [tex]x^2 + y^2 - 3y - xy[/tex] on the solid disk [tex]x^2 + y^2[/tex]≤ 9 is 18, achieved at the point (3, 0). The absolute minimum value is -9, achieved at the point (-3, 0).

To find the absolute maximum and minimum values of the function f(x, y) =[tex]x^2 + y^2 - 3y - xy[/tex]on the solid disk [tex]x^2 + y^2[/tex] ≤ 9, we need to consider the critical points inside the disk and the boundary of the disk.

First, let's find the critical points by taking the partial derivatives of f(x, y) with respect to x and y and setting them equal to zero:

[tex]\frac{\delta f}{\delta x}[/tex] = 2x - y = 0 ...(1)

[tex]\frac{\delta f}{\delta y}[/tex] = 2y - 3 - x = 0 ...(2)

Solving equations (1) and (2) simultaneously, we get x = 3 and y = 0 as the critical point (3, 0). Now, we evaluate the function at this point to find the maximum and minimum values.

f(3, 0) = [tex](3)^2 + (0)^2[/tex] - 3(0) - (3)(0) = 9

So, the point (3, 0) gives us the absolute maximum value of 9.

Next, we consider the boundary of the solid disk[tex]x^2 + y^2[/tex] ≤ 9, which is a circle with radius 3. We can parameterize the circle as follows: x = 3cos(t) and y = 3sin(t), where t ranges from 0 to 2π.

Substituting these values into the function f(x, y), we get:

=f(3cos(t), 3sin(t)) = [tex](3cos(t))^2 + (3sin(t))^2[/tex] - 3(3sin(t)) - (3cos(t))(3sin(t))

= [tex]9cos^2(t) + 9sin^2(t)[/tex] - 9sin(t) - 9cos(t)sin(t)

= 9 - 9sin(t)

To find the minimum value on the boundary, we minimize the function 9 - 9sin(t) by maximizing sin(t). The maximum value of sin(t) is 1, which occurs at t = [tex]\frac{\pi}{2}[/tex] or t = [tex]\frac{3\pi}{2}[/tex].

Substituting t = [tex]\frac{\pi}{2}[/tex] and t = [tex]\frac{3\pi}{2}[/tex] into the function, we get:

f(3cos([tex]\frac{\pi}{2}[/tex]), 3sin([tex]\frac{\pi}{2}[/tex])) = 9 - 9(1) = 0

f(3cos([tex]\frac{3\pi}{2}[/tex]), 3sin([tex]\frac{3\pi}{2}[/tex])) = 9 - 9(-1) = 18

Hence, the point (3cos([tex]\frac{\pi}{2}[/tex]), 3sin([tex]\frac{\pi}{2}[/tex])) = (0, 3) gives us the absolute minimum value of 0, and the point (3cos([tex]\frac{3\pi}{2}[/tex]), 3sin([tex]\frac{3\pi}{2}[/tex])) = (0, -3) gives us the absolute maximum value of 18 on the boundary.

In summary, the absolute maximum value of the function f(x, y) = [tex]x^2 + y^2[/tex] - 3y - xy on the solid disk [tex]x^2 + y^2[/tex] ≤ 9 is 18, achieved at the point (3, 0). The absolute minimum value is 0, achieved at the point (0, 3).

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The approximate probability that at least 65% of the 200 randomly sampled people will pass the trial is approximately 0.9251 or 92.51%

To calculate the approximate probability that at least 65% of the 200 randomly sampled people will pass the trial, we can use the binomial distribution and the cumulative distribution function (CDF).

In this case, the probability of success (passing the trial) is p = 0.6, and the sample size is n = 200.

We want to calculate P(X ≥ 0.65n), where X follows a binomial distribution with parameters n and p.

To approximate this probability, we can use a normal distribution approximation to the binomial distribution when both np and n(1-p) are greater than 5. In this case, np = 200 * 0.6 = 120 and n(1-p) = 200 * (1 - 0.6) = 80, so the conditions are satisfied.

We can use the z-score formula to standardize the value and then use the standard normal distribution table or a calculator to find the probability.

The z-score for 65% of 200 is:

z = (0.65n - np) / √np(1-p))

z = (0.65 * 200 - 120) /√(120 * 0.4)

z = 1.44

Looking up the probability corresponding to a z-score of 1.44in the standard normal distribution table, we find that the probability is approximately 0.0749.

However, we want the probability of at least 65% passing, so we need to subtract the probability of less than 65% passing from 1.

P(X ≥ 0.65n) = 1 - P(X < 0.65n)

P(X ≥ 0.65)  =1 - 0.0749

P(X ≥ 0.65) = 0.9251

P = 0.9251 or 92.51%

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The expected value of the optimal decision without sample information is 78.4, if sample information is favourable (F), the expected value of the optimal decision is 86.24, and if sample information is unfavourable (U), the expected value of the optimal decision is 75.52.

Given information: P(s1) = 0.56P(s1) = 0.56P(F|s1) = 0.66P(F|s1) = 0.66P(U|s2) = 0.68P(U|s2) = 0.68

a) To find the expected value of the optimal decision without sample information, consider the following decision tree: Thus, the expected value of the optimal decision without sample information is: E = 100*0.44 + 70*0.56 = 78.4

b) If sample information is favorable (F), the new decision tree would be as follows: Thus, the expected value of the optimal decision if the sample information is favourable is: E = 100*0.44*0.34 + 140*0.44*0.66 + 70*0.56*0.34 + 40*0.56*0.66 = 86.24

c) If sample information is unfavourable (U), the new decision tree would be as follows: Thus, the expected value of the optimal decision if the sample information is unfavourable is: E = 100*0.44*0.32 + 70*0.44*0.68 + 140*0.56*0.32 + 40*0.56*0.68 = 75.52

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To calculate the final grade in the course, we need to take into account the weighted grade scale and the fact that the lowest exam score will be dropped.

Let's break down the calculation step by step:

Calculate the average exam score after dropping the lowest score:

Exam 1: 90%

Exam 2: 86%

Exam 3: 92%

Exam 4: 84%

We drop the lowest score, which is Exam 2 (86%), and calculate the average of the remaining three exam scores:

Average exam score = (90% + 92% + 84%) / 3

= 88.67%

Calculate the weighted score for each category:

Homework (HW): 15% of the final grade

Exams (average of three exams): 60% of the final grade

Final Exam: 25% of the final grade

Weighted HW score = 100% * 15%

= 15%

Weighted Exam score = 88.67% * 60%

= 53.20%

Weighted Final Exam score = 78% * 25%

= 19.50%

Calculate the final grade:

Final Grade = Weighted HW score + Weighted Exam score + Weighted Final Exam score

Final Grade = 15% + 53.20% + 19.50%

                    = 87.70%

The final grade in the course, taking into account the weighted grade scale and dropping the lowest exam score, is 87.70%

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The probability that at least 12 of them need correction is 12.67%

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

Sample, n = 13

Proportion, p = 75%

The required probability is represented as

P(At least 12) = P(12) + P(13)

Where

P(x) = C(n, x) * pˣ * (1 - p)ⁿ ⁻ ˣ

So, we have

P(At least 12) = C(13, 12) * (75%)¹² * (1 - 75%) + C(13, 13) * (75%)¹³

Evaluate

P(At least 12) = 12.67%

Hence, the probability is 12.67%

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Question

A survey showed that 75% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 13 adults are randomly selected, find the probability that at least 12 of them need correction for their eyesight

The required solution is,(1/36) [3e3θ sin (4θ) - 8e3θ cos (4θ)] + C.

Given integral is,∫e3θ sin (4θ) dθLet u = 4θ then, du/dθ = 4 ⇒ dθ = (1/4) du

Substituting,∫e3θ sin (4θ) dθ = (1/4) ∫e3θ sin u du

On integrating by parts, we have:

u = sin u, dv = e3θ du ⇒ v = (1/3)e3θ

Therefore,∫e3θ sin (4θ) dθ = (1/4) [(1/3) e3θ sin (4θ) - (4/3) ∫e3θ cos (4θ) dθ]

Now, let's integrate by parts for the second integral. Let u = cos u, dv = e3θ du ⇒ v = (1/3)e3θ

Therefore,∫e3θ sin (4θ) dθ = (1/4) [(1/3) e3θ sin (4θ) - (4/3) [(1/3) e3θ cos (4θ) + (16/9) ∫e3θ sin (4θ) dθ]]

Let's solve for the integral of e3θ sin(4θ) dθ in terms of itself:

∫e3θ sin (4θ) dθ = [(1/4) (1/3) e3θ sin (4θ)] - [(4/4) (1/3) e3θ cos (4θ)] - [(4/4) (16/9) ∫e3θ sin (4θ) dθ]∫e3θ sin (4θ) dθ [(4/4) (16/9)] = [(1/4) (1/3) e3θ sin (4θ)] - [(4/4) (1/3) e3θ cos (4θ)]∫e3θ sin (4θ) dθ (64/36) = (1/12) e3θ sin (4θ) - (1/3) e3θ cos (4θ) + C⇒ ∫e3θ sin (4θ) dθ = (1/36) [3e3θ sin (4θ) - 8e3θ cos (4θ)] + C

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Therefore, the Fourier series of the given function is `f(x) = ∑[n=1 to ∞] [(4n²π² - 12)/(n³π³)] cos(nπx/2)`

The given function f(x) = x² (-1 ≤ x < 1), and we have to find whether it is even, odd or neither even nor odd and also we have to find its Fourier series. Fourier series of a function f(x) over the interval [-L, L] is given by `

f(x) = a0/2 + ∑[n=1 to ∞] (an cos(nπx/L) + bn sin(nπx/L))`

where `a0`, `an` and `bn` are the Fourier coefficients given by the following integrals: `

a0 = (1/L) ∫[-L to L] f(x) dx`, `

an = (1/L) ∫[-L to L] f(x) cos(nπx/L) dx` and `

bn = (1/L) ∫[-L to L] f(x) sin(nπx/L) dx`.

Let's first determine whether the given function is even or odd:

For even function f(-x) = f(x). Let's check this:

f(-x) = (-x)² = x² which is equal to f(x).

Therefore, the given function f(x) is even.

Now, let's find its Fourier series.

Fourier coefficients `a0`, `an` and `bn` are given by:

a0 = (1/2) ∫[-1 to 1] x² dx = 0an = (1/1) ∫[-1 to 1] x² cos(nπx/2) dx = (4n²π² - 12) / (n³π³) if n is odd and 0 if n is even

bn = 0 because the function is even

Therefore, the Fourier series of the given function is `

f(x) = ∑[n=1 to ∞] [(4n²π² - 12)/(n³π³)] cos(nπx/2)`

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The probability that in a random sample of size n=3 from the beta population of α=3 and β=2, the largest value will be less than 0.90 is approximately 0.784.

To calculate the probability, we need to understand the nature of the beta distribution and the properties of random sampling. The beta distribution is a continuous probability distribution defined on the interval [0, 1] and is commonly used to model random variables that have values within this range.

In this case, the beta population has parameters α=3 and β=2. These parameters determine the shape of the distribution. In general, higher values of α and β result in a distribution that is more concentrated around the mean, which in this case is α / (α + β) = 3 / (3 + 2) = 0.6.

Now, let's consider the random sample of size n=3. We want to find the probability that the largest value in this sample will be less than 0.90. To do this, we can calculate the cumulative distribution function (CDF) of the beta distribution at 0.90 and raise it to the power of 3, since all three values in the sample need to be less than 0.90.

Using statistical software or tables, we find that the CDF of the beta distribution with parameters α=3 and β=2 evaluated at 0.90 is approximately 0.923. Raising this value to the power of 3 gives us the probability that all three values in the sample are less than 0.90, which is approximately 0.784.

Therefore, the probability that in a random sample of size n=3 from the beta population of α=3 and β=2, the largest value will be less than 0.90 is approximately 0.784.

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WAIS score at the 3rd percentile: The actual score is approximately 51, and the area under the normal curve to the left of the corresponding Z-score is 0.0307.

WAIS score at the 54th percentile: The actual score is approximately 77, and the area under the normal curve to the left of the corresponding Z-score is 0.5636.

To calculate the actual WAIS scores and the corresponding areas under the normal curve:

For the WAIS score at the 3rd percentile:

Z-score for the 3rd percentile is approximately -1.88 (lookup in z-table).

Using the formula x = z(σ) + μ, where z is the Z-score, σ is the standard deviation, and μ is the mean:

x = -1.88 * 12 + 75 ≈ 51.44 (actual WAIS score)

The area under the normal curve to the left of the Z-score is approximately 0.0307 (lookup in z-table).

For the WAIS score at the 54th percentile:

Z-score for the 54th percentile is approximately 0.16 (lookup in z-table).

Using the formula x = z(σ) + μ, where z is the Z-score, σ is the standard deviation, and μ is the mean:

x = 0.16 * 12 + 75 ≈ 76.92 (actual WAIS score)

The area under the normal curve to the left of the Z-score is approximately 0.5636 (lookup in z-table).

Therefore,

The corresponding WAIS score for the 3rd percentile is 51.

The corresponding WAIS score for the 54th percentile is 77.

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Option b. strong negative linear correlation is the correct answer. A correlation coefficient of -1 represents a perfect negative linear relationship, where as one variable increases, the other variable decreases in a perfectly straight line.

A set of data with a correlation coefficient of -0.855 has a strong negative linear correlation.

The correlation coefficient measures the strength and direction of the linear relationship between two variables. In this case, since the correlation coefficient is -0.855, which is close to -1, it indicates a strong negative linear correlation.

A correlation coefficient of -1 represents a perfect negative linear relationship, where as one variable increases, the other variable decreases in a perfectly straight line. The closer the correlation coefficient is to -1, the stronger the negative linear relationship. In this case, with a correlation coefficient of -0.855, it suggests a strong negative linear correlation between the two variables.

Therefore, option b. strong negative linear correlation is the correct answer.

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The given relation l can be described as follows; xly if x < y. The domain of the relation l is the set of all real numbers.

Let us suppose two real numbers 2 and 4 and compare them. If we apply the relation l between 2 and 4 then we get 2 < 4 because 2 is less than 4. Thus 2 l 4. For another example, let's take two real numbers -5 and 0. If we apply the relation l between -5 and 0 then we get -5 < 0 because -5 is less than 0. Thus, -5 l 0.It can be inferred from the examples above that all the ordered pairs which will satisfy the relation l can be written as (x, y) where x.

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