A plane is headed due south at a speed of 298mph. A wind from direction 51 degress is blowing at 18 mph. Find the bearing pf the plane

There is a range of 0-4 and 5-9 with each stem repeated for leaves .

A split stem-and-leaf plot, in which each stem appears twice (once for leaves 0-4 and again for leaves 5-9), is shown below:

Stem/Leaf

5/ 00016668888999

6/ 000011112344

7/ 0000234458

8/ 01124588

9/ 1

This table presents the amounts of particulate emissions for 50 vehicles, and we constructed a split stem-and-leaf plot to display the data.

The plot shows the stems followed by the corresponding leaves, with each stem repeated for leaves in the range of 0-4 and 5-9.

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The probabilities are (a) P(X = 2) = 3/64, (b) P(X ≤ 2) = 9/16, (c) P(X > 2) = 0, and (d) P(X ≥ 1) = 3/8.

Given a function:

f(x) = (3/4)(1 / 4)*, x = 0,1,2.

Let's find the probability of f(x).

The formula for finding probability is given below:

∑ f(x) = 1

From the above formula, we have 3 equations:(

3/4)(1/4) + (3/4)(1/4) + (3/4)(1/4) = 1(3/16) + (3/16) + (3/16)

= 1(9/16)

= 1

So, it is a probability mass function. Now, let's determine the probabilities.

(a) P(X = 2)f(x) = (3/4)(1 / 4)*,

for x = 2= (3/4)(1/16)

= 3/64(b) P(X ≤ 2)P(X ≤ 2)

= f(0) + f(1) + f(2)= (3/4)(1/4) + (3/4)(1/4) + (3/4)(1/4)

= 3/16 + 3/16 + 3/16

= 9/16(c) P(X > 2)P(X > 2)

= f(0) = 0(d) P(X ≥ 1)P(X ≥ 1)

= f(1) + f(2)= (3/4)(1/4) + (3/4)(1/4)

= 3/16 + 3/16

= 6/16

= 3/8

Therefore, the probabilities are (a) P(X = 2) = 3/64,

(b) P(X ≤ 2) = 9/16,

(c) P(X > 2) = 0, and (d) P(X ≥ 1) = 3/8.

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The standard deviation of the average (X) amount of sunscreen from a sample of 6 tubes is approximately 1.29 mL. The probability that the average amount of sunscreen from 6 tubes will be less than 338 mL is about 0.9999.

To calculate the standard deviation of the average X, we can use the formula for the standard deviation of the sample mean:

σ(X) = α / √n,

where α is the standard deviation of the population, and n is the sample size. In this case, α = 5 mL and n = 6. Plugging in these values, we get:

σ(X) = 5 / √6 ≈ 1.29 mL.

This tells us that the average amount of sunscreen from a sample of 6 tubes is expected to vary by about 1.29 mL.

To find the probability that the average amount of sunscreen from 6 tubes will be less than 338 mL, we need to standardize the value using the formula for z-score:

z = (x - μ) / α,

where x is the value we want to find the probability for, μ is the mean of the population, and α is the standard deviation of the population. In this case, x = 338 mL, μ = 298 mL, and α = 5 mL. Plugging in these values, we get:

z = (338 - 298) / 5 = 8,

which means that the average amount of sunscreen from 6 tubes is 8 standard deviations above the mean. Since we are dealing with a normal distribution, the probability of being less than 8 standard deviations above the mean is extremely close to 1, or about 0.9999.

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Two sample means (Group A mean = 20, Group B mean = 24) are represented in the bar graph along with their standard errors. We can conclude that Group B has a bigger sample size than Group A since Group B's error bar displays a larger standard error than does Group A's.

This is because the standard error of the mean decreases as sample size increases. Consequently, the statistics that most closely match the bar chart are s-8, Group A n=16, and Group B n=30-50.The only set of statistics from the options provided that accounts for Group B having a larger sample size than Group A is s-8, Group A n=16, and Group B n=30-50. The offered bar chart and this set of statistics match each other.

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False. In hypothesis testing, we make inferences about population parameters based on sample statistics.

False. In hypothesis testing, the objective is not to estimate the true mean or true proportion of a population directly. Instead, it focuses on making statistical inferences about population parameters based on sample data.

Hypothesis testing involves formulating null and alternative hypotheses, collecting a sample, calculating test statistics, and determining the likelihood of observing the sample data under the null hypothesis. The goal is to assess the evidence against the null hypothesis and make a decision about its validity.

Estimating population parameters is typically done through point estimation or interval estimation techniques, such as calculating sample means or proportions to estimate the true population mean or proportion. However, hypothesis testing and estimation are distinct concepts in statistical analysis.

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In the given problem, random samples of four different models of cars were selected and the gas mileage of each car was measured. The results are shown below:21 226 22 725 21
Given that,The null hypothesis H0: All the population means are equal. The alternative hypothesis H1: At least one population mean is different from the others .

To find the hypothesis test, we will use the one-way ANOVA test. We calculate the grand mean (X-bar) and the sum of squares between and within to obtain the F-test statistic. Let's find out the sample size (n), the total number of samples (N), the degree of freedom within (dfw), and the degree of freedom between (dfb).
Sample size (n) = 4 Number of samples (N) = n × 4 = 16 Degree of freedom between (dfb) = n - 1 = 4 - 1 = 3 Degree of freedom within (dfw) = N - n = 16 - 4 = 12 Total sum of squares (SST) = ∑(X - X-bar)2
From the given data, we have X-bar = (21 + 22 + 26 + 25) / 4 = 23.5
So, SST = (21 - 23.5)2 + (22 - 23.5)2 + (26 - 23.5)2 + (25 - 23.5)2 = 31.5 + 2.5 + 4.5 + 1.5 = 40.0The sum of squares between (SSB) is calculated as:SSB = n ∑(X-bar - X)2
For the given data,SSB = 4[(23.5 - 21)2 + (23.5 - 22)2 + (23.5 - 26)2 + (23.5 - 25)2] = 4[5.25 + 2.25 + 7.25 + 3.25] = 72.0 The sum of squares within (SSW) is calculated as:SSW = SST - SSB = 40.0 - 72.0 = -32.0
The mean square between (MSB) and mean square within (MSW) are calculated as:MSB = SSB / dfb = 72 / 3 = 24.0MSW = SSW / dfw = -32 / 12 = -2.6667
The F-statistic is then calculated as:F = MSB / MSW = 24 / (-2.6667) = -9.0
Since we are testing whether at least one population mean is different, we will use the F-test statistic to test the null hypothesis. If the p-value is less than the significance level, we will reject the null hypothesis. However, the calculated F-statistic is negative, and we only consider the positive F-values. Therefore, we take the absolute value of the F-statistic as:F = |-9.0| = 9.0The p-value corresponding to the F-statistic is less than 0.01. Since it is less than the significance level (α = 0.05), we reject the null hypothesis. Therefore, we can conclude that at least one of the population means is different from the others.

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The distance between the two planes at the end of 5 hours is approximately 3364.6 miles.

The question is asking for the distance between two planes, one flying due north at 464 mph for 5 hours and the other flying southeast at an angle 146° clockwise from due north at 405 mph for 5 hours.

To solve this, we can use the Law of Cosines.

The formula for the Law of Cosines is:

c² = a² + b² - 2ab cos(C), where a and b are the side lengths and C is the included angle of the triangle we are solving. In this case, the distance between the two planes is the side length we are solving for.

We can use the given velocities and times to calculate the distances each plane travels, and we can use the given angle to calculate the included angle between the two paths.

Then we can apply the Law of Cosines to find the distance between the two planes.

Distance of the first plane = 464 mph × 5 hours = 2320 miles

Distance of the second plane = 405 mph × 5 hours = 2025 miles

The angle between the two paths is 360° - 90° - 146° = 124°.

Now we can plug in the values into the formula:

c² = a² + b² - 2ab cos(C)

c² = 2320² + 2025² - 2(2320)(2025) cos(124°)

c² = 11320520.03

c ≈ 3364.6

Therefore, the distance between the two planes at the end of 5 hours is approximately 3364.6 miles.

Rounding this to the nearest whole number gives us the answer of 3365 miles.

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the dose (in mg) that yields the maximum reaction is 1800 mg (rounded off to the nearest integer).

The given equation for the reaction r(x) to an injection of a drug related to the dose x (in milligrams) is:

r(x) = x²⁷⁰⁰ − x³

The dose (in mg) that yields the maximum reaction is to be determined from the given equation.

To find the dose (in mg) that yields the maximum reaction, we need to differentiate the given equation w.r.t x as follows:

r'(x) = 2x(2700) - 3x² = 5400x - 3x²

Now, we need to equate the first derivative to 0 in order to find the maximum value of the function as follows:

r'(x) = 0

⇒ 5400x - 3x² = 0

⇒ 3x(1800 - x) = 0

⇒ 3x = 0 or 1800 - x = 0

⇒ x = 0

or x = 1800

The above two values of x represent the critical points of the function.

Since x can not be 0 (as it is a dosage), the only critical point is:

x = 1800

Now, we need to find out whether this critical point x = 1800 is a maximum point or not.

For this, we need to find the second derivative of the given function as follows:

r''(x) = d(r'(x))/dx= d/dx(5400x - 3x²) = 5400 - 6x

Now, we need to check the value of r''(1800).r''(1800) = 5400 - 6(1800) = -7200

Since the second derivative r''(1800) is less than 0, the critical point x = 1800 is a maximum point of the given function. Therefore, the dose (in mg) that yields the maximum reaction is 1800 mg (rounded off to the nearest integer).

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a) Assuming the numbers must be distinct, there are 36 choices for the first number, 35 choices for the second number, and 34 choices for the third number. Therefore, there are 36 x 35 x 34 = 42,840 possible lock combinations.
b) Assuming the numbers may be the same, there are still 36 choices for each number, so the number of possible lock combinations remains the same, which is 36 x 36 x 36 = 46,656.

a) When the numbers must be distinct, we can choose any number between 0 and 35 for the first number. Once the first number is chosen, we have 35 remaining inging choices for the second number, since it cannot be the same as the first number. Similarly, we have 34 choices for the third number, as it cannot be the same as the first or second number. Therefore, the total number of possible lock combinations is given by the product of the choices for each number: 36 x 35 x 34 = 42,840.
b) When the numbers may be the same, we still have 36 choices for each number, including the possibility of choosing the same number multiple times. Therefore, the number of possible lock combinations remains the same as in case a), which is 36 x 36 x 36 = 46,656.

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The value of dt/dy is -1/12.

We are given the equation dy/dt = -3, and we need to find dt/dy. To do this, we can use the chain rule. We start with the given equation:

dt/dx * dx/dy * dy/dt = 1

Rearranging the equation, we have:

dt/dy = 1 / (dt/dx * dx/dy)

Next, we differentiate the given equation with respect to t using the chain rule. We have:

2y * (dy/dt) - 4x^3 * (dx/dt) = 0

Substituting the values dy/dt = -3, x = 2, and y = 6, we get:

12 - 32 * (dx/dt) = 0

Simplifying further, we have:

32 * (dx/dt) = 12

Solving for dx/dt, we find:

dx/dt = 12/32 = 3/8

Substituting this value and dx/dy = 1/dy/dx = 1/(dt/dx), we can evaluate dt/dy:

dt/dy = 1 / (3/8) = 8/3 = -1/12

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A histogram is used to display the distribution of continuous data while a stem-and-leaf plot is used to display the distribution of small data set.There are three numbers in bin 1, two numbers in bin 2, four numbers in bin 3, six numbers in bin 4, and one number in bin 5.

Here is the histogram and stem-and-leaf plot with five bins for the given 16 randomly selected numbers from 2 to 200:HISTOGRAM:

There are five bins, with intervals 20: 1. 5-24 2. 25-44 3. 45-64 4. 65-84 5. 85-104

There are three numbers in bin 1, two numbers in bin 2, four numbers in bin 3, six numbers in bin 4, and one number in bin 5.  STEM-AND-LEAF:    5| 5  5|    6| 5  6  6|    7| 8  |  9| 9  9|     10| 0  0|     11| 7 |     12| 0  0  0  0  |     13|  |     14|  |     15|  |     16|  |     17|  |     18|  |     19| 1There are three numbers in the 50s, six numbers in the 60s, one number in the 70s, four numbers in the 80s, and two numbers in the 90s.

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To determine the values of b for which the given vectors are orthogonal, we need to find the dot product of the two vectors and set it equal to zero.

The dot product of two vectors (a1, a2, a3) and (b1, b2, b3) is given by:

Dot product = a1 * b1 + a2 * b2 + a3 * b3

In this case, the given vectors are:

Vector A = (-29, b, 4)

Vector B = (b, b^2, b)

The dot product of Vector A and Vector B is:

Dot product = (-29) * b + b * b^2 + 4 * b

Setting the dot product equal to zero, we have:

(-29) * b + b * b^2 + 4 * b = 0

Simplifying the equation:

b^3 - 25b = 0

Factoring out b:

b(b^2 - 25) = 0

Setting each factor equal to zero, we have two cases:

Case 1: b = 0

Case 2: b^2 - 25 = 0

For Case 2, we solve for b:

b^2 - 25 = 0

(b - 5)(b + 5) = 0

So, we have two additional solutions:

b - 5 = 0 => b = 5

b + 5 = 0 => b = -5

Therefore, the values of b for which the given vectors (-29, b, 4) and (b, b^2, b) are orthogonal are:

b = 0, 5, -5

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ANOVA is a method for determining whether group means differ more than group means do. It lets us see if the means of two or more groups differ significantly. If the null hypothesis is rejected, it suggests that at least one group is distinct from the others.

An analysis of variance (ANOVA) method is used to determine whether two or more population means are equal. The variability within and between the various samples is compared using the ANOVA method. It is more likely that the population means are equal when the variability within the samples is comparable to the variability between them.

When the examples' changeability is greater than their variation, the populace means almost certainly are not equivalent. ANOVA is used to test the hypothesis that the method for at least two populaces is equivalent. It indicates that the means of more than two populations are not equal if the null hypothesis is rejected.

However, the null hypothesis suggests that the means of multiple populations are identical if it is not ruled out. To put it another way, the purpose of ANOVA is to ascertain whether group means differ more than group means do. It lets us see if there is a significant difference in the means of two or more groups. It suggests that at least one group is distinct from the others if the null hypothesis is rejected.

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The dot product of (3, -1) and (1, 5) is 8.

The dot product, also known as the scalar product, is a mathematical operation performed on two vectors to yield a scalar value. In order to calculate the dot product of two vectors, we multiply their corresponding components and then sum up the results.

For the given vectors (3, -1) and (1, 5), we can calculate their dot product as follows:

(3 * 1) + (-1 * 5) = 3 - 5 = -2

Therefore, the dot product of (3, -1) and (1, 5) is -2.

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The Taylor series for f(x) centered at a=8 for f(x) = ln(x) is given by:f(x) = ln(8) + (1/8)(x-8) - (1/64)(x-8)² + (1/192)(x-8)³ - (1/768)(x-8)⁴ + ...

To find the Taylor series for f(x) centered at a=8 for f(x) = ln(x), first, we need to find the values of f, f′, f″, f‴, ... at x=a. Then use them to construct the series.

The first several derivatives of f(x) = ln(x) are:

f(x) = ln(x)f′(x) = 1/xf″(x) = -1/x²f‴(x) = 2/x³f⁴(x) = -6/x⁴

The general formula for the Taylor series expansion of ln(x) about a=8 is:

f(x) = f(a) + f′(a)(x-a) + (1/2!) f″(a)(x-a)² + (1/3!) f‴(a)(x-a)³ + ... + (1/n!) fⁿ(a)(x-a)^ⁿ

The term f(a) is simply ln(8).

Since the derivatives of f(x) are equal to 1/x, -1/x², 2/x³, and so on, we can simplify the series to:

f(x) = ln(8) + (1/8)(x-8) - (1/64)(x-8)² + (1/192)(x-8)³ - (1/768)(x-8)⁴ + ...

The Taylor series for f(x) centered at a=8 for f(x) = ln(x) is given by:f(x) = ln(8) + (1/8)(x-8) - (1/64)(x-8)² + (1/192)(x-8)³ - (1/768)(x-8)⁴ + ...

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In this experiment, the fluidized bed drying system was used to dry Material A. The experiment was conducted to study the relationship between the drying time and the grams of solids removed from Material A.

The experiment resulted in ten observations, which were recorded as follows: x  2.0  3.0  4.0  5.0  6.0  7.0  8.0  9.0  10.0  11.0y  27.0  38.0  52.0  65.0  81.0  98.0  118.0  136.0  160.0  180.0.

The data obtained from the experiment is given in the table above. The next step is to plot the data on a scatter plot. The scatter plot helps us to visualize the relationship between the two variables, i.e., drying time (x) and the grams of solids removed from Material A (y).

The scatter plot for this experiment is shown below: From the scatter plot, it is evident that the relationship between the two variables is linear, which means that the grams of solids removed from Material A are directly proportional to the drying time.

The next step is to find the equation of the line that represents this relationship. The equation of the line can be found using linear regression analysis. The regression equation is as follows:[tex]y = 12.48x + 3.086[/tex]

The regression equation tells us that for every unit increase in drying time, the grams of solids removed from Material A increase by 12.48.

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The standard deviation of the given data can be calculated using the formula for the population standard deviation:

Standard deviation = √[∑(X - μ)² * f / N]

where X is the data value, μ is the mean, f is the frequency, and N is the total number of observations.

Given the data:

X: 3 4 5 6 7 8 9

f: 37 8 10 12 4 3 2

Assumed mean (μ) = 6

To calculate the standard deviation, we need to calculate the squared difference between each data value and the mean, multiply it by the frequency, and sum up these values. Then divide the sum by the total number of observations (N) and take the square root of the result.

Let's calculate it step by step:

(X - μ)² * f:

(3 - 6)² * 37 = 111

(4 - 6)² * 8 = 32

(5 - 6)² * 10 = 10

(6 - 6)² * 12 = 0

(7 - 6)² * 4 = 4

(8 - 6)² * 3 = 12

(9 - 6)² * 2 = 18

Sum of (X - μ)² * f = 187

Now divide the sum by the total number of observations (N = 37 + 8 + 10 + 12 + 4 + 3 + 2 = 76) and take the square root of the result:

Standard deviation = √(187 / 76) ≈ 1.82

Therefore, the standard deviation of the given data is approximately 1.82.

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Z - score of tallest man is more , his height was more extreme .  

Here, we have,

Average height = 176.55 cm

Height of tallest man = 249 cm

Standard deviation = 7.23

z score of tallest man

= (249 - 176.55) / 7.23

= 10.02

Average height = 176.55 cm

Height of shortest  man = 120.2 cm

Standard deviation = 7.23

z score of smallest  man

= ( 176.55 - 120.2 )  / 7.23

= 7.79

Since Z - score of tallest man is more , his height was more extreme .  

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complete question:

Use z scores to compare the given values. The tallest living man at one time had a height of 249 cm. The shortest living man at that time had a height of 120.2 cm. Heights of men at that time had a mean of 176.55 cm and a standard deviation of 7.23 cm. Which of these two men had the height that was more​ extreme?

The y-intercept of the linear equation represented by the table is 8, so the correct option is C.

Here we have a function represented by the table:

x     y

-2    16

1      4

2      0

4      -8

7      -20

This seems to be a linear function, such that each time we increase the value of x by one unit, the value of y decreases by 4.

Then the equation is something like:

y = -4x + b

b is the y-intercept.

We can replace the values of a known point like (2, 0) to get:

0 = -4*2 + b

0 = -8 + b

8 = b

Then the line is:

y = -4x + 8

The y-intercept is 8, the correct option is C.

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the estimated standard error of the sampling distribution of the sample proportion is 0.0455.

A researcher found that in a random sample of 111 people, 55 stated that they owned a laptop. The estimated standard error of the sampling distribution of the sample proportion is 0.0455. Standard error is defined as the standard deviation of the sampling distribution of the mean. It provides a measure of how much the sample mean is likely to differ from the population mean. The formula for the standard error of the sample proportion is given as:SEp = sqrt{p(1-p)/n}

Where p is the sample proportion, 1-p is the probability of the complement of the event, and n is the sample size. We are given that the sample size is n = 111, and the sample proportion is:p = 55/111 = 0.495To find the estimated standard error, we substitute these values into the formula:SEp = sqrt{0.495(1-0.495)/111}= sqrt{0.2478/111} = 0.0455 (rounded to 4 decimal places).Therefore, the estimated standard error of the sampling distribution of the sample proportion is 0.0455.

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The error in finding LM is that the formula should be LM = LN + LP, not LM = LN • LP.

The correct formula for finding LM is LM = LN + LP, not LM = LN • LP. In the given calculation, the multiplication symbol (•) is used instead of the addition symbol (+). The correct formula indicates that LM is the sum of LN and LP, not the product.

To correct the error, we need to replace the multiplication symbol with the addition symbol:

LM = LN + LP

Given the values LN = 7 and LP = 15, we substitute these values into the corrected formula:

LM = 7 + 15

Now we can calculate the sum:

LM = 22

Therefore, the corrected value of LM is 22, not 105 as initially calculated.

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Based on a survey of 400 non-fatal accidents, where 173 involved faulty equipment, the point estimate for the population proportion (p) of accidents that involved faulty equipment is 173/400 = 0.4325.

To calculate the point estimate for the population proportion, we divide the number of accidents involving faulty equipment (173) by the total number of accidents surveyed (400).

This gives us a ratio of 0.4325, which represents the estimated proportion of accidents involving faulty equipment in the population.

A point estimate is a single value that serves as an approximation or best guess for an unknown population parameter.

In this case, the population proportion (p) represents the proportion of all accidents that involved faulty equipment. The point estimate of 0.4325 suggests that approximately 43.25% of non-fatal accidents may involve faulty equipment based on the sample data.

It's important to note that this point estimate is subject to sampling variability and may not perfectly reflect the true population proportion. To obtain a more precise estimate with a measure of uncertainty, one would need to consider confidence intervals or conduct hypothesis testing using statistical methods.

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The regression equation of the given bivariate set of data with y as the response variable is y = 10.9 + 0.98x.

Given, the bivariate set of data with y as the response variable X y50.2 21.214.3 82.542.6 27.530 61.727.1 56.16.6 79.112.9 63.936.1 25.623.5 27.145.5 20.83

We have to perform regression analysis by the given data set.

In order to find the regression equation, we need to calculate the following terms:

∑X∑Y∑X²∑Y²∑XYN,

where N = number of data points

∑X = sum of all X values

∑Y = sum of all Y values

∑X² = sum of squares of all X values

∑Y² = sum of squares of all Y values

∑XY = sum of products of corresponding X and Y values

Now we will compute the values of the above terms and find the regression equation

∑X = 329.7

∑Y = 463.9

∑X² = 10733.19

∑Y² = 35562.69

∑XY = 12607.67N = 20Now, using the above formula we have:

Regression equation: y = 10.9 + 0.98x

Hence, the conclusion is that the regression equation of the given bivariate set of data with y as the response variable is y = 10.9 + 0.98x.

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Given, probability distribution function of continuous random variables X and Y is given as [tex]f(x, y) = {kxy, (x, y)[/tex] E D Others D y=2 y=xTo find: The constant [tex]k, P(X > 1.5). x=1We[/tex] know that, for a function f(x,y) to be probability density function, it must satisfy the following conditions.

1[tex]. f(x,y) ≥ 0 for all (x,y)2. ∫∫ f(x,y) dx dy = 1[/tex] Where D is the domain of (x,y) such that [tex]D={(x,y): y = 2, y=x}[/tex]

Given, the probability distribution function of continuous random variables X and Y is given as [tex]f(x, y) = {kxy, (x, y) E D Others D y=2 y=x[/tex]

The domain is given by [tex]{(x,y): y = 2, y=x} and f(x,y)=kxy[/tex]

[tex]∫∫ f(x,y) dx dy = ∫∫ kxy dx dy = k ∫∫ xy dx dy-----------------(1)[/tex]To find the value of constant k, we will use the above equation.

[tex]∫∫ xy dx dy = ∫2x x x²/2 dy = ∫2x x³/2 dy[limits: x to 2x] = x³(y/2) [limits: x to 2x]= 3/4 x³ = 3/4x[/tex]

using equation (1),[tex]∫∫ f(x,y) dx dy = k ∫∫ xy dx dy = k(3/4x³)[/tex]

Since, [tex]∫∫ f(x,y) dx dy = 1k(3/4x³) = 1∴ k = 4/3x³∴ k = 4/3[/tex]

Also, [tex]P(X > 1.5, x=1) is given by ∫1.5^2 4/3 * xy dy[/tex]

Now, putting [tex]P(X > 1.5, x=1) is given by ∫1.5^2 4/3 * xy dy[/tex]

[tex]P(X > 1.5, x=1) = 0.30556[/tex],

when x = 1

The value of constant k is 4/3 and the value of [tex]P(X > 1.5, x=1) is 0.30556.[/tex]

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The distribution represents a probability distribution. X 3 6 9 1 P(X) 0.3 0.4 0.3  is b. No

For a distribution to represent a probability distribution, the probabilities for each outcome must be non-negative and sum to 1. In this case, the sum of the probabilities is 0.3 + 0.4 + 0.3 = 1, which satisfies the second condition.

However, the first condition is not satisfied because the probability for the outcome X = 1 is given as 0, which is not non-negative. Therefore, this distribution does not represent a probability distribution.

In a probability distribution, the probabilities assigned to each outcome must meet certain criteria. Firstly, the probabilities must be non-negative, meaning they cannot be negative values. Secondly, the sum of all probabilities in the distribution must equal 1, indicating that the total probability across all possible outcomes is complete.

In the given distribution, the probabilities assigned to the outcomes are 0.3, 0.4, and 0.3 for X = 3, 6, and 9, respectively. However, the probability for X = 1 is given as 0, which violates the requirement of non-negativity. Since one of the probabilities is not non-negative, the distribution does not meet the criteria of a probability distribution.

Therefore, the distribution does not represent a probability distribution, and the correct answer is b. No.

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The probability of the complement of event A, given the complement of event B, denoted as [tex]P(A^c|B^c)[/tex], cannot be determined based on the information provided.

To find [tex]P(A^c|B^c)[/tex], we need to know the conditional probability of the complement of event A given the complement of event B.

However, the information provided only includes the probabilities of events A, B, and their union.

The complement of event A, denoted as [tex]A^c[/tex], represents all outcomes that are not in event A. Similarly, the complement of event B, denoted as [tex]B^c[/tex], represents all outcomes that are not in event B.

To find [tex]P(A^c|B^c)[/tex], we would need additional information about the conditional probabilities or the intersection of [tex]A^c[/tex] and[tex]B^c[/tex].

Without this additional information, it is not possible to determine the value of [tex]P(A^c|B^c)[/tex] based solely on the given probabilities. Therefore, the probability of the complement of event A given the complement of event B cannot be determined.

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The image of the point (-3, 0) under the same translation is (-2, 3).

Left/Right movement: -3 + 1 = -2 Up / Down movement: 0 + 3 = 3. The image of the point (-3, 0) under the same translation is (-2, 3).Given the point (2, 4) is translated to (1, 1) after translation. Therefore, The distance moved left/right = 2 - 1 = 1 and the distance moved up/down = 4 - 1 = 3.

Using the same distances to translate point (-3, 0), we get the new coordinates:

Left/Right movement: -3 + 1 = -2 Up / Down movement: 0 + 3 = 3

The image of the point (-3, 0) under the same translation is (-2, 3).

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To compute Empirical Probability, you must observe the outcomes of the variable over a period of time.

Empirical probability is the probability that comes from actual experiments or observations. Empirical probability is calculated by counting the number of times an event of interest occurs in an experiment or observation, then dividing by the total number of trials or observations. Empirical probability is an estimate based on observed data. The larger the number of trials or observations, the closer the empirical probability is to the true probability. To find empirical probability, follow the below steps: Count the number of times the event of interest happened. (The event can be the result of a coin toss, the number on a dice, or any other simple occurrence.)Divide that by the total number of trials or observations. (The sample space, in other words.)Express this ratio as a decimal or a fraction. This is the empirical probability.

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To solve this problem, we must use the limit comparison test. To begin, we must determine if the series is convergent or divergent. We know that the denominator of this series is 9n, which is always greater than 1. So, we can write 1/9n < 1/n.

The given series is [infinity] n = 1 sin(n) / 9n. We will discuss the convergence of the given series below:

To solve this problem, we must use the limit comparison test. To begin, we must determine if the series is convergent or divergent. We know that the denominator of this series is 9n, which is always greater than 1. So, we can write 1/9n < 1/n. So, we can say that 1/9n is a convergent series. Now, we need to find out whether the given series is convergent or divergent. To find out if the given series is convergent or divergent, we must first calculate the limit of the following expression :lim n → ∞ (sin n)/(9n).

Using the limit comparison test, we compare the given series with the convergent series 1/9n:lim n → ∞ (sin n)/(9n) ÷ 1/9nlim n → ∞ (sin n)/(9n) × 9n/1lim n → ∞ sin n

Thus, using the limit comparison test, we see that the given series is divergent. The series is neither absolutely convergent nor conditionally convergent. Therefore, the series is simply divergent.Note: The series is not absolutely convergent because | sin(n)/(9n) | is not convergent. The series is not conditionally convergent because the series itself is not convergent.

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Answer : 1) Exactly two breakdowns is 0.084.2) At most one breakdown is 0.047.3) At least four breakdowns is 0.729.

Explanation : Given that a washing machine in a laundromat breaks down an average of five times per month.

Let X be the number of breakdowns in a month. Then X follows the Poisson distribution with mean µ = 5.So, P(X = x) = (e-µ µx) / x!Where e = 2.71828 is the base of the natural logarithm.

Exactly two breakdowns

Using the Poisson distribution formula, P(X = 2) = (e-5 * 52) / 2! = 0.084

At most one breakdown

Using the Poisson distribution formula,P(X ≤ 1) = P(X = 0) + P(X = 1)P(X = 0) = (e-5 * 50) / 0! = 0.007 P(X = 1) = (e-5 * 51) / 1! = 0.04 P(X ≤ 1) = 0.007 + 0.04 = 0.047

At least four breakdowns

P(X ≥ 4) = 1 - P(X < 4) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]P(X = 0) = (e-5 * 50) / 0! = 0.007 P(X = 1) = (e-5 * 51) / 1! = 0.04 P(X = 2) = (e-5 * 52) / 2! = 0.084 P(X = 3) = (e-5 * 53) / 3! = 0.14

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.007 + 0.04 + 0.084 + 0.14 = 0.271P(X ≥ 4) = 1 - 0.271 = 0.729

Therefore, the probability that during the next month the machine will have:1) Exactly two breakdowns is 0.084.2) At most one breakdown is 0.047.3) At least four breakdowns is 0.729.

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