what is the answer to this question -5(x+2)=5

The probability that the mean weight of 12 fruit will be between 681 and 682 grams is 0.0184.

We can solve this problem by using the central limit theorem, which tells us that the distribution of sample means will be approximately normal if the sample size is sufficiently large.

First, we need to calculate the standard error of the mean:

standard error of the mean = standard deviation / sqrt(sample size)

= 23 / sqrt(12)

= 6.639

Next, we can standardize the sample mean using the formula:

z = (x - mu) / (standard error of the mean)

where x is the sample mean, mu is the population mean, and the standard error of the mean is calculated above.

z1 = (681 - 692) / 6.639 = -1.656

z2 = (682 - 692) / 6.639 = -1.506

Using a standard normal distribution table or calculator, we can find the probabilities corresponding to these z-scores:

P(z < -1.656) = 0.0484

P(z < -1.506) = 0.0668

The probability of the sample mean being between 681 and 682 grams is the difference between these probabilities:

P(-1.656 < z < -1.506) = P(z < -1.506) - P(z < -1.656)

= 0.0668 - 0.0484

= 0.0184

Therefore, the probability that the mean weight of 12 fruit will be between 681 and 682 grams is 0.0184.

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The number of hectares that the plane search is 129450 hectares

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

A search plane covers 50 square miles of countryside

This means that

Area = 50 square miles of countryside

As a general rule

1 square miles = 258.999 hectares

Substitute the known values in the above equation, so, we have the following representation

50 * 1 square miles = 258.999 hectares * 50

Evaluate

50 square miles = 129450 hectares

Hence, the number of hectares is 129450 hectares

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The quantity q(t) of a radioactive substance left at time t > 0 with a half-life of 3200 years can be found using the formula: q(t) =

[tex]q(0) * 0.5^(t/3200)[/tex]

after 6400 years, only 10 grams of the substance will be left. where q(0) is the initial quantity of the substance.

Given q(0) = 20 g, we can find q(t) for any time t > 0 using the formula above. For example, if we want to find q(6400) - the quantity of the substance left after 6400 years - we can substitute t = 6400 in the formula and get: q(6400) =

[tex]20 * 0.5^(6400/3200)[/tex]

= 10 g.

After 6400 years, only 10 grams of the substance will be left. It is important to note that the half-life of a radioactive substance is the time it takes for half of the substance to decay.

After one half-life (3200 years), the initial quantity of the substance will be reduced to half (10 g). After two half-lives (6400 years), it will be reduced to one-fourth (5 g), and so on.

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Step-by-step explanation:

the main triangle is an isoceles triangle (both legs are equally long). that means that the height y bergen the 2 legs splits the baseline in half.

therefore,

x = 10/2 = 5

Pythagoras gives us y.

c² = a² + b²

c being the Hypotenuse (the side opposite of the 90° angle). in our case 10.

a and b are the legs. in our case x and y.

10² = 5² + y²

100 = 25 + y²

75 = y²

y = sqrt(75) = 8.660254038...

The times it rotates is given by 207 rotations, the weapon that will do the more damage is sword and percent probability of rolling a six on a single six sided die is 17%.

Probability refers to potential. A random event's occurrence is the subject of this area of mathematics. The range of the value is 0 to 1. Mathematics has included probability to forecast the likelihood of certain events. The degree to which something is likely to happen is basically what probability means. You will understand the potential outcomes for a random experiment using this fundamental theory of probability, which is also applied to the probability distribution.

a) Number of rotation in 1min = 143

No of rotation in 60 seconds = 143

No. of rotation in 1 seconds = 143/60

number of rotation in 87 seconds = 143/60 x 87 = 207 rotations.

b) Sword damage 14 in 1 seconds

Axe damage is 25 in 3 seconds

so in 1 seconds it is 25/3

Sword damage in 1 min = 14 x 60 = 840 units

Axe damage in 1 min = 25/3 x 60 = 500 units

Swords will do more damage in 1 min .

c) Probability = No of favorable outcome / Total number of outcome x 100

= Total outcomes = {1, 2, 3, 4, 5, 6}

= 1/6 = 100

= 17%.

Therefore, percent probability is 17%.

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The test statistic for this sample is z = (0.254 - 0.203) / (p_hat * (1 - p_hat) * (1/n1 + 1/n2))

Based on the given information, we can set up the hypotheses as follows:

Null hypothesis: p1 - p2 = 0
Alternative hypothesis: p1 - p2 > 0

where p1 represents the proportion of successes in the first population and p2 represents the proportion of successes in the second population.

Since the sample sizes are large (n1 and n2 are not given, but we can assume they are large enough for the normal approximation to hold), we can use the normal distribution to approximate the sampling distribution of the difference in sample proportions.

The test statistic for this sample can be calculated as follows:

z = (p1 - p2) / sqrt(p_hat * (1 - p_hat) * (1/n1 + 1/n2))

where p_hat = (x1 + x2) / (n1 + n2), x1 and x2 are the number of successes in the two samples respectively.

Plugging in the given values, we get:

p_hat = (0.254n1 + 0.203n2) / (n1 + n2)
z = (0.254 - 0.203) / sqrt(p_hat * (1 - p_hat) * (1/n1 + 1/n2))

Since the significance level is not given, we cannot determine the critical value for the test. However, we can use the test statistic to calculate the p-value for the test, which is the probability of observing a difference in sample proportions as extreme as the one we observed (or more extreme) under the null hypothesis.

Once we have the p-value, we can compare it to the significance level to make a decision about whether to reject or fail to reject the null hypothesis.

Note: It is important to mention that using the normal approximation without the continuity correction may not always be accurate, especially when the sample sizes are small or the proportion of successes is close to 0 or 1. In such cases, it is recommended to use other methods (such as exact tests or simulation) that do not rely on the normal approximation.

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Step-by-step explanation:

which of the principles and the question is not clear i saw something different before i clicked on it

Answer:

[tex] \dfrac{x^2 + 2x}{(x + 1)^2} [/tex]

Step-by-step explanation:

[tex] f(x) = \dfrac{x^2}{x + 1} [/tex]

[tex] \dfrac{d}{dx} \dfrac{x^2}{x + 1} = [/tex]

[tex] = \dfrac{d}{dx} [(x^2)(x + 1)^{-1}] [/tex]

[tex]= (x^2)(-1)(x + 1)^{-2} + (x + 1)^{-1}(2x)[/tex]

[tex] = \dfrac{-x^2}{(x + 1)^{2}} + \dfrac{2x}{x + 1} [/tex]

[tex] = \dfrac{-x^2}{(x + 1)^{2}} + \dfrac{2x^2 + 2x}{(x + 1)^2} [/tex]

[tex] = \dfrac{x^2 + 2x}{(x + 1)^2} [/tex]

a. Yes. Nonlinear Solver will be guaranteed to identify the level of training that maximizes productivity b. If training is set to 5, the resulting level of productivity is 62.

a. Yes, Nonlinear Solver will be guaranteed to identify the level of training that maximizes productivity.

This is because the formula given is a quadratic equation with a positive coefficient for the x-squared term (2x2), indicating a concave upward curve. The maximum point of a concave upward curve is always at the vertex, which can be found using the Nonlinear Solver.

b. If training is set to 5, the resulting level of productivity can be found by substituting x=5 into the equation:

y = -3x + 2x^2 + 27
y = -3(5) + 2(5)^2 + 27
y = -15 + 50 + 27
y = 62

Therefore, the resulting level of productivity when training is set to 5 is 62 (rounded to the nearest whole number).

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The inverse function of the function f(x) = -3x/8 is f⁻¹(x) = -8x/3

To find the inverse of a function, we need to switch the roles of x and y and then solve for y.

Let's begin by rewriting the function f(x) in terms of y:

y = f(x) = -3x/8

Now, let's switch x and y:

x = -3y/8

Next, we'll solve for y:

x = -3y/8

8x = -3y

y = -8x/3

So the inverse function of f(x) = -3x/8 is f⁻¹(x) = -8x/3

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Answer:

Vertical angles are congruent, so the two triangles are congruent by AAS (Angle-Angle-Side).

None of the theorems listed can be used to show congruence.

The equation that represents the length of the completed tunnel based on the number of days is y = 45x + 140.

Option A is the correct answer.

We have,

From the table,

We take two ordered pairs:

(15, 815) and (20, 1040)

Now,

The equation can be written as y = mx + c.

And,

m = (1040 - 815) / (20 - 15)

m = 225/5

m = 45

And,

(15, 815) = (x, y)

815 = 15 x 45 + c

c = 815 - 675

c = 140

Now,

y = mx + c

y = 45x + 140

Thus,

The equation that represents the length of the completed tunnel based on the number of days is y = 45x + 140.

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A linear or exponential model would not model the relationship between the number of employees and number of customers

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

number of employees 1 2 3 4 10

number of customers 8 4 13 17 39

Testing a linear model

To do this, we calculate the difference between the y values

So, we have

13 - 4 = 4 - 8

9 = -4 ---- this is false

So, the function is not a linear function

Testing an exponential model

To do this, we calculate the ratio of the y values

So, we have

13/4 = 4/8

3.25 = 1/2 ---- this is false

So, the function is not an exponential function

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We can use the formula:

distance = rate × time

where distance is the total length of the road built, rate is the speed at which the road was built, and time is the number of weeks it took to build the road.

Substituting the given values, we get:

7.68 kilometers = 6 kilometers/week × time

To solve for time, we can divide both sides of the equation by 6 kilometers/week:

7.68 kilometers ÷ 6 kilometers/week = time

Simplifying the left-hand side, we get:

1.28 weeks = time

Therefore, it took the construction crew approximately 1.28 weeks to build the road.

The statements that is false concerning the hypothesis testing procedure for a regression model is "The null hypothesis is rejected if the adjusted r2 is above the critical value".

The statement that the null hypothesis is rejected if the adjusted r2 is above the critical value is false concerning the hypothesis testing procedure for a regression model.

The F-test statistic is used to test the overall significance of the regression model, and an α level must be selected to determine the level of significance.

The null hypothesis is that the true slope coefficient is equal to zero, which means that there is no linear relationship between the dependent variable and the independent variable.

The alternative hypothesis is that the true slope coefficient is not equal to zero, which means that there is a linear relationship between the dependent variable and the independent variable.

The adjusted R-squared value is a measure of the goodness of fit of the regression model and represents the proportion of the variation in the dependent variable that is explained by the independent variable(s) in the model.

The null hypothesis is rejected if the F-test statistic is above the critical value, which indicates that the regression model is statistically significant and the independent variable(s) have a significant linear relationship with the dependent variable.

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The equation of the graph is determined as y - 1 = 5x/3.

The equation of the graph is calculated by applying the general equation of a line form.

y = mx + c

where;

The slope of the graph is calculated as follows;

m = Δy/Δx

m = (y₂ - y₁ ) / (x₂ - x₁ )

m = ( -4 - 1 ) / (3 - 0)

m = -5/3

y = -5x/3 + 1

y - 1 = 5x/3

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The television researcher watched the Simpsons for 3 and 1/4 hours.

To find the number of hours the researcher watched the Simpsons, we need to use the given information that Bart makes a bad decision every 1/4 of an hour. This means that in one hour (or 4/4 of an hour), Bart makes 4 bad decisions.

To find how many hours the researcher watched, we can divide the number of bad decisions by 4:

13 bad decisions ÷ 4 bad decisions per hour = 3.25 hours

Therefore, the researcher watched the Simpsons for 3 and 1/4 hours.

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The end behavior of the rational function is described as follows:

As x -> ∞, f(x) -> 1.

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

For this problem, we have that both when x goes to negative infinity and when x goes to positive infinity, the graph of the function goes to y = 1, hence the end behavior of the function is defined by the horizontal asymptote as follows:

As x -> ∞, f(x) -> 1.

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Answer:

1. Based on the grades, it is likely that Class 1 was the lecture class and Class 2 was the small group class. This is because the grades in Class 1 have a wider range (70-90) and a larger variance, while the grades in Class 2 are more tightly clustered together (82-90) and have a smaller variance.

2. Histograms or box plots could be drawn to compare the shapes of the distributions, but we cannot do this through text.

3. The most appropriate measure of center for these data sets is the mean, since the distributions are approximately symmetric. The most appropriate measure of spread for these data sets is the standard deviation, since the distributions are not strongly skewed and there are no extreme outliers.

Step-by-step explanation:

The correct values are,

                     Q1            Q2       IQR   Mean      Median         MAD

Class 1            76.25         81.75    5.5      79.35         79.50       3.12

Class 2             84             88        4           84.60         86           3.85

Subtraction in mathematics means that is taking something away from a group or number of objects. When you subtract, what is left in the group becomes less.

Now, The first step is to arrange the grades in the classes in ascending order.

Class 1: 70, 72, 74, 75, 76, 77, 77,77, 78, 79, 80, 80, 80, 81, 81, 82, 84, 86, 88, 90

Class 2: 70, 77, 79, 82, 84, 84, 84, 84, 86, 86, 86, 86, 86, 87, 88, 88, 88, 88, 89, 90

Hence, We get;

Q1 for class 1= 1/4(n + 1) = 21/4 = 5.25 = 76.25

Q2 for class 2 = 1/4(n + 1) = 5.25 = 84

Q3 for class 1= 3/4(n + 1) = 15.75 = 81.75

Q3 for class 2 = 3/4(n + 1) = 15.75 = 88

And,

IQR for class 1 = Q3 - Q1 = 81.75 - 76.25 = 5.50

IQR for class 2 = Q3 - Q1 = 88 - 84 = 4

Mean for class 1 = sum of grades / total number of grades = 1587 / 20 = 79.35

Mean for class 2 = sum of grades / total number of grades= 1692 / 20 = 84.6

Median for class 1 = (n + 1) / 2 = 21/2 = 10.5 = 79.50

Median for class 1 = (n + 1) / 2 = 21/2 = 10.5 = 86

Since, We know that;

MAD = 1/n ∑ l x - m(x) l

Where: n = number of observations

x = number in the data set

m = mean

Hence,

Mean absolute deviation for class 1 = 62. 3/ 20 = 3.12

Mean absolute deviation for class 2. = 77/ 20 = 3.85

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To find the orthogonal trajectories of the given family of curves, we first need to understand what the term "orthogonal" means. In simple terms, two lines or curves are said to be orthogonal if they intersect at a right angle. Now, coming back to the problem, the given family of curves can be written as x^2 - 2y^2 = c, where c is a constant.

To find the orthogonal trajectories, we need to differentiate this equation with respect to y, treating x as a constant. This gives us:
-4xy = dy/dx

Now, we need to find the equation of the curves that intersect the given family of curves at a right angle, i.e., the slopes of the curves must be negative reciprocals of each other. Therefore, we can write:
dy/dx = 4xy/k

where k is a constant. To solve this differential equation, we can separate the variables and integrate:

∫dy/4xy = ∫dx/k

ln|y| - ln|x^2| = ln|c| + ln|k|

ln|y/x^2| = ln|ck|

y/x^2 = ±ck

Therefore, the orthogonal trajectories of the given family of curves are given by y = ±kx^2/c, where k is a constant. These curves intersect the original family of curves at right angles.

1. Identify the family of curves: The given equation is x^2 + 2y^2 = c, where c is a constant. This represents a family of ellipses with different sizes depending on the value of c.

2. Calculate the derivative: To find the orthogonal trajectories, we first need to find the derivative of the given equation with respect to x. Differentiate both sides with respect to x:

d/dx(x^2) + d/dx(2y^2) = d/dx(c)
2x + 4yy' = 0

3. Find the orthogonal slope: The slope of the orthogonal trajectory is the negative reciprocal of the original slope. Since the original slope is y', the orthogonal slope is -1/y':

Orthogonal slope = -1/y'

4. Replace the original slope with the orthogonal slope:

2x + 4y(-1/y') = 0

5. Solve for y':

y' = -2x/(4y)

6. Solve the differential equation: Now we have a first-order differential equation to find the equation of the orthogonal trajectories:

dy/dx = -2x/(4y)

Separate variables and integrate both sides:

∫(1/y) dy = ∫(-2x/4) dx

ln|y| = -x^2/4 + k

7. Solve for y:

y = e^(-x^2/4 + k) = C * e^(-x^2/4), where C is a new constant.

The orthogonal trajectories of the given family of curves are represented by the equation y = C * e^(-x^2/4).

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The approximate height of the cliff, given the angle of elevation would be 480. 09 meters.

One approach to determining the cliff's height involves applying trigonometry's tangent function . In this instance, a given angle of elevation (58°) and 300 meters' distance from its base are known.

The tangent function is defined as:

tan (angle) = opposite side / adjacent side

tan ( 58 ° ) = h / 300

h = 300 x tan(58°)

h = 300 x 1. 6003

=  480. 09 meters

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the answer to your question is  130

The probability that exactly 12 of the selected people belong to the labor union is approximately 0.167 or 16.7%.

To solve this problem, we can use the binomial probability formula:
[tex]P(X = k) = (n choose k)p^{k}(1 - p)^{(n - k)}[/tex]

where:
- P(X = k) is the probability of getting k successes (12 in this case)
- n is the number of trials (20 in this case)
- p is the probability of success (belonging to the labor union, which is 3600/8000 or 0.45)
- (n choose k) is the number of ways to choose k items from a set of n items, which is given by the binomial coefficient formula (n! / (k! (n-k)!))

So, plugging in the values we get:

[tex]P(X = 12) = (20 choose 12) (0.45)^{12} (1 - 0.45)^{(20 - 12)}\\ = (167,960)(0.45)^{12}(0.55)^{8}\\ = 0.167[/tex]
Therefore, the probability that exactly 12 of the selected people belong to a labor union is approximately 0.167 or 16.7%.

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Answer:

44/35

Step-by-step explanation:

this answer cannot be further simplified*

Answer:

To find the distance between two points, we can use the distance formula:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

where (x1, y1) and (x2, y2) are the coordinates of the two points.

In this case, the two points are (3, 4) and (6, 8), so we have:

d = sqrt((6 - 3)^2 + (8 - 4)^2)

d = sqrt(3^2 + 4^2)

d = sqrt(9 + 16)

d = sqrt(25)

d = 5

Therefore, the distance between the points (3, 4) and (6, 8) is 5 units.

It's worth noting that the values 5√22 and √7 do not match the above

f(x) is concave down on the interval -5 ≤ x ≤ -6 and 0 ≤ x ≤ 6.

f(x) is concave up on the interval -6 ≤ x ≤ 0.

To determine where f(x) is concave up or concave down, we need to calculate the second derivative of f(x):

f(x) = x √([tex]x^2[/tex] + 36)

f'(x) = √[tex]x^2[/tex] + 36) + [tex]x^2[/tex] √([tex]x^2[/tex] + 36)

f''(x) = (x ([tex]x^2[/tex] +72) )/(([tex]x^2[/tex]+36)[tex]^(3[/tex]/2))

To find where f(x) is concave up or concave down, we need to find where f''(x) > 0 (concave up) or f''(x) < 0 (concave down).

f''(x) = 0 when x = 0 or x = +/-6.

Thus, f(x) is concave down on the interval -5 ≤ x ≤ -6 and 0 ≤ x ≤ 6, and concave up on the interval -6 ≤ x ≤ 0.

The inflection point for this function is at x = 0.

To find the minimum and maximum for this function, we need to look at the endpoints and critical points of the interval -5 ≤ x ≤ 6.

f(-5) = -5√61 and f(6) = 6√72, so the minimum occurs at x = -5 and the maximum occurs at x = 6.

Therefore:

f(x) is concave down on the interval -5 ≤ x ≤ -6 and 0 ≤ x ≤ 6.

f(x) is concave up on the interval -6 ≤ x ≤ 0.

The inflection point for this function is at x = 0.

The minimum for this function occurs at x = -5.

The maximum for this function occurs at x = 6.

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42 inches divided by 1 gives the measurement 3 feet and 6 inches.

We have to find what number divides the number 42 inches to 3 feet and 6 inches.

We know that the conversion of measurement units,

1 foot = 12 inches

3 feet = 3 × 12 inches = 36 inches

3 feet 6 inch = 36 + 6 = 42 inches

So the required number divides 42 inches in to 42 inches itself.

Any number divided by 1 gives the same number.

So the required number is 1.

Hence the unknown number which divide 42 inches to 3 feet and 6 inches is 1.

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a. The probability of getting a z-score less than -2.67 is 0.0038

b. The range to capture the middle 95% of averages is 66.56 mph to 69.44 mph.

c. The range to capture the middle 90% of averages is 66.77 mph to 69.23 mph.

d. the probability of having an average exceeding 67 mph is 0.9082.

a. The sampling distribution of the mean of a sample of 16 cars is normally distributed with a mean of 68 mph and a standard deviation of 3/√16 = 0.75 mph. The shape of the distribution is normal, the center is 68 mph, and the standard deviation is 0.75 mph. To find the probability that the mean speed of a random sample of 16 cars is less than 66 mph, we need to calculate the z-score:

z = (66 - 68) / 0.75 = -2.67

Using a z-table, we find that the probability of getting a z-score less than -2.67 is 0.0038.

b. To capture the middle 95% of averages, we need to find the z-scores that correspond to the 2.5th and 97.5th percentiles of the normal distribution. Using a z-table, we find that these z-scores are -1.96 and 1.96, respectively. Then we can use the formula:

68 + (-1.96)(0.75) < μ < 68 + (1.96)(0.75)

which gives us the range of 66.56 mph to 69.44 mph.

c. To capture the middle 90% of averages, we need to find the z-scores that correspond to the 5th and 95th percentiles of the normal distribution. Using a z-table, we find that these z-scores are -1.645 and 1.645, respectively. Then we can use the formula:

68 + (-1.645)(0.75) < μ < 68 + (1.645)(0.75)

which gives us the range of 66.77 mph to 69.23 mph.

d. To find the probability of having an average exceed 67 mph, we need to find the z-score that corresponds to 67 mph:

z = (67 - 68) / 0.75 = -1.33

Using a z-table, we find that the probability of getting a z-score less than -1.33 is 0.0918. Therefore, the probability of having an average exceed 67 mph is 1 - 0.0918 = 0.9082.

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Answer:

0.6

Step-by-step explanation:

WE KNOW THAT

P(E)+P(F)=1

P(E)=0.4

NOW

P(E)+P(F)=1

0.4+P(F)=1

P(F)=0.6

HENCE THE PROBABILITY OF NOT SPINNING A BLUE COLOUR IS 0.6

Probability of not spinning a blue colour is 0.6

We know that sum of all Probability is 1,

So the probability of not spinning a blue is = 1 - Probability of  spinning a blue colour.

Putting values we get, = 1 - 0.4 = 0.6

Hence the probability of not spinning a blue colour is 0.6

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The probability that the mean weight of 20 randomly selected men is between 170 and 185 lbs is approximately 0.7189 or approximately 72%.

To solve this problem, we need to use the formula for the sampling distribution of the mean, which states that the mean of a sample of size n drawn from a population with mean μ and standard deviation σ is normally distributed with a mean of μ and a standard deviation of σ/sqrt(n).

In this case, we have a population of men with a mean weight of 178 lbs and a standard deviation of 26 lbs. We want to know the probability that 20 randomly selected men will have a mean weight between 170 and 185 lbs.

First, we need to calculate the standard deviation of the sampling distribution of the mean. Since we are taking a sample of size 20, the standard deviation of the sampling distribution is:

σ/sqrt(n) = 26/sqrt(20) = 5.82

Next, we need to standardize the interval between 170 and 185 lbs using the formula:

z = (x - μ) / (σ/sqrt(n))

For x = 170 lbs:

z = (170 - 178) / 5.82 = -1.37

For x = 185 lbs:

z = (185 - 178) / 5.82 = 1.20

Now we can use a standard normal distribution table (or a calculator) to find the probability of the interval between -1.37 and 1.20:

P(-1.37 < z < 1.20) = 0.8042 - 0.0853 = 0.7189

Therefore, the probability that 20 randomly selected men will have a mean weight between 170 and 185 lbs is 0.7189 or approximately 72%.

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The HR administrator must collect a sample size of 108 employees to be accurate within 0.09 at the 95% confidence interval.

To determine the necessary sample size, we need to use the formula:

[tex]n = \frac{(z^2 )(p) (1-p)}{E^2}[/tex]

Where:
- n = sample size
- z = the z-score for the desired level of confidence (in this case, 1.96 for 95%)
- p = the estimated proportion of employees using the benefit (since we have no preliminary notion, we will use 0.5 as the most conservative estimate)
- E = the desired margin of error (0.09)

Plugging in these values, we get:

[tex]n = \frac{(1.96^2 )(0.5) (1-0.5)}{0.09^2}[/tex]
n = 107.92 = 108

We round up to the next whole number since we can't have a fraction of a person in our sample.

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