Which of the following equations represents a function (15 points)

Answer:

the answer is $598

Step-by-step explanation:

299×2=598

The actual area of the closet's floor is 3.52 square meters.

To find the actual area of the closet's floor, we first need to convert the dimensions from the scale drawing to their actual dimensions.

According to the given scale, 2 cm on the scale drawing represents 1 meter in reality. Thus, to convert the dimensions from the scale drawing to actual dimensions, we need to multiply each dimension by the conversion factor:

Actual length = 4.4 cm × (1 m / 2 cm) = 2.2 m

Actual width = 3.2 cm × (1 m / 2 cm) = 1.6 m

Now that we have the actual dimensions, we can calculate the actual area of the closet's floor:

Actual area = Actual length × Actual width

= 2.2 m × 1.6 m

= 3.52 square meters

Therefore, the actual area of the closet's floor is 3.52 square meters.

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A)The pdf assigns a positive probability to each possible value of the random variable

B)The height of the graph of the equation must be greater than or equal to 0 for all possible values of the random variable.

D)The pdf represents a valid probability distribution, where the probabilities sum up to 1.

What is probability density?

Probability density refers to a concept in probability theory that is used to describe the likelihood of a continuous random variable taking on a particular value within a given range. It is associated with continuous probability distributions, where the random variable can take on any value within a specified interval.

A probability density function (pdf) is a function that describes the likelihood of a random variable taking on a specific value within a certain range. The properties of a pdf are as follows:

A. The probability that X takes on any single individual value is greater than 0. This means that the pdf assigns a positive probability to each possible value of the random variable.

B. The height of the graph of the equation must be greater than or equal to 0 for all possible values of the random variable. This ensures that the pdf is non-negative over its entire range.

C. The values of the random variable must be greater than or equal to 0. This property is not necessarily true for all pdfs, as some may have support on negative values or extend to negative infinity.

D. The total area under the graph of the equation over all possible values of the random variable must equal 1. This property ensures that the pdf represents a valid probability distribution, where the probabilities sum up to 1.

E. The graph of the probability density function may or may not be symmetric. Symmetry is not a universal property of pdfs and depends on the specific distribution.

F. The high point of the graph is not necessarily at the value of the population standard deviation, \(\sigma$.\) The location of the high point is determined by the specific distribution and is not directly related to the standard deviation.

Therefore, the correct options are A, B, and D.

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Answer:

Yes

Step-by-step explanation:

Yes because if you insert -39 for x and 42 for y then your equation is:

42 = -39 + 81

If you were to solve this equation it would be true.

The car will be worth $2,000 between years 4 and 5. To find the years when the car is worth $2,000, we need to solve the equation f(x) = 2,000. Solving 15500(0.68)^x = 2000 gives x ≈ 4.56. Since x represents years, the car will be worth $2,000 between years 4 and 5, specifically in the fifth year.

To explain further, we are given the depreciation rate function of the used car, f(x) = 15500(0.68), where x represents the number of years since the car was purchased. We need to determine when the car's value reaches $2,000, which means we need to find the value of x that satisfies the equation f(x) = 2,000.

By substituting 2,000 into the equation, we get 15500(0.68)^x = 2,000. To solve for x, we need to isolate the exponential term. Dividing both sides by 15,500, we have (0.68)^x = 2,000/15,500. Taking the logarithm of both sides (base 0.68), we can solve for x. However, since the base is less than 1, we'll take the natural logarithm (ln) instead.

\(ln(0.68)^x = ln(2,000/15,500)x ln(0.68) = ln(2,000/15,500)x = ln(2,000/15,500) / ln(0.68)x ≈ 4.56\)

Since x represents years, we can conclude that the car will be worth $2,000 between years 4 and 5, specifically in the fifth year.

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The number of nonnegative integer solutions to the inequality x1 + x2 + x3 ≤ 11 is C(14,3) = 364.

We can solve this inequality by introducing an auxiliary variable x4, such that x1 + x2 + x3 + x4 = 11. Here, x1, x2, x3, and x4 are all nonnegative integers.

We can interpret this equation as follows: imagine we have 11 identical objects and we want to distribute them among four boxes (x1, x2, x3, and x4). Each box can contain any number of objects, including zero. The number of solutions to this equation will give us the number of nonnegative integer solutions to the original inequality.

We can use a technique known as stars and bars to count the number of solutions to this equation. Imagine we represent the 11 objects as stars: ***********.

We can then place three bars to divide the stars into four groups, each group representing one of the variables x1, x2, x3, and x4. For example, if we place the first bar after the first star, the second bar after the third star, and the third bar after the fifth star, we get the following arrangement:

| ** | * | ****

This arrangement corresponds to the solution x1=1, x2=2, x3=1, and x4=7. Notice that the number of stars to the left of the first bar gives the value of x1, the number of stars between the first and second bars gives the value of x2, and so on.

We can place the bars in any order, so we need to count the number of ways to arrange three bars among 14 positions (11 stars and 3 bars). This is equivalent to choosing 3 positions out of 14 to place the bars, which can be done in C(14,3) ways.

Therefore, the number of nonnegative integer solutions to the inequality x1 + x2 + x3 ≤ 11 is C(14,3) = 364.

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Answer: m<ABE = 115.5, m<DBE = 25.5

Step-by-step explanation:

I said that DBE is 25.5 because I know that <DBC is 90 degrees so I substracted 90 - 64.5 which gave me 25.5.

Since we know that DBC is 90 degrees that means DBA is also 90 degrees since ABC is a supplementary angle (180 degrees) (90 + 90 = 180)
Therefore, I did 90 + 25.5 to find m<ABE

Hope this helps.

Answer:

yes, -30 is greater than -90

Step-by-step explanation:

Answer:

Yes, its correct

Step-by-step explanation:

Answer:

A = 3 + 9/10

B = -(3 + 3/10)

Step-by-step explanation:

We have two numbers, A and B, such that:

A + B = 3/5

A - B = 7 + 1/5

To solve this, we first need to isolate one of the variables in one of the equations, I will isolate A in the first equation:

A = 3/5 - B

Now we can replace this in the other equation:

A - B = 7 + 1/5

(3/5 - B) - B = 7 + 1/5

3/5 - 2*B = 7 + 1/5

-2*B = 7 + 1/5 - 3/5

-2*B = 7 - 2/5 = 35/5 - 2/5 = 33/5

B = -(33/5)/2 = -33/10 = - 30/10 - 3/10 = -3 - 3/10 = -(3 + 3/10)

Now that we know the value of B, we can use the equation

A = 3/5 - B

To find the value of A:

A = 3/5 - ( -(3 + 3/10))

A = 3/5 + (3 + 3/10)  = 6/10 + 3 + 3/10 = 3 + 9/10

The numbers are:

A = 3 + 9/10

B = -(3 + 3/10)

Based on the given data, we can calculate the control limits using 3-sigma for the diameter of the auto pistons.

a) The value of x is 155.56 mm (rounded to two decimal places).

b) The value of R is 4.44 mm (rounded to two decimal places).

c) Using 3-sigma, the Upper Control Limit (UCL) for the diameter is calculated as:

UCL = x + 3R = 155.56 + 34.44 = 156.93 mm (rounded to two decimal places)

The Lower Control Limit (LCL) for the diameter is calculated as:

LCL = x - 3R = 155.56 - 34.44 = 154.19 mm (rounded to two decimal places)

d) Using 3-sigma, the Upper Control Limit for the Range (UCLR) is calculated as:

UCLR = D4 * R = 2.115 * 4.44 = 7.89 mm (rounded to two decimal places)

The Lower Control Limit for the Range (LCLR) is always 0 in this case since negative ranges are not possible.

e) If the true diameter mean should be 155 mm, the new centerline (nominal line) would be 155 mm. In this case, the UCL and LCL would be calculated using 3-sigma as follows:

UCL = Nominal + 3R = 155 + 34.44 = 156.37 mm (rounded to two decimal places)

LCL = Nominal - 3R = 155 - 34.44 = 153.63 mm (rounded to two decimal places)

Please note that the control limits calculated using 3-sigma assume a normal distribution and the data follows the same pattern in the future.

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The complete question is:

What is the value of x? = x= 155.56 mm (round your response to two decimal places). b) What is the value of R? R= 4.44 mm (round your response to two decimal places). c) What are the UCL, and LCL; using 3-sigma? Upper Control Limit (UCL) = 156.93 mm (round your response to two decimal places). Lower Control Limit (LCL) = 154.19 mm (round your response to two decimal places). d) What are the UCLR and LCLR using 3-sigma? Upper Control Limit (UCLR)= 7.89 mm (round your response to two decimal places). Lower Control Limit (LCLR)= 0.99 mm (round your response to two decimal places). e) If the true diameter mean should be 155 mm and you want this as your center (nominal) line, what are the new UCL and LCL? Upper Control Limit (UCL)= 156.37 mm (round your response to two decimal places). Lower Control Limit (LCL;)= 153.63 mm (round your response to two decimal places). Refer to Table 56.1-Factors for Computing Control Chart Limits (3 sigma) for this problem Auto pistons at Wemming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: Day 1 2 3 4 5 Mean x (mm) 150.9 153.2 153.6 153.5 154.6 Range R (mm) 4.0 4.8 4.1 4.8 4.5

Finite-state machines for given inputs: (a) 0s multiple of 3: 3-state machine. (b) At least four 1s: 4-state machine. (c) Exactly one 1: 2-state machine. (d) Begins with 000: 3-state machine. (e) Second is 0, fourth is 1: 4-state machine. (f) 01 pairs or 2 1s + 0s: 3-state machine. (g) Ends in 110: 3-state machine.

To construct finite-state machines that act as recognizers for the given inputs, we can follow these guidelines:

(a) For the set of all strings where the number of 0s is a multiple of 3, we can use a finite-state machine with three states. Start with the initial state, and transition to the next state whenever a 0 is encountered. After three transitions, go back to the initial state. If the machine ends in the accepting state, output 1.
(b) For the set of all strings containing at least four 1s, we can use a finite-state machine with four states. Start with the initial state, and transition to the next state whenever a 1 is encountered. If the machine enters the final state after four transitions, output 1.

(c) For the set of all strings containing exactly one 1, we can use a finite-state machine with two states. Start with the initial state and transition to the final state when the first 1 is encountered. Output 1 only if the final state is reached.

(d) For the set of all strings beginning with 000, we can use a finite-state machine with three states. Start with the initial state and transition to the next state whenever a 0 is encountered. If the machine reaches the final state after three transitions, output 1.

(e) For the set of all strings where the second input is 0 and the fourth input is 1, we can use a finite-state machine with four states. Start with the initial state and transition to the next state based on the inputs. Output 1 only if the machine reaches the final state.

(f) For the set of all strings consisting entirely of any number (including none) of 01 pairs or consisting entirely of two 1s followed by any number (including none) of 0s, we can use a finite-state machine with three states. Start with the initial state and transition based on the inputs. Output 1 only if the final state is reached.

(g) For the set of all strings ending in 110, we can use a finite-state machine with three states. Start with the initial state and transition based on the inputs. Output 1 only if the final state is reached.

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Finite-state machines (FSMs) can be constructed to act as recognizers for specific patterns in input strings. These are examples of how to construct FSMs as recognizers for different patterns in input strings. Each FSM is designed to produce an output of 1 when the input received matches the description provided.

Let's consider the given cases and construct FSMs for each one.

(a) The set of all strings where the number of Os is a multiple of 3:
To construct an FSM for this, we can keep track of the number of Os encountered so far. Initially, set the count to zero. When an O is encountered, increment the count by one. If the count becomes a multiple of 3, the FSM outputs 1; otherwise, it outputs 0. Reset the count to zero whenever a 1 is encountered.

(b) The set of all strings containing at least four 1s:
To create an FSM for this, we can keep track of the number of 1s encountered so far. Initially, set the count to zero. When a 1 is encountered, increment the count by one. If the count becomes equal to or greater than four, the FSM outputs 1; otherwise, it outputs 0.

(c) The set of all strings containing exactly one 1:
To build an FSM for this, we can have two states: a "no 1 encountered" state and a "1 encountered" state. Initially, start in the "no 1 encountered" state. Whenever a 1 is encountered, transition to the "1 encountered" state. If another 1 is encountered in the "1 encountered" state, transition to a third "more than one 1 encountered" state. In this case, the FSM outputs 0. Otherwise, if no additional 1s are encountered, the FSM outputs 1.

(d) The set of all strings beginning with 000:
To create an FSM for this, start in an initial state. When a 0 is encountered, transition to a second state. If two consecutive 0s are encountered in the second state, transition to a third state. Finally, if a third 0 is encountered in the third state, the FSM outputs 1; otherwise, it outputs 0.

(e) The set of all strings where the second input is 0 and the fourth input is 1:
To construct an FSM for this, start in an initial state. When the first input is read, transition to a second state. In the second state, transition to a third state if the second input is 0. In the third state, transition to a fourth state if the third input is not 0. Finally, in the fourth state, if the fourth input is 1, the FSM outputs 1; otherwise, it outputs 0.

(f) The set of all strings consisting entirely of any number (including none) of 01 pairs or consisting entirely of two Is followed by any number (including none) of Os:
To create an FSM for this, we can have multiple states to represent different scenarios. We start in an initial state and transition to a second state when a 0 is encountered. In the second state, transition back to the initial state if a 1 is encountered. If a 1 is encountered in the initial state, transition to a third state. In the third state, transition to a fourth state if an O is encountered. Finally, if an O is encountered in the fourth state, the FSM outputs 1; otherwise, it outputs 0.

(g) The set of all strings ending in 110:
To construct an FSM for this, start in an initial state. Transition to a second state if a 1 is encountered. In the second state, transition to a third state if a 1 is encountered again. Finally, if a 0 is encountered in the third state, the FSM outputs 1; otherwise, it outputs 0.

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Answer:

x = \(\frac{4}{5},5\)

Step-by-step explanation:

If both the triangles ΔABC and ΔBCD are congruent,

Corresponding sides of both the triangles will be proportional.

\(\frac{AB}{BD}=\frac{AC}{CD}\)

\(\frac{5x}{3x+10}=\frac{5x-2}{4x+3}\)

5x(4x + 3) = (5x - 2)(3x + 10)

20x² + 15x = 15x² + 50x - 6x - 20

20x² + 15x = 15x² + 44x - 20

20x² - 15x² = 44x - 15x - 20

5x² = 29x - 20

5x² - 29x + 20 = 0

5x² - 25x - 4x + 20 = 0

5x(x - 5) - 4(x - 5) = 0

(5x - 4)(x - 5) = 0

\(x=\frac{4}{5},5\)

Answer: 66 2/3% decrease because you got less this week and if you do what you got this week over what you got last week then you get the percent dicrease

Step-by-step explanation:

The speed of the water wave is 2.0 meters per second.

The speed of a wave is calculated by dividing the distance traveled by the time it takes to travel that distance. In this case, the distance traveled by the water wave is 10.0 meters, and the time taken is 5.0 seconds.

To determine the speed, we use the formula:

Speed = Distance / Time

Substituting the given values, we have:

Speed = 10.0 meters / 5.0 seconds = 2.0 meters per second

Therefore, from the given information, we can determine that the speed of the water wave is 2.0 meters per second.

This information about the speed of the water wave is useful for various purposes. It allows us to understand how quickly the wave is propagating through the medium. It also helps in analyzing wave behavior, such as interference, reflection, or refraction, and studying the characteristics of the medium through which the wave is traveling. Additionally, the speed of the wave can be used in calculations involving wave frequencies, wavelengths, and periods.

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Answer:

-7/9

Step-by-step explanation:

To solve this equation you must isolate the variable.

First, add like terms (terms with the same variable and degree)

2x+2=11x+9

Then, subtract 2x from both sides

2=9x+9

Next, subtract 9 from both sides

-7=9x

Finally, divide both sides by 9

\(\frac{-7}{9}=x\)

The value of the function sin 120 is  0.866025403784. So, the correct answer is A).

We can find the value of sin 120 using the unit circle or a calculator.

Using the unit circle, we draw a radius from the origin to the point on the circle where the angle measure of 120 degrees intersects.

This forms a right triangle with the x-axis and a line from the origin to the point. The length of the side opposite the angle is √3/2, and the length of the hypotenuse is 1. Therefore, sin 120 = √3/2 ≈ 0.866025403784.

Using a calculator, we can directly evaluate sin 120. Make sure the calculator is set to degree mode, and then enter sin(120) or sin(120°). The result is approximately 0.580611184212.

Therefore, the answer is (a) 0.866025403784.

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For the student the lift increased for the weight will be 50%.

What is mean by Percentage?

A number or ratio that can be expressed as a fraction of 100 or a relative value indicating hundredth part of any quantity is called percentage.

Given that;

A student doing resistance exercise.

After one week, She can lift the weight = 2 kg

And, After four weeks, she can lift the weight = 3 kg

Since, We know that;

Percent = (increased weight - Actual weight) / Actual weight  x 100

Substitute all the values, we get;

Increased percent = (3 - 2) / 2 x 100

Increased percent = 1 / 2 x 100

Increased percent = 100 / 2

Increased percent = 50%

Therefore,

For the student the lift increased for the weight will be 50%.

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We need to add the first 4 terms of the series in order to find the sum to the indicated accuracy of 0.00005.

The series ∑n=1[infinity]​(−1)n−1n49​ is an alternating series, which means that the terms alternate in sign and decrease in size.

This type of series converges,and the error   in approximating the sum with the first n terms is less than or equal to the absolute value of the (n+1)th term.

In this case, we are given that the desired accuracy is 0.00005.

The (n+1)th   term of the series is (-1)^n / n⁴⁹, so we need to find the smallest n such that (-1)^n / n⁴⁹ <0.00005.

Using a calculator, we can find that n = 4   satisfies this condition. Therefore, we need to add the first 4 terms of the series in order to find the sum to the indicated accuracy.

The first 4 terms of the series are -

1/1⁴⁹ = 1

-1/2⁴⁹ = -1/1610612736

1/3⁹ = -1/29859862048

-1/4⁴⁹ = 1/209227898880

The sum of these 4 terms is 0.12345, which is accurate to within 0.00005

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c=1/2  which makes fx|y (x, y) a valid conditional probability mass function.

What is conditional probability?

The possibility of an event or outcome occurring based on the existence of a prior event or outcome is known as conditional probability. It is determined by multiplying the likelihood of the earlier event by the increased likelihood of the later, or conditional, event.

Given that;

p(x) = \(c (\frac{2}{3} )^n\)   where x = 1,2,3,........

for getting value of c we are using the expression :

\(\sum_{n=1}^{\infty} c(\frac{2}{3})^n = 1\)

\(c\sum_{n=1}^{\infty} (\frac{2}{3})^n = 1\)

Since we want p(x) to be a Probability Mass Function, your approach is correct, since for any such function

fX:A→[0,1],X:S→A⊆R, it is :

\(\sum _{x\in A} \ fX(x)=1\)

As long as you've correctly understood that condition, you can proceed with your specific exercise as follows:

\(\sum_{n=1}^{\infty} c(\frac{2}{3})^n = 1\)

⇔ \(c\sum_{n=1}^{\infty} (\frac{2}{3})^n = 1\)

⇔2c = 1

⇔c = 1/2

Hence c =  1/2 is the valid conditional probability mass function.

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Answer:

The answer is C.

Step-by-step explanation:

The truck can hold up to 22 boxes.

Answer:C

Step-by-step explanation:

2.3 : 2.2

3x : 5y and 7 : 11

3x : 5y

7 : 11

To find x:

3x = 7

x = 7/3 (Bring 3 over to 7 to divide)

x = 2.3 OR 2/1/3

To find y:

5y = 11

y = 11/5 (Bring 5 over to 11 to divide)

y = 2.2 OR 2/1/5

So,

x = 2.3

y = 2.2

Therefore, the answer is:

x : y = 2.3 : 2.2

The IQR becomes a 1.5.

IQR stands for Interquartile Range, which is a measure of variability in a set of data. It is calculated as the difference between the upper quartile (Q3) and the lower quartile (Q1) of the data.

To determine how the IQR changes, we first need to find the quartiles of the data set.

Arranging the data set in ascending order:

5.5, 6, 6.5, 7, 7.5, 8, 8, 8.5

The median is 7, which is the second quartile (Q2).

To find the first quartile (Q1), we find the median of the lower half of the data set:

5.5, 6, 6.5, 7

The median is 6.25.

To find the third quartile (Q3), we find the median of the upper half of the data set:

7.5, 8, 8, 8.5

The median is 8.

Therefore, the IQR = Q3 - Q1 = 8 - 6.25 = 1.75.

If we add a shoe size of 7 to the data set, the new data set becomes:

5.5, 6, 6.5, 7, 7, 7.5, 8, 8, 8.5

Arranging the new data set in ascending order:

5.5, 6, 6.5, 7, 7, 7.5, 8, 8, 8.5

The median is 7, which is still the second quartile (Q2).

To find the first quartile (Q1), we find the median of the lower half of the data set:

5.5, 6, 6.5, 7, 7

The median is 6.5.

To find the third quartile (Q3), we find the median of the upper half of the data set:

7.5, 8, 8, 8.5

The median is 8.

Therefore, the IQR = Q3 - Q1 = 8 - 6.5 = 1.5.

So, the answer is: the IQR becomes a 1.5.

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Answer:

The slope of the line is \(\frac{5}{2}\)

Step-by-step explanation:

Hi there!

We are given the following points: (-2, 4) and (0, 9)

We want to find the slope (m) of these 2 lines, and the problem asks us to use the slope formula

The slope formula is given as \(\frac{y_2-y_1}{x_2-x_1}\), where \((x_1, y_1)\) and \((x_2, y_2)\) are points

We have everything we need to calculate the slope, but let's label the values of the points to avoid any confusion.

\(x_1=-2\\y_1=4\\x_2=0\\y_2=9\)

Now substitute these values into the formula to find the slope.

m=\(\frac{y_2-y_1}{x_2-x_1}\)

m=\(\frac{9-4}{0--2}\)

Simplify.

m=\(\frac{9-4}{0+2}\)

Add (or subtract) the numbers together

m=\(\frac{5}{2}\)

The slope of the line is \(\frac{5}{2}\)

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The difference between a type II error and a type I error is that a type I error rejects the null hypothesis when it is true (i.e., a false positive). The probability of committing a type I error is equal to the level of significance that was set for the hypothesis test.

The null hypothesis in inferential statistics is that two possibilities are equal. The underlying assumption is that the observed difference is just the result of chance. It is feasible to determine the probability that the null hypothesis is correct using statistical testing.

A statistical hypothesis known as a null hypothesis asserts that no statistical significance can be found in a collection of provided observations. Using sample data, hypothesis testing is performed to judge a theory' veracity. It is sometimes referred to as just "the null," and its symbol is H0.

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Answer:

No Solution

Step-by-step explanation:

As the two lines have the identical slope (-1/2x), that means these two lines are parallel, and therefore they have no solution.

Answer:

$ 360

Step-by-step explanation:

Puppy is on sale for = $ 324

It is marked down by 10% from original selling price of $ 360

Markup from the original buying cost = $ 54

Original Cost was = 360 - 54 = $ 306

$ 360 was the original price which was marked down by 10%  i.e. $ 36 to the sale price of $ 324

If the first number is not allowed to be "0", we will have that the aviable numbers will be:

So, we will have 9 posibilites, now:

First, we will have that the biggest number of 7 digits that we will be able to write will be 9 999 999 and the smallest is 1 000 000.

Now, we will have that the total number of aviable values is:

Then, the total number of possible is:

The radius of the pool is 9.01 m.

Given,

In the question:

A circular pool is surrounded by a brick walkway 3 m wide.

The area of the walk- way is 198 m^2.

To find the Radius of the pool.

Now, According to the question:

"Area of the circle bounded by the outside edge of the walkway" minus "area of the pool" = "area of the walkway".

Let R = Radius of the pool

Area of the circle bounded by the outside edge of the walkway is:

\(\pi\)(R +3)^2

Area of the pool is:

\(\pi R^2\)

Now, Our equation is:;

\(\pi\)(R +3)^2 - \(\pi R^2\) = 198

\(\pi\)((R+3)^2 - \(R^2\)) = 198

Open the inner bracket :

\(\pi\)(\(R^2+6R+9-R^2\)) = 198

\(\pi\)(6R +9) = 198

6R+9 = 198/\(\pi\)

6R = 198/\(\pi\) - 9

R = (198/\(\pi\) - 9)/6

R = (198/(3.14) - 9)/6

R = (63.057 - 9)/6

R = 54.057/6

R = 9.01 meters

Hence, The radius of the pool is 9.01 m.

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