In 1979 topical storm Claudette produced torrential rains when it hit Texas. The highest one-day total was reported in Alvin, Texas where a record breaking 42 inches of rain fell in a single day. This remains the 24 hour record for any location in the United States. A rectangular region R of a National Weather Service isohyet map has been subdivided into grid areas, each 5 miles by 5 miles. The isohyets show levels of rainfall in inches within the 3 day period July 24-27, 1979. If the accumulated rain water somehow didn't flow away and formed a watery surface in the region R, isohyets will be the level sets of that surface.

The closest answer choice to the probability that a patient has the disease given a positive test result is 0.16.

To determine the probability that a patient has the disease given a positive test result, we need to use Bayes' theorem:

[tex]P_{(disease| positive test)} = P_{(positive test)} \times P_{(disease)} / P_{(positive test)}[/tex]

where,

[tex]P_{(disease |positive test)}[/tex] = probability of having the disease given a positive test result

[tex]P_{(positive test|disease)[/tex] = sensitivity = 0.7

[tex]P_{ (disease)[/tex]  = unconditional probability of having the disease = 0.04

[tex]P_{(positive test)} = probability of testing positive = (P_{(positive test |disease)}\times P_{(disease)}) + (P_{(positive test |no disease)}\times P_{(no disease)})[/tex]

To calculate P(positive test |no disease), we need to use the specificity of the test, which is:

[tex]P_{(negative test |no disease)}[/tex]  = specificity = 0.85

Therefore,

[tex]P_{(positive test |no disease)} = 1 - P_{(negative test |no disease)} = 1 - 0.85 = 0.15[/tex]

And,

[tex]P_{(positive test)} = (0.7 \times 0.04) + (0.15 \times 0.96) = 0.0676[/tex].

Now we can calculate the probability of having the disease given a positive test result as follows:

[tex]P_{(disease |positive test)} = 0.7 \times 0.04 / 0.0676 = 0.413.[/tex]

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Answer:

the median is D) 12

Step-by-step explanation:

First of all what it median

median is the value in the middle of a data set.

For example:- The median of 2,3,4 is 3. In Maths, the median is also a type of average, which is used to find the centre value.

Answer:

Step-by-step explanation:

Stem leaf plots are read from top to bottom

The center columned number is the first digit in the number, your 10's place (stem)

The other numbers to right and left are the leaves. and will be your ones place.

So the list of numbers for seaside would be

05, 08

10, 11, 12, 15, 16, 18

25, 25, 27, 27, 28

30 and 36

Put them in a line and find the middle number  I counted,  on the chart to 7.  I counted the (5, 8, 0, 1, 2, 5, 6)   6 was my 7th number with a one in front making it 16

Numbers for Bayside (reads somewhat backwards) since leaves go towards left

05, 06, 08

10, 12, 14, 15, 16, 18

20, 20, 22, 23, 25

42

no leaves in front of 3 on this side so no numbers for 30's

Count 7 and that's 15

Since there are 15 for each school, count 7 numbers and that's your middle number for each

Bayside 15

Seaside 16

Variability is range of numbers

Bayside goes from 5 to 42 so 42-5=38

Seaside 36-5=31

Answer:

31

Step-by-step explanation:

Since there are 15 for each school, count 7 numbers and that's your middle number for each

Bayside 15

Seaside 16

Variability is range of numbers

Bayside goes from 5 to 42 so 42-5=38

Seaside 36-5=31

Answer:

--------------------------

Area of a trapezoid formula:

We have the following values as per picture:

Substitute the given values into formula to get the area:

Hence the area of the trapezoid is 15h cm².

Answer:

15h cm²

Step-by-step explanation:

<3

Jonty is correct as the original cost of paint is £196 which is less than £200 as said by him.

The area of part of cuboid to be painted will be the sum of all the unpainted areas. So, the area remaining to be painted will be = (2 × 3 × 2.5) + (2 × 3 × 12) + (12 × 2.5)

Remaining area = 15 + 72 + 30

Remaining area = 117 m²

Let us assume the original cost of paint be x. So,

x + 10%x = 26.95

110x = 26.95 × 100

110x = 2695

x = £24.5

Now, number of required tins = total unpainted area/area covered by one tin

Number of required tins = 117/15

Number of required tins = 7.8 tins

Taking it as 8 tins.

Previous cost of tins = 8 × 24.5

Previous cost = £196

Since the original cost is less than £200, Jonty is stating truth.

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The complete question is attached in figure.

80% of the time it takes more than 7.52 minutes to find a parking space at 9 A.M.

We can solve this problem by using the inverse normal distribution. We want to find the value of x such that P(X > x) = 0.8, where X is the time it takes to find a parking space.

First, we standardize the distribution: Z = (X - μ) / σ, where μ = 5 and σ = 3. Thus, we want to find the value of z such that P(Z > z) = 0.8.

Using a standard normal distribution table or a calculator, we can find that the z-value corresponding to a cumulative probability of 0.8 is approximately 0.84.

So, we have:

0.84 = (X - 5) / 3

Solving for X, we get:

X = 0.84(3) + 5 = 7.52

Therefore, 80% of the time it takes more than 7.52 minutes to find a parking space at 9 A.M.

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It's important to note that individual results may vary, and it's advisable to consult with a healthcare professional or nutritionist before starting any weight loss program.

(a) To conduct the relevant tests to determine the differences in average weight loss among the four groups, we can use Analysis of Variance (ANOVA) test. ANOVA compares the means of multiple groups to determine if there are significant differences.

After performing the ANOVA test, if we find that there is a significant difference among the means of the four groups, we can conclude that there are differences in average weight loss between the groups. This indicates that the different approaches (keto diet, gym, no exercise, intermittent) have a significant impact on weight loss outcomes.

On the other hand, if the ANOVA test does not reveal a significant difference, we would conclude that there is no evidence of a difference in average weight loss between the groups. In this case, we would not have enough evidence to conclude that the different approaches have distinct effects on weight loss.

The specific conclusions of the study would depend on the results of the ANOVA test and the significance level chosen for the study.

(b) In our study, we investigated the differences in average weight loss among four groups of individuals: those following a keto diet, those going to the gym, those not engaging in any exercise, and those practicing intermittent fasting. We wanted to understand if these different approaches had an impact on weight loss outcomes.

After analyzing the data using statistical methods, we found that there were significant differences in average weight loss among the four groups. This means that the approach individuals take to weight loss does make a difference.

To help visualize the results, we have prepared a bar chart (Figure 1) displaying the average weight loss for each group. As you can see, the group following the keto diet achieved the highest average weight loss, followed by the intermittent fasting group. The gym group also showed significant weight loss compared to the no exercise group.

Overall, our findings suggest that selecting the right approach, such as a keto diet or intermittent fasting, can lead to more substantial weight loss compared to no exercise or less structured methods. It's important to note that individual results may vary, and it's advisable to consult with a healthcare professional or nutritionist before starting any weight loss program.

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The value of system of equations are,

⇒ x = - 1 and y = 1

We have to given that;

The system of equations are,

x + 2y = 1  .. (i)

x = y - 2  .. (ii)

Now, We can plug the value of x in (i);

x + 2y = 1

(y - 2) + 2y = 1

3y - 2 = 1

3y = 3

y = 1

And, From (ii);

x = y - 2

x = 1 - 2

x = - 1

Thus, The value of system of equations are,

⇒ x = - 1 and y = 1

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As per the unitary method, it would cost $33 for Kelly and her two friends to bowl one game before 4 pm and one game after 4 pm each at Super Strike Bowling.

Bowling is a popular recreational activity enjoyed by many people around the world. Super Strike Bowling charges different rates for games played at different times of the day. To model the cost for one game, we can use a step function, where the value of x represents the number of hours after 12 pm. This function is defined as follows:

Cost per game (C) =

$5 if 1 pm ≤ x < 4 pm

$6 if 4 pm ≤ x < 8 pm

$8 if 8 pm ≤ x ≤ 12 am

Now, let's apply this step function to the scenario of Kelly and her two friends bowling. They each played one game before 4 pm, which cost $5 per game, and one game after 4 pm, which cost $6 per game. Therefore, the total cost for one person to play two games is:

Total cost = ($5 per game) + ($6 per game)

= $11

Since there were three people bowling, we can multiply the total cost by 3 to get the cost for all three to bowl:

Cost for all three to bowl = 3 × $11

= $33

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From the side limits of function, f(x), [tex]\lim_{x→ 6^{+}} f(x)= \lim_{ x→6^{- }} f(x) = 2, the limit value of function f(x) when x approaches to 6, [tex] \lim_{x →6} f(x) \\ [/tex] is equals to 2. So, option (d) is right one.

In Calculus a part of mathematics, a limit is the value that a function approaches when its input approaches some other value. That f(x) be approaches L when x approaches 0 then L is called limit of f(x).

Also, limit of a function f(x) if and only if the one sided limits of the function are equal, [tex] \lim_{ x → c} f(x) = L \\ [/tex] iff

[tex] \lim_{ x → c^- } f(x) = \lim_{ x → c^+} f(x) = L \\ [/tex]. We have a limit function f(x) the right hand and left hand limits are defined as, [tex] \lim_{x → 6^{-}} f(x) = 2 \\ [/tex], [tex] \lim_{ x → 6^{+}} f(x) = 2\\ [/tex]

but f( 6) does not exist.

We have to determine the value [tex] \lim_{ x → 6} f(x) \\ [/tex]. From above definition of limit of a function exist, if and only if RHS and LHS limits exist and equal. Here, both RHS and LHS limits are exist and equal so, [tex]\lim_{x→ 6} f(x) = lim_{x→ 6 ^{+}}f(x) = \lim_{x → 6^{-}} f(x) = 2.\\ [/tex] Hence, required value is equals to 2.

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Complete question:

[tex] \lim_{ x -> 6^{ - } } f(x) = 2 \\ [/tex]

and

[tex]\lim_{x --> 6 ^{ + } } f(x) = 2, \\ [/tex]

but f(6) does not exist. What can you say about

[tex] \lim_{x--> 6} f(x) \\ [/tex]

?

A. Is - 2

B. Does not exist

C. Is oo

D. Is 2

The lists that contain only rational numbers is 4/3, -12/13 ,9/4 , -5/7 , 3/4

A rational number can be described as the number which can be expressed in the form of p/q where p and q are integers  when writing this number, q  must not equal to 0 .

Examples of rational numbers are , however in mathematics, a rational number i can be seen as one that can be expressed as the quotient or fraction  which can involves  two integers,  wherby one will be the  numerator p and a non-zero  as well as the denominator q.

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Answer:Step 1:

Enter the expression you want to simplify into the editor.

The simplification calculator allows you to take a simple or complex expression and simplify and reduce the expression to it's simplest form. The calculator works for both numbers and expressions containing variables.

Step-by-step explanation: I really hope this helps

The  area of the lateral face of cylinder = 75.4 in²

The  area of the two bases of the cylinder = 25.13 in²

The total surface area of the cylinder =  100.53 in²

We know that the formula for the surface area of cylinder is:

A = 2πrh + 2πr²

where r is the radius of the cylinder

and h is the height of the cylinder

Here, r = 2 in and h = 6 in

The area of the lateral face of cylinder is given by,

A₁ = 2 × π × r × h

A₁ = 2 × π × 2 × 6

A₁ = 24 × π

A₁ = 75.4 sq. in.

And the area of two base is,

A₂ = 2πr²

A₂ = 2 × π × 2²

A₂ = 8 × π

A₂ = 25.13 sq. in.

The total surface area of cylinder would be,

A = A₁ + A₂

A = 75.4 + 25.13

A = 100.53 sq. in.

Therefore, the required area = 100.53 in²

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Answer:olution:. Given data:. Answer:. sum_(n=4)^10 15(3/10)^(n-1)= sum_(n=4)^10 15(0.3)^(n-1) = 15 [(0.3)^3 + (0.3)^4 + (0.3)^5+ (0.3)^6 + (0.3)^7+ (0.3)^8 + ...

Doesn’t include: 0.58 ‎b ‎1.92 ‎c ‎6.42 ‎d ‎9.43

Evaluate: summation from n equals 4 to 10 of 15 times 3 tenths to the n minus 1 power period Round to the nearest hundredth.

Step-by-step explanation:Example

Evaluate X

4

r=1

r

3

.

Solution

This is the sum of all the r

3

terms from r = 1 to r = 4. So we take each value of r, work out

r

3

in each case, and add the results. Therefore

X

4

r=1

r

3 = 13 + 23 + 33 + 43

= 1 + 8 + 27 + 64

= 100 .

Example

Evaluate X

5

n=2

n

2

.

Solution

In this example we have used the letter n to represent the variable in the sum, rather than r.

Any letter can be used, and we find the answer in the same way as before:

X

5

n=2

n

2 = 22 + 32 + 42 + 52

= 4 + 9 + 16 + 25

= 54 .

Example

Evaluate X

5

k=0

2

k

.

The solution to the recurrence relation is:

[tex]Qn = (1/2)(2^n) - (1/2)[/tex]

To find an explicit formula for the Fibonacci sequence, we first write out the first few terms:

[tex]f0 = 0[/tex]

[tex]f1 = 1[/tex]

[tex]f2 = 1[/tex]

[tex]f3 = 2[/tex]

[tex]f4 = 3[/tex]

[tex]f5 = 5[/tex]

[tex]f6 = 8[/tex]

...

We notice that each term is the sum of the two preceding terms. Therefore, we can write:

[tex]fn = fn-1 + fn-2[/tex]

Let's solve this recurrence relation to find an explicit formula for the nth term. First, we write out the first few terms in terms of f1 and f0:

[tex]f2 = f1 + f0[/tex]

[tex]f3 = f2 + f1 = f1 + f0 + f1 = 2f1 + f0[/tex]

[tex]f4 = f3 + f2 = 3f1 + 2f0[/tex]

[tex]f5 = f4 + f3 = 5f1 + 3f0[/tex]

[tex]f6 = f5 + f4 = 8f1 + 5f0[/tex]

We can see that the coefficients of f1 and f0 are the Fibonacci numbers themselves (1, 1, 2, 3, 5, 8, ...). Therefore, we can write the explicit formula:

[tex]fn = (1/√5) [(1+√5)/2]^n - (1/√5) [(1-√5)/2]^n[/tex]

(a) To solve the recurrence relation [tex]On = 7On-1 - 10On-2[/tex], we first find the roots of the characteristic equation:

[tex]r^2 = 7r - 10[/tex]

[tex]r = (7 ± √(7^2 + 40))/2[/tex]

[tex]r1 = 5, r2 = -2[/tex]

Therefore, the general solution to the recurrence relation is:

[tex]On = c1(5^n) + c2(-2^n)[/tex]

We can find the values of c1 and c2 by using the initial conditions:

[tex]O0 = 1, O1 = 5[/tex]

[tex]c1 + c2 = 1[/tex]

[tex]5c1 - 2c2 = 5[/tex]

Solving these equations, we get:

[tex]c1 = 1, c2 = -1/3[/tex]

Therefore, the solution to the recurrence relation is:

On = 5^n - (1/3)(-2)^n

(b) To solve the recurrence relation Qn [tex]= 2Qn-1 + 1[/tex], we first find the root of the characteristic equation:

[tex]r - 2 = 0[/tex]

[tex]r = 2[/tex]

Therefore, the general solution to the recurrence relation is:

Qn = c(2^n) + d

We can find the values of c and d using the initial conditions:

[tex]Q0 = 0, Q1 = 1[/tex]

[tex]c + d = 0[/tex]

[tex]2c + d = 1[/tex]

Solving these equations, we get:

[tex]c = 1/2, d = -1/2[/tex]

Therefore, the solution to the recurrence relation is:

Qn [tex]= (1/2)(2^n) - (1/2)[/tex]

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The Laplace transform of tJo(t) is s / (s^2 - 1)^(3/2).

To find the Laplace transform of tJo(t), we can use the following formula:

L{t^n f(t)} = (-1)^n F^(n)(s)

where F(s) is the Laplace transform of f(t) and F^(n)(s) denotes the nth derivative of F(s) with respect to s.

Using this formula with f(t) = Jo(t), we have:

L{tJo(t)} = -d/ds [ L{Jo(t)} ]

We can find L{Jo(t)} by using the formula for the Laplace transform of Jo(t):

L{Jo(t)} = 1 / sqrt(s^2 - 1)

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

d/ds [ L{Jo(t)} ] = d/ds [ 1 / sqrt(s^2 - 1) ]

= (-1/2) (s^2 - 1)^(-3/2) (2s)

= -s / (s^2 - 1)^(3/2)

Substituting this result back into our original equation, we get:

L{tJo(t)} = -d/ds [ L{Jo(t)} ]

= -d/ds [ 1 / sqrt(s^2 - 1) ]

= s / (s^2 - 1)^(3/2)

Therefore, the Laplace transform of tJo(t) is s / (s^2 - 1)^(3/2).

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a) The light flashes as the circuit leaves the field at a speed of 16 ms.

b) The current flow as the loop exits the field in the clockwise direction.

c) The power dissipated in the bulb during the flash is 0.04 W. 

To reply to these questions, we will utilize Faraday's Law, which states that a changing attractive field actuates an electromotive drive (EMF) in a circuit, and the initiated EMF is rise to the rate of alter of attractive flux through the circuit.

a) The attractive flux through the circle is given by the item of the attractive field, region of the circle, and cosine of the point between the attractive field and the ordinary to the plane of the circle.

As the circle is pulled out of the attractive field, the magnetic flux through the circle diminishes, and thus, an EMF is actuated within the circle. This initiated EMF drives a current through the light bulb, causing it to light up.

The time term of the streak of light can be decided from the time taken by the circle to move out of the attractive field.

The removal voyage by the circle is 40 cm, and the speed is 25 m/s, so the time taken is:

t = d/v = 0.4 m / 25 m/s = 0.016 s = 16 ms

Subsequently, the streak of light endures for 16 ms.

b) Concurring to Lenz's Law, the course of the initiated current is such that it contradicts the alter within the attractive flux that produces it. As the circle is pulled out of the attractive field, the attractive flux through the circle diminishes.

Hence, the actuated current flows in a course that makes a magnetic field that restricts the initial attractive field. This could be accomplished by the induced current streaming clockwise as seen from above. Hence, the reply is clockwise.

c) The control scattered within the light bulb can be calculated utilizing the equation P = V²/R, where V is the voltage over the bulb and R is its resistance.

The voltage over the bulb is break even with to the initiated EMF, which can be calculated from Faraday's Law. The attractive flux through the circle changes at a rate of (40 cm) x (25 m/s) = 1 T.m²/s.

The region of the circle is (40 cm) x (10 cm) = 0.04 m². The cosine of the point between the attractive field and the ordinary plane of the circle is 1 (since the circle is opposite to the field). Subsequently, the induced EMF is:

EMF = -d(phi)/dt = -NA(dB/dt)

= -(1)(0.04 m²)(1 T.m²/s)/0.016 s

= -1 V

The negative sign indicates that the actuated EMF is within the inverse course of the current stream. Subsequently, the voltage over the light bulb is:

V = -EMF = 1 V

The power dissipated within the bulb is:

P = V²/R = (1 V)²/25 ohm = 0.04 W

Subsequently, the control scattered within the bulb during the streak is 0.04 W. 

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Answer: Use the table to identify values of p and q that can be used to factor x2 + x – 12 as (x + p)(x + q).A. –2 and 6. B. 3 and –4. C. –3 and 4. D. 2 and –6.

Step-by-step explanation:

Answer:

Step-by-step explanation:

2

The shorter sides of heptagon as 9 feet each based on the relation, length of longer sides and total length.

Let the three shorter sides of heptagon (with seven sides) be of x feet. Hence, the remaining four sides will be of 2x feet. Now, sum of their lengths is stated thus, representing them as equation

(4 × 2x) + 3x = 99

Solving the bracket first

8x + 3x = 99

Adding the values on Left Hand Side of the equation

11x = 99

Rewriting the equation in terms of x

x = 99/11

Performing division on Right Hand Side of the equation

x = 9

Hence, the length of shorter sides is 9 feet each.

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The measure of <BAE is 130 degree.

We have,

m <C = 65

As, opposite sides are parallel then ABCD is a parallelogram.

Then, opposite angles of a parallelogram are congruent

∠ BAD = ∠ BCD = 65°

∠ ADE = ∠ BAD = 65° ( alternate angles )

Since, DE = AE then Δ ADE is isosceles Triangle.

So, ∠ DAE = ∠ ADE = 65°

We can write,

∠ BAE = ∠ BAD + ∠ DAE = 65° + 65° = 130°

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To determine whether a data set is approximately periodic, we need to look for patterns that repeat over time. If we see a consistent pattern in the data that repeats with some regularity, then we can say that the data set is approximately periodic.

If the data set is approximately periodic, we also need to determine its period and amplitude. The period is the time it takes for the pattern to repeat, while the amplitude is the distance between the highest and lowest points of the pattern.

Without more information about the data set, it's difficult to say for certain whether it's approximately periodic. However, if we assume that it is, we can make some educated guesses about its period and amplitude.

Based on the information given, it's possible that the data set has a period of either 3 or 4, and an amplitude of about 5 or 7.5. It's difficult to be more precise without seeing the data itself.

In summary, the data set may be approximately periodic with a period of either 3 or 4, and an amplitude of about 5 or 7.5. However, without more information, we can't say for certain whether it's truly periodic.
To determine if the data set is approximately periodic, you need to look for repeating patterns in the data. A periodic data set will have a constant period and amplitude throughout.

Period refers to the interval between repetitions, while amplitude is the maximum value of the fluctuation from the average value.

Unfortunately, you didn't provide a specific data set for me to analyze. However, I can provide you with a general explanation of how to identify periodicity and determine the period and amplitude.

1. Observe the data set to see if there are any repeating patterns.
2. If a pattern is present, find the interval between repetitions - this is the period.
3. Determine the difference between the maximum and minimum values in the pattern.
4. Divide this difference by 2 to find the amplitude.

Once you have applied these steps to your data set, you can compare your results to the provided options to find the best match.

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Let 0-1 represent students who participate in sports and let 2-9 represent students who do not participate in sports.

Let 0-9 represent students selected for the sample, with digits 0-8 representing students who participate in sports and digit 9 representing a student who does not participate in sports.

So, an appropriate assignment of digits for this simulation would be: Let 0-1 represent students who participate in sports and let 2-9 represent students who do not participate in sports.

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There must be a mistake or typo in the expression.

We can evaluate the expression 7/2 = 4 + sin0 as follows:

sin0 is the sine of 0 degrees, which is equal to 0. Therefore, the expression simplifies to:

Now, we can compare this value to the left-hand side of the equation, which is 7/2. Since 7/2 is not equal to 4, we can conclude that the equation is not true.

7/2 = 4 + 0

Next, we can simplify the right side of the equation by combining like terms:

4 + 0 = 4

So, the equation becomes:

7/2 = 4

To check if this is true, we can multiply both sides by 2:

7/2 × 2 = 4 × 2

Simplifying:

7 = 8

Since 7 is not equal to 8, we know that the original equation 7/2 = 4 + sin0 is not true. Therefore, there must be a mistake or typo in the expression.

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The value of the test statistic is approximately 1.73.

To find the value of the test statistic, we can use a one-sample t-test.

The null hypothesis is that the true mean of the population is equal to the normal level of ozone, 5.7 ppm. The alternative hypothesis is that the true mean is not equal to 5.7 ppm.

We can calculate the t-value using the formula:

t = (sample mean - hypothesized mean) / (standard deviation / √(sample size))

Substituting in the given values:

t = (6.1 - 5.7) / (0.7 / √(8))

t = 1.73

To determine if this t-value is significant at a level of significance of 0.02, we need to compare it to the critical t-value from the t-distribution with 7 degrees of freedom (8 samples - 1). Using a t-table or calculator, the critical t-value is 2.998.

Since our calculated t-value of 1.73 is less than the critical t-value of 2.998, we fail to reject the null hypothesis. There is not enough evidence to conclude that the ozone level is not at a normal level.

Therefore, the value of the test statistic is t = 1.73.

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Hawthorne Effect

You are conducting a research study and giving a group of participants an accelerometer. When analyzing the data, you notice that all participants had the highest levels of physical activity on Day 1. You decide to exclude this data. Excluding this data is an example of trying to avoid: Hawthorne Effect

The Hawthorne Effect refers to the phenomenon where participants modify their behavior in response to being observed or aware of being part of a study. By excluding the data from Day 1, you are trying to avoid the potential influence of this effect on the study results.

You collect your data by watching the employees during their work breaks. If employees are aware that you are observing them, this can affect your study's results. For example, you may record higher or lower smoking rates than are genuinely representative of the population under study.

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The radius of the circle is 4 units

A circle is simply a round shape that has no corners or line segments. The body of a circle is called the circumference and a cut out of circumference is called an arc.

The distance from the centre of a circle to any part of its circumference is called a radius. Twice of a radius is called the diameter.

In the circle, the distance between the center of the circle and it's circumference is ;

4-0 = 4 units

Therefore the radius of the circle is 4 units.

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The coordinates of the original figure are (-2, 4), (4, 4), (-2, 1), and (4, 1).

The coordinates of the final transformed figure are (-1, 2), (2, 2), (-1, 0.5), and (-2, 0.5).

In Mathematics and Geometry, a dilation simply refers to a type of transformation which typically changes the size of a geometric figure, but not its shape.

This ultimately implies that, the size of the geometric figure would be increased (stretched or enlarged) or decreased (compressed or reduced) based on the scale factor applied.

Next, we would apply a dilation to the coordinates of the pre-image by using a scale factor of 0.5 centered at the origin as follows:

(-2, 4) → (-2 × 1/2, 4 × 1/2) = (-1, 2).

(4, 4) → (4 × 1/2, 4 × 1/2) = (2, 2).

(-2, 1) → (-2 × 1/2, 1 × 1/2) = (-1, 0.5).

(-4, 1) → (-4 × 1/2, 1 × 1/2) = (-2, 0.5).

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