which value is used in the denominator of the formula for the chi-square test statistic?total number of valuessum of the values expected countobserved count

The possible dimensions of the plot are:

Length = 9.2 meters, width = 25.6 meters

Length = 25.6 meters, width = 9.2 meters

What is quadratic equation?

It's a second-degree quadratic equation which is an algebraic equation in x.

Let the length of the rectangular plot be L meters and the width be W meters. Then the perimeter of the plot is:

P = 2L + W

Since there are only three sides to the fence, we can set the perimeter equal to the length of the fencing:

2L + W = 44

Solving for W, we get:

W = 44 - 2L

The area of the plot is given as 210 square meters:

L * W = 210

Substituting for W, we get:

L * (44 - 2L) = 210

Simplifying and rearranging, we get a quadratic equation:

2L² - 22L + 105 = 0

Using the quadratic formula, we can solve for L:

L = (22 ± √(22² - 4(2)(105))) / (2(2))

L = (22 ± 2√34) / 4

L = 11/2 ± √34/2

We discard the negative solution, so:

L = 11/2 + √34/2 ≈ 9.2 meters

Now we can use the equation W = 44 - 2L to find the corresponding width:

W = 44 - 2(9.2) = 25.6 meters

Therefore, the possible dimensions of the plot are:

Length = 9.2 meters, width = 25.6 meters

Length = 25.6 meters, width = 9.2 meters

Note that both sets of dimensions result in the same area of 210 square meters.

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The sum of the expression is 3.5x+6.8.

Given that are two expressions, we need to find the sum of the expressions,

The expression is a mathematical statement which is constructed by putting arithmetical operations, variables and numbers together.

So, finding the sum.

(-4.25x + 9) and (8x - 2.2)

Putting the sign of addition,

= (-4.25x + 9) + (8x - 2.2)

Opening the brackets,

= -4.5x + 9 + 8x - 2.2

Combining the like terms.

= 8x-4.5x + 9-2.2

Operating the like terms,

= 3.5x+6.8

Hence, the sum of the expression is 3.5x+6.8.

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the distance between point A and point B is approximately 777.9 feet.

what is  distance ?

Distance is a numerical measurement of how far apart two objects or points are from each other. It is a fundamental concept in mathematics and physics, and it can be defined in different ways depending on the context.

In the given question,

Let's denote the distance between Jeriel and point A as x, and the distance between Jeriel and point B as y. We want to find the distance between point A and point B, which we can call d.

From the first measurement, we can draw a right triangle with one leg of length x, another leg of length 113, and an acute angle of 8 degrees. The angle opposite the leg of length 113 is 90 degrees minus 8 degrees, or 82 degrees. We can use tangent to find x:

tan(8) = 113/x

x = 113/tan(8)

x =802.5

From the second measurement, we can draw another right triangle with one leg of length y, another leg of length 113, and an acute angle of 52 degrees. The angle opposite the leg of length 113 is 90 degrees minus 52 degrees, or 38 degrees. We can use tangent to find y:

tan(52) = 113/y

y = 113/tan(52)

y =72.4

Now we can use the Law of Cosines to find d:

d² = x² + y² - 2xy cos(130)

where 130 degrees is the angle between the sides of length x and y. We can simplify this equation and substitute the values we found for x and y:

d² = (802.5)² + (72.4)² - 2(802.5)(72.4) cos(130)

d =777.9

Therefore, the distance between point A and point B is approximately 777.9 feet.

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The area of the square is 16in².

The area of the circle is 12.56in².

The area of the shaded region is 3.44in².

What is circle?

A circle is a geometric shape that consists of all points in a plane that are equidistant from a fixed point called the center. It can also be defined as the set of points that are a fixed distance (called the radius) away from the center point.

To find the area of the shaded region, we need to first find the area of the square and the area of the circle.

The diameter of the circle is equal to the diagonal of the square, which is 2 times the length of a side of the square. Therefore, the length of a side of the square is:

2 x radius = 2 x 2in = 4in

The area of the square is then:

Area of square = side x side = 4in x 4in = 16in²

The area of the circle is:

Area of circle = π x radius² = 3.14 x 2in x 2in = 12.56in²

Now, to find the area of the shaded region, we need to subtract the area of the circle from the area of the square:

Area of shaded region = Area of square - Area of circle = 16in² - 12.56in² = 3.44in²

Therefore, the area of the shaded region is 3.44in².

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The results are:

7a) Explicit = h(n) = [tex]0.87^{n-1}[/tex] * 200.

Recursive = h(0) = 200 and h(n) = 0.87 * h(n-1)

7b) Rebound height = 99.76 cm (rounded to the nearest hundredth).

8a) Explicit formula: A(n) = 575 + 47.50n

Recursive formula: A(0) = 575, A(n) = A(n-1) + 47.50

8b) 30 weeks to save $2000'

7a) Explicit: h(n) = [tex]0.87^{n-1}[/tex] * 200 (

where h(n) = height of the golf ball after n bounces,

0.87 is the rebound factor,

200 is the initial height)

h(n) = [tex]0.87^{n-1}[/tex] * 200

Recursive:  h(0) = 200

h(n) = 0.87 * h(n-1)

where h(n) = the height after n bounces,

h(n-1) = height after (n-1) bounces,

0.87 = rebound factor.

Apply given values:

h(0) = 200

h(n) = 0.87 * h(n-1)

7b) Rebound height, using recursive formula:

h(0) = 200

h(1) = 0.87 * 200 = 174

h(2) = 0.87 * 174 = 151.38

h(3) = 0.87 * 151.38 = 131.70

h(4) = 0.87 * 131.70 = 114.71

h(5) = 0.87 * 114.71 = 99.76 (rounded to the nearest hundredth)

8a)Explicit: A(n) = 575 + 47.50n,

where A(n) = amount of money Jessica has after n weeks,

575 = initial amount given by grandfather,

47.50 = amount Jessica adds each week.

Recursive: A(0) = 575, A(n) = A(n-1) + 47.50,

where A(n) = the amount of money Jessica has after n weeks,

A(n-1) = amount Jessica has after (n-1) weeks

47.50 = amount Jessica adds each week.

8b) Weeks to save $2000:

A(n) = 575 + 47.50n = 2000

Solving for n:

575 + 47.50n = 2000

47.50n = 2000 - 575

47.50n = 1425

n = 1425 / 47.50

= 30 weeks.

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If you are given a sample percentage of 43%, you would need to know the sample size in order to convert this percentage to a proportion.

The conversion formula is proportion = percentage/100. However, the proportion alone does not give information about the sample size, which is necessary for inference and hypothesis testing. The other statements are not true.

The test statistic is not affected by the sample size, but its value can be used to determine the significance of a hypothesis test. A larger p-value indicates weaker evidence against the null hypothesis, not stronger evidence. Finally, we assume the null hypothesis is true until we have sufficient evidence to reject it.

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Answer:

7

(

x

−

2

)

−

7

x

−

14

=

0

7

(

x

−

2

)

−

7

x

−

14

=

07

(

x

−

2

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−

7

x

−

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=

0

Step-by-step explanation:

Answer: 0 = 0

Step-by-step explanation: i showed the steps with these screen shots

The surface area of this solid would be; 127.

To Find the surface area of the composite solid, we have;

Sides (a) area

4 * 3 = 12

12 * 2 = 24

Now Sides (b) area

7 * 3 = 21

21 * 3 = 63

Then Face (c) area

3 * 4 / 2 = 6

6 * 2 = 12

Base area

4 * 7 = 28

Total surface area

To calculate the total surface area we have to add the value of each face :

28 + 12 + 63 + 24 = 127

Thus, the surface area of this solid is 127.

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The paired t-test results suggest that there is no significant improvement in students' learning after the new unit on factoring, based on a 90% confidence level.

To determine if there has been an improvement in students' learning after a new unit on factoring, we can perform a paired t-test on the pre-test and post-test scores of the five randomly selected students.

The differences between the pre-test and post-test scores for each student are: 3, 3, 4, 8, 0.

The mean of the differences is 3.6, and the sample standard deviation is 2.39.

The t-value for a 90% confidence interval with 4 degrees of freedom is 2.776. Since the calculated t-value of 2.68 (calculated as the mean of the differences divided by the standard error of the mean of the differences) is less than the critical t-value of 2.776, we fail to reject the null hypothesis that there is no improvement in students' learning.

Therefore, there is not enough evidence to conclude that the new unit on factoring has helped students learn based on the sample data.

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to get the equation of any straight line, we simply need two points off of it, let's use those two in the picture below

[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{3}-\stackrel{y1}{2}}}{\underset{\textit{\large run}} {\underset{x_2}{3}-\underset{x_1}{0}}} \implies \cfrac{ 1 }{ 3 }\\[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{ \cfrac{ 1 }{ 3 }}(x-\stackrel{x_1}{0})\implies y-2=\cfrac{ 1 }{ 3 }x \\\\\\ {\Large \begin{array}{llll} y=\cfrac{ 1 }{ 3 }+2\implies y=\cfrac{2}{6}x+2 \end{array}}[/tex]

The statement "Both IPv4 and IPv6 provide for subnetting, but the much larger IPv6 address field makes this a lot simpler to design and manage" is correct for subnetting.

Hence, the correct option is B.

In IPv4, subnetting involves dividing a network into smaller subnetworks using the Classless Inter-Domain Routing (CIDR) protocol at Layer 3, which allows for more efficient use of IP addresses. The subnet mask determines the network and host portions of the IP address.

IPv6 also supports subnetting, but uses a different approach called prefix notation, where the network and subnet are represented by a prefix length rather than a subnet mask. Since the address field in IPv6 is much larger than IPv4, there are more available addresses and therefore it is easier to subnet.

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Answer:   C

Step-by-step explanation:

hope this helped

The square to make it a perfect square trinomial is 1/8. The perfect square trinomial is as follows:

(x + 1/6)² + 1/8

A trinomial is an algebraic expression with three terms.

To finish the square of the given quadratic expression, add and subtract the square of half of the linear term's coefficient (x-term), which is (1/6)x, inside the parentheses:

[tex]x^{2} + (1/3)x + c = x^{2} + (1/6)x + (1/6)x + c = (x + 1/6)^{2} + (3/4) c[/tex]

The constant term must be equal to half the coefficient of the linear term squared for the quadratic equation to be a perfect square trinomial. In other terms, we need:

[tex](3/4) c = (1/6)^{2} \\\\Solving for c, we get: c = (1/6)^{2} / (3/4) = 1/8[/tex]

As a result, the value of c in the provided expression that completes the square to make it a perfect square trinomial is 1/8. The perfect square trinomial is as follows:

(x + 1/6)² + 1/8

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The probability that a New Yorker who possesses a cat also owns a dog is 0.103. if a new yorker is chosen at random.

New Yorkers own a dog (D) = 30%

New Yorkers own a cat (C) =  25%

New Yorkers own a cat and dog = 15%

This problem can be calculated using Bayes' theorem, which notes that the possibility of an event A given event B is equal to the possibility of event B given A times the probability of A, divided by the probability of event B.

P(D) = 0.30

P(C) = 0.25

P(C|D) = 0.15

Using Bayes' theorem:

P(D|C) = P(C|D) * P(D) / P(C)

P(C) = [tex]P(C|D) * P(D) + P(C|not D) * P(not D)[/tex]

P(C) = [tex]P(C|D) * P(D) + P(C) * P(not D)[/tex]

P(C) =[tex]P(C|D) * P(D) / (1 - P(D) * P(C|not D))[/tex]

Now we can substitute these values into the Bayes' theorem formula:

P(D|C) = P(C|D) * P(D) / P(C)

P(D|C) = [tex]0.15 * 0.3 / (P(C|D) * P(D) + P(C) * P(not D))[/tex]

P(D|C) = [tex]0.15 * 0.3 / (0.15 * 0.3 + P(C) * 0.7)[/tex]

P(C) = [tex]0.15 * 0.3 / (1 - 0.3 * 0.25)[/tex]

P(C) = 0.16

P(D|C) = [tex]0.15 * 0.3 / (0.15 * 0.3 + 0.16 * 0.7)[/tex]

P(D|C) = 0.103

Therefore, we can conclude that the probability that a New Yorker who owns a cat also owns a dog is 0.103.

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Answer: ~$0.99 (0.9875)

Just divide 3.95 by 4

To create a histogram, we need to divide the range of the data into a set of intervals or "bins" and count the number of data points that fall in each bin. We can then represent this information using a bar chart, where the height of each bar corresponds to the frequency (i.e., number of data points) in that bin.

To determine appropriate bin intervals, we can use the range of the data and the number of bins to create equal-width intervals. For example, we could create 5 bins of width 5:

Bin 1: 2 - 6 Bin 2: 7 - 11 Bin 3: 12 - 16 Bin 4: 17 - 21 Bin 5: 22 - 26

We can then count the number of data points that fall in each bin:

Bin 1: 2 Bin 2: 4 Bin 3: 3 Bin 4: 0 Bin 5: 3

Therefore, the intervals and their corresponding frequencies are:

Interval: 2 - 6 Frequency: 2

Interval: 7 - 11 Frequency: 4

Interval: 12 - 16 Frequency: 3

Interval: 17 - 21 Frequency: 0

Interval: 22 - 26 Frequency: 3

We can then use this information to create a histogram, where the x-axis represents the intervals and the y-axis represents the frequencies, with each bar corresponding to an interval and having a height equal to the frequency in that interval.

Answer:

2x-13

Step-by-step explanation:

First flip the operation to be (-10x^2+2x-10) + (10x^2-3).

Then you simplify it. The 10x^2 would get canceled out. You would then get 2x-13.

Answer:150°

Step-by-step explanation:

The angle measures of the central angle GFE and the circumscribed angle GCE in circle C are 83° and 79° respectively.

Angle is a geometric term used to describe the measure between two lines or surfaces that intersect each other. It is measured in degrees, and is typically represented by the Greek letter θ (theta). Angles can be acute, obtuse, right, reflex, or straight. Acute angles are angles that are less than 90°, obtuse angles are angles that are greater than 90°, and right angles are angles that measure exactly 90°. Reflex angles are angles that measure greater than 180°, and straight angles measure exactly 180°.

The two correct answers are 83° and 79°. The angle measure of the central angle GFE is 3x + 11. Therefore, 3x + 11 = 83°, and x = 26. The angle measure of the circumscribed angle GCE is 5x - 23. Therefore, 5x - 23 = 79°, and x = 26. As both equations have the same value for x, the two angle measures are 83° and 79°.

In conclusion, the angle measures of the central angle GFE and the circumscribed angle GCE in circle C are 83° and 79° respectively.

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The interval best describes the range of the transformed function as [0,70].

What is a function?

Each element of X receives exactly one element of Y when a function from one set to the other is used. The sets X and Y are collectively referred to as the function's domain and codomain, respectively. Initially, functions represented the idealized relationship between two changing quantities.

Here, we have

Given: When f(x) = 0, the swing is at starting position. When f(x) > 0, the swing is in front of the starting position, and when f(x) < 0, the swing is behind the starting position. The function describing this graph is a transformation of the parent sine function, y = sin x.

To begin with, the amplitude of the sine curve, according to the graph, is 70 at the maximum.

y = 70 sin(something)

Next, a full sine wave occurs after 15 seconds approximately. That means after 15 seconds, you should get 2π. Call the values along the x-axis = t

y = 70*sin(t/15 ×2π)

It is just under 4 seconds. It's an estimation.

y = 70×sin(3.9/15 ×2π) Multiply 2 × 3.9

y = 70× sin(7.8/15π) Multiply 7.8π : Use 3.14 for π

y = 70 × sin(24.49/ 15)

y = 70× sin(1.633) Make sure you are in radians for the next step.

y = 70×0.99806 = 69.86 inches. which is pretty close to 70.

Hence, the interval best describes the range of the transformed function as [0,70].

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Answer:

The value of 3⁴ is:

3⁴ = 3 × 3 × 3 × 3 = 81

Therefore, 3⁴ is equal to 81.

Scenic Cinemas should offer discounted weekday matinee showings for seniors and continue to focus on weekend showings for their broader audience.

Based on the survey results, it would be wise for Scenic Cinemas to tailor their offerings to accommodate both preferences.

One option would be to offer discounted weekday matinee showings targeted toward seniors. This could incentivize them to visit on weekdays while also offering them a more affordable option. At the same time, Scenic Cinemas could continue to focus on weekend showings to cater to their broader audience.

Another approach would be to offer more diverse programming during the weekdays, such as classic films or independent movies that may appeal more to seniors. This could create a niche for Scenic Cinemas and attract a more loyal customer base.

Ultimately, it's important for Scenic Cinemas to balance the needs of their different audience segments to maximize revenue and customer satisfaction. By offering targeted promotions and programming, they can ensure that both seniors and other moviegoers feel valued and have a reason to visit the theatre.

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the volume of the apple would increase by ΔVtotal over the five-week period

The volume of the apple is directly proportional to the cube of its radius. Therefore, if the radius increases by 1/8 inch per week for five weeks, the total increase in radius would be 5/8 inches.

To calculate the change in volume, we can use the formula V = (4/3)πr^3.

Initially, let's assume the radius of the apple is r.

After the first week, the radius would be r + 1/8 inches. Therefore, the new volume would be:
[tex]V_1 = (4/3)\pi(r + 1/8)^3[/tex]

To calculate the change in volume, we can subtract the initial volume from the new volume:

[tex]\DeltaV_1 = V_1 - V_0[/tex]

where V0 is the initial volume of the apple.

Similarly, after the second week, the radius would be r + 2/8 inches, and the new volume would be:

[tex]V2 = (4/3)\pi(r + 2/8)^3[/tex]

The change in volume after the second week would be:

ΔV2 = V2 - V1

We can repeat this process for the remaining three weeks and find the total change in volume by adding all the individual changes:

ΔVtotal = ΔV1 + ΔV2 + ΔV3 + ΔV4 + ΔV5

By simplifying this equation, we get:

[tex]\Delta V_{total }=  (4/3) \pi r^3[(r + 1/8)^3 + (r + 2/8)^3 + (r + 3/8)^3 + (r + 4/8)^3 + (r + 5/8)^3 - 5r^3][/tex]

Therefore, the volume of the apple would increase by ΔVtotal over the five-week period.

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The volume of the apple will increase by [tex](4/3)π[(r0 + 5/8)^3 - r0^3][/tex] during the five-week period.

Assuming that the apple is a perfect sphere, we can use the formula for the volume of a sphere:

[tex]V = (4/3)πr^3[/tex]

where V is the volume of the sphere and r is the radius of the sphere.

Let's start by finding the initial volume of the apple. We are not given the initial radius, so let's assume it to be r0.

[tex]V0 = (4/3)πr0^3[/tex]

After one week, the radius increases by 1/8 inch, so the new radius is r1 = r0 + 1/8. The new volume is:

[tex]V1 = (4/3)π(r0 + 1/8)^3[/tex]

After two weeks, the radius increases by another 1/8 inch, so the new radius is r2 = r0 + 2/8. The new volume is:

[tex]V2 = (4/3)π(r0 + 2/8)^3[/tex]

Similarly, after three weeks, the new radius is r3 = r0 + 3/8 and the new volume is:

[tex]V3 = (4/3)π(r0 + 3/8)^3[/tex]

After four weeks, the new radius is r4 = r0 + 4/8 and the new volume is:

[tex]V4 = (4/3)π(r0 + 4/8)^3[/tex]

Finally, after five weeks, the new radius is r5 = r0 + 5/8 and the new volume is:

[tex]V5 = (4/3)π(r0 + 5/8)^3[/tex]

To find how the volume changes during the five-week period, we need to subtract the initial volume from the final volume:

ΔV = V5 - V0

[tex]ΔV = (4/3)π(r0 + 5/8)^3 - (4/3)πr0^3[/tex]

Simplifying this expression, we get:

[tex]ΔV = (4/3)π[(r0 + 5/8)^3 - r0^3][/tex]

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The circumferences and areas of the circles are:

a)

C =  18.84cm

A = 28.26 cm²

b)

C =  62.8 in

A = 314 in²

c)

C =  75.36 m

A = 452.16 m²

Remember that for a circle of radius R, the circumference is:

C = 2pi*R

The area is:

A = pi*R²

Where pi = 3.14

a) Here we have R = 3cm, replacing that in the formulas above we will get.

C = 2*3.14*3cm = 18.84cm

A = 3.14*(3cm)² = 28.26 cm²

Now let's do that for some more.

b) The radius is 10 inches.

C = 2*3.14*10 in = 62.8 in

A = 3.14*(10in)² = 314 in²

c) The radius is 12m

C = 2*3.14*12m = 75.36 m

A = 3.14*(12 m)² = 452.16 m²

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The area of a circle of radius of 0.5 meters is 0.785 square meters.

Remember that for a circle of radius r, the area is:

A = pi*r²

Where pi = 3.14

Here we know that r = 0.5m, then we can input that in the formula for the area that is above, we will get.

A = 3.14*(0.5m)²

A = 3.14*0.25 m²

A = 0.785  m²

That is the area of the circle.

Complete question: Let's say that r is the radius of a circle and A is its area, then: If r=0.5 m, A = ?

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The factorial expressions after solving are as follows:

(A) 6! = 720

(B) 2!5! = 240

(C) 10!/7! = 720

The product of all positive integers less than or equal to n in mathematics is known as the factorial of a non-negative integer, indicated by the symbol n!.

Additionally, the factorial of n is equal to the sum of n and the subsequent smaller factorial:

For instance, According to the convention for an empty product, the value of 0! is 1.

So, solve as follows:

6!:
6*5*4*3*2*1 = 720

2!5!:
2*1 * 5*4*3*2*1 = 240

10!/7!

10*9*8*7*6*5*4*3*2*1/7*6*5*4*3*2*1 = 720

Therefore, the factorial expressions after solving are as follows:

(A) 6! = 720

(B) 2!5! = 240

(C) 10!/7! = 720

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The approximate surface area of the sphere is 616 square centimeters. The answer is (C) 616 square centimeters.

What is an algebraic expression?

An algebraic expression is a mathematical phrase that contains variables, constants, and mathematical operations. It may also include exponents and/or roots.

The formula for the surface area of a sphere is given by:

A = 4πr²

Substituting r = 7 cm into this formula, we get:

A = 4π(7)²

A = 4π(49)

A = 196π

Now, we can use the given expression to approximate the value of π:

4.2%2 = 0.042

π ≈ 3.14

Substituting this approximate value of π into our equation for A, we get:

A ≈ 196(3.14)

A ≈ 615.44

Rounding to the nearest whole number, we get:

A ≈ 616

Therefore, the approximate surface area of the sphere is 616 square centimeters. The answer is (C) 616 square centimeters.

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When six integers are chosen from the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, there must be at least one pair of integers that adds up to 11. This can be proven using a proof by contradiction.

To solve this problem, we can use a proof by contradiction. We assume that there are six integers chosen from A such that no two integers add up to 11.

If there is no pair of integers in the six chosen that sum up to 11, then we can consider the pairs of integers (1,10), (2,9), (3,8), (4,7), and (5,6). These are the only possible pairs of integers in A that add up to 11.

However, since there are five pairs and we can choose at most one integer from each pair, we can choose at most five integers in total. This is a contradiction since we were asked to choose six integers from A. Therefore, our assumption that there are no pairs of integers that add up to 11 is false.

Hence, we conclude that there must be at least one pair of integers among the six chosen that adds up to 11.

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