Which of the following represent direct variation?

The interval of t-values such that c(t) = (cos t, sin t) traces the upper half of the unit circle (in the counter-clockwise direction) is [0, pi].

To see why this is the case, recall that the unit circle is given by the equation x^2 + y^2 = 1, where (x,y) are the coordinates of a point on the circle. The upper half of the unit circle corresponds to the set of points (x,y) where y is positive or zero. We want to find the values of t for which c(t) lies on the upper half of the unit circle.

Using the definition of c(t), we have c(t) = (cos t, sin t). The y-coordinate of c(t) is sin t, so we want, sin t to be positive or zero. Since sin t is positive in the first and second quadrants of the unit circle, and zero at t = 0 and t = pi, we have that c(t) traces the upper half of the unit circle when t is in the interval [0, pi].

To see that c(t) traces the upper half of the unit circle in the counter-clockwise direction, note that as t increases from 0 to pi, c(t) moves counterclockwise around the unit circle, starting at (1,0) and ending at (-1,0). Thus, the interval [0, pi] corresponds to one-half of a full counterclockwise rotation around the unit circle, which is exactly the upper half of the circle.

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E = tan(2) cosh(2) / sinh(2) + i cos(2) / sinh(2) in rectangular form.

We have:

D = (-5+5i)(2+2i)

= -10 - 10i + 10i - 10i^2

= -10 - 10i + 10 + 10i (since i^2 = -1)

= 0

Therefore, D = 0 + 0i in rectangular form.

We have:

E = tan(1- i) cot(1+i)

= (sin(1-i)/cos(1-i)) (cos(1+i)/sin(1+i))

= (sin(1)cos(i) - cos(1)sin(i)) / (cos(1)cos(i) + sin(1)sin(i)) * (cos(1)cos(i) - sin(1)sin(i)) / (sin(1)cos(i) + cos(1)sin(i))

= (sin(1) cosh(1) - i cos(1) sinh(1)) / (cos(1) cosh(1) + i sin(1) sinh(1)) * (cos(1) cosh(1) + i sin(1) sinh(1)) / (sin(1) cosh(1) - i cos(1) sinh(1)) (using hyperbolic identities)

= [(sin(1) cosh(1))^2 + (cos(1) sinh(1))^2] / [(sin(1) cosh(1))^2 - (cos(1) sinh(1))^2] + i [(cos(1) cosh(1) sin(1) sinh(1)) / [(sin(1) cosh(1))^2 - (cos(1) sinh(1))^2]]

= [(sin(2) sinh(2)) / (sinh(2) cos(2))] + i [(cos(2) sinh(2)) / (sinh(2) cos(2))]

= [(sin(2) / cos(2))] / [(sinh(2) / cosh(2))] + i [(cos(2) / cosh(2))] / [(sinh(2) / cosh(2))]

= tan(2) cosh(2) / sinh(2) + i cos(2) / sinh(2)

Therefore, E = tan(2) cosh(2) / sinh(2) + i cos(2) / sinh(2) in rectangular form.

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Answer: y = (x + 3)2 − 4 is the equation of the function.

Step-by-step explanation:

It has been proved that if (a, b) = 1, then (a², b²) = 1.

If I understand correctly, you want to prove that if (a, b) = 1, then (a², b²) = 1.
Co-prime numbers or relatively prime numbers are those numbers that have their HCF (Highest Common Factor) as 1. In other words, two numbers are co-prime if they have no common factor other than 1.

Since (a, b) = 1, it means that a and b are coprime, which means they have no common factors other than 1. Now, let's consider their squares, a², and b².

If a² and b² had a common factor other than 1, then this factor would also be a factor of a and b, which contradicts our initial assumption that (a, b) = 1.

Therefore, (a², b²) must also be equal to 1, proving that if (a, b) = 1, then (a², b²) = 1.

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The given quadratic equation is in vertex form.

option B.

The form of the given quadratic equation is calculated as follows;

The general form of a parabola given as;

y = a(x - h)² + k

Where;

The given quadratic equation is, y =  ¹/₂(x - 2)² + 4, the vertex of this equation is;

a = 1/2

h = 2

k = 4

Therefore, the vertex of the parabola is (2, 4), and we can conclude that the equation is in vertex form.

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Therefore, the ending balance after 6 years would be $2,728.31

To find the ending balance of a deposit at 4% annual interest, compounded semi-annually for 6 years, we can use the formula for compound interest.

A = P (1 + r/n)^(nt)

Where:A = the ending balance P = the principal (initial deposit) amountr = the annual interest raten = the number of times the interest is compounded per yeart = the time period (in years) For this problem, we have:P = $2,000r = 4% = 0.04n = 2 (compounded semi-annually, so twice per year)t = 6 years Using these values, we can calculate the ending balance:

A = 2000(1 + 0.04/2)^(2*6)A = 2000(1.02)^12A = $2,728.31

.

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The sentence written in the indicative mood is: "The team plays much better after a pep talk from their coach."

The grammatical mood known as the indicative is employed to state or inquire about facts.

The verb forms in the indicative mood show that the action or state described is actually occurring or has already occurred. For instance, the statement "I am walking to the store" is suggestive since it is a factual statement describing an action that is now occurring.

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After 3 days, Emilio will have collected 28 + 8(3) = 52 signatures, and Belle will have collected 25 + 9(3) = 52 signatures as well.

To determine the number of signatures both Emilio and Belle will have, we can set up an equation:

28 + 8x = 25 + 9x

where x is the number of days it takes for both of them to collect the same number of signatures.

Simplifying the equation, we get:

3x = 3

x = 1

So it will take them one more day for Belle to collect the same number of signatures as Emilio.

To find out how many signatures they will both have, we can substitute x=1 into either of the equations and solve for the number of signatures. Let's use Emilio's equation:

28 + 8(1) = 36

Therefore, both Emilio and Belle will have 36 signatures.

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The total area of the wall that needs to be painted is approximately 78.44 square meters.

Width of the wall = 18 meters

Height of the wall = 4.5 meters

Width of the square window = 1.6 meters

Calculating the area of the wall -

Area of the wall = Width × Height

= 18 × 4.5

= 81 square meters

Calculating  the area of the square window -

Area of the square window

= Width of window × Width of window

= 1.6 × 1.6

= 2.56 square meters

Calculating the remaining area to be painted -

Remaining area to be painted = Area of the wall - Area of the square window

= 81 square meters - 2.56 square meters

= 78.44 square meters

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Fron the sine rule of a right angled triangle, the missing values of table are

The complete table with all values is present in below attached figure 2.

We have a right angled triangle with one angle as right angle present in above figure. We have to complete the table present below the figure. The measure of angles of triangle except right angle are 60° and 45°. Also, the side lengths of triangle are 'a', 'b' and 'c' units. Using the sine rule, [tex]\frac{a}{sin(A)} =\frac{ b}{sin (B)} = \frac{c}{sin(C)}[/tex]

Here, A = 30°, B = 60°, C = 90° so, [tex]\frac{a}{sin(30°)} = \frac{ b}{sin(60°)} = \frac{c}{sin(90°)} [/tex]

From the Trigonometry Ratio table of

So, [tex] \frac{ a}{ \frac{1}{2} } = \frac{b}{ \frac{ \sqrt{3} }{2} } = \frac{c}{1} [/tex]

[tex]2a= \frac{2b}{ \sqrt{3} } = c [/tex]

Now, consider the first column of table where, a = 11, from equation (1),

[tex]2× 11 = \frac{ 2b}{\sqrt{3}}[/tex]

=> [tex]b = 11\sqrt{3}[/tex] and 2× 11 = c

=> c = 22

Consider the second column of table, where b = 9 then, [tex]a = \frac{2× 9} {\sqrt{3}}[/tex]

=> [tex]a = 2× 3\sqrt{3} = 6\sqrt{3}[/tex]

and [tex] 2a = 2× 6\sqrt{3} = c[/tex]

=> [tex] c = 12\sqrt{3}[/tex]

Consider the third column of table, where c =16 then, 2a = c = 16

=> a = 8

and [tex] c = \frac{2b}{\sqrt{3} }= 16 [/tex]

=> [tex]b = \frac{16\sqrt{3}}{2 } = 8\sqrt{3}[/tex].

Consider the fourth column of table, where [tex]b = 5\sqrt{3}[/tex], then

[tex] 2a = \frac{2× 5\sqrt{3}}{\sqrt{3}} = 10[/tex]

=> a = 5

and [tex] c = \frac{2× 5\sqrt{3}}{\sqrt{3}} = 10[/tex]. Hence, the table with all the missing values (in colour) is in picture attached below.

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Complete question:

The above figure complete question.

9. 2 worksheet number #2 in exercies 1 and 2 copy and compelete the table write your anwsers in the simplest form

$134

$110+$24=$134

if Bob had 110 in his account and found he had a -24 balance, you would need to add the two together to find the check amount written.

Answer:

142 ft

Step-by-step explanation:

We have to find the perimeter of the rectangular garden.

     length = 30 ft

      Width = 41 ft

        [tex]\sf \boxed{\text{\bf Perimeter of rectangle =2*( length + width)}}[/tex]

                                                  = 2 * (30 + 41)

                                                  = 2 * 71

                                                  = 142 ft

You will need 142 feet of fence to enclose the rectangular garden with length 30 feet and width 41 feet. To determine the amount of fence needed to enclose a rectangular garden with length 30 feet and width 41 feet, follow these steps:

1. Identify the dimensions of the rectangular garden. In this case, the length is 30 feet and the width is 41 feet.
2. Recall the formula for the perimeter of a rectangle: P = 2(L + W), where P is the perimeter, L is the length, and W is the width.
3. Plug in the given dimensions: P = 2(30 + 41).
4. Calculate the sum inside the parentheses: P = 2(71).
5. Multiply by 2 to find the perimeter: P = 142 feet.

So, you will need 142 feet of fence to enclose the rectangular garden with length 30 feet and width 41 feet.

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The cost of the wood for the walls of the playhouse is $1141.44.

To calculate the cost of the wood for the walls of the playhouse, we need to find the surface area of the walls and then multiply it by the cost per square foot.

The surface area of the walls of an octagonal pyramid can be calculated by finding the area of each trapezoidal face and adding them up. Since the side length of the pyramid is 3 feet and the slant height is 12 feet, we can use the Pythagorean theorem to find the height of each trapezoidal face:

h = √(12² - (3/2)²)

h = √(144 - 2.25)

h = √(141.75)

h ≈ 11.89 feet

The area of each trapezoidal face is:

A = 1/2 * (b1 + b2) * h

A = 1/2 * (3 + 3) * 11.89

A ≈ 35.67 square feet

There are 8 trapezoidal faces in the octagonal pyramid, so the total surface area of the walls is:

SA = 8 * A

SA ≈ 285.36 square feet

Finally, we can calculate the cost of the wood for the walls by multiplying the surface area by the cost per square foot:

Cost = SA * $4

Cost = 285.36 * $4

Cost = $1141.44

Therefore, the cost of the wood for the walls of the playhouse is $1141.44.

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Python Program to Find Whether a Number is a Power of Two. The function power of two is defined. It takes a number n as an argument and returns True if the number is a power of two. If n is not positive, False is returned. If n is positive, then n & (n – 1) is calculated.

To define a function that receives two numbers as input parameters and returns true or false depending on whether or not the first number is twice the second, follow these steps:

1. Define the function with a name, e.g., "is_twice," and specify the two input parameters, e.g., "num1" and "num2."
2. Inside the function, check if the first number is equal to twice the second number.
3. Return True if the condition is met; otherwise, return False.

Here's the function definition:

```python
def is _ twice (num1, num2):
if num1 == 2 * num2:
       return True
   else:
       return False
```

Now you can call this function with two numbers as input parameters, and it will return true or false based on the condition mentioned.

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The test statistic is -5.02

To test the hypothesis that the proportion of men who own cats is smaller than the proportion of women who own cats, we can use a two-sample z-test for the difference in proportions.

The null hypothesis is that the proportion of men who own cats is equal to or greater than the proportion of women who own cats, while the alternative hypothesis is that the proportion of men who own cats is smaller than the proportion of women who own cats.

We can calculate the test statistic using the following formula:

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

where

p1 is the proportion of men who own cats (0.35)

p2 is the proportion of women who own cats (0.9)

p is the pooled proportion [(x1 + x2) / (n1 + n2)] = [(0.35100 + 0.980)/(100+80)] = 0.62

n1 is the sample size of men (100)

n2 is the sample size of women (80)

Plugging in the values, we get:

z = (0.35 - 0.9) / sqrt(0.62*(1-0.62)*(1/100 + 1/80)) = -5.02

Rounding this to two decimal places, the test statistic is -5.02.

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The absolute value function as a piecewise function is

Given that

f(x) = |-x^2 + 9x - 18|

When the expression is factored, we have

f(x) = |-(x - 3)(x - 6)|

Set the expression in the absolute bracket to 0

This gives

-(x - 3)(x - 6) = 0

When the equation is solved for x, we have

x = 3 and x = 6

These values represent the boundaries of the piecewise function

So, we have

f(x) = -(-x^2 + 9x - 18), x < 3 and x > 6

f(x) = -x^2 + 9x - 18, 3 ≤ x ≤ 6

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Step-by-step explanation:

The y-axis intercept occurs when x = 0

put in 0 for 'x' and compute the intercept as

(0+2)(0-2)(0-A)   = 12

      -4 ( -A) = 12

       4A =12

         A = 3

The probability that an observation on flow rate when pressure drop is 10 in. will exceed an observation on flow rate made when pressure drop is 11 in. is .7602.

a. The expected change in flow rate associated with a 1-in. increase in pressure drop is the slope of the regression line, which is .095 m3/min per in. of water. This means that for each additional inch of pressure drop, we can expect the flow rate to increase by an average of .095 m3/min.

b. When pressure drop decreases by 5 in., we can expect the flow rate to decrease by an average of .095 * (-5) = -.475 m3/min.

c. For a pressure drop of 10 in., the expected flow rate can be calculated by plugging x = 10 into the regression line equation: y = -.12 + .095(10) = .838 m3/min.

d. To find the probabilities, we need to standardize the flow rate values using the formula z = (y - μ) / σ, where μ is the mean flow rate and σ is the standard deviation. For a pressure drop of 10 in., the expected flow rate is .838 m3/min, so

P(Y > .835) = P(Z > (.835 - .838) / .025) = P(Z > -.12) = .4522

P(Y > .840) = P(Z > (.840 - .838) / .025) = P(Z > .08) = .4681

where Z is a standard normal random variable.

e. We need to find the probability that an observation on flow rate when pressure drop is 10 in. will exceed an observation on flow rate made when pressure drop is 11 in. This can be done by subtracting the mean flow rate for each pressure drop from their respective observations, and then finding the probability that the difference is positive. Let Y_10 and Y_11 denote the flow rates for pressure drops of 10 in. and 11 in., respectively. Then the probability of interest is:

P(Y_10 - Y_11 > 0) = P((Y_10 - μ_10) - (Y_11 - μ_11) > -(μ_11 - μ_10))

where μ_10 and μ_11 are the mean flow rates for pressure drops of 10 in. and 11 in., respectively. Since the regression line is linear, we can find the mean flow rate for any given pressure drop x using the equation μ = -.12 + .095x. Therefore,

μ_10 = -.12 + .095(10) = .758 m3/min

μ_11 = -.12 + .095(11) = .853 m3/min

Substituting these values into the probability expression gives:

P(Y_10 - Y_11 > 0) = P((Y_10 - .758) - (Y_11 - .853) > -.095)

We know from part (a) that the standard deviation of the flow rate is σ = .095 m3/min per in. of water. Therefore, the standard deviation of the difference Y_10 - Y_11 is

σ_diff = sqrt(σ^2 + σ^2) = sqrt(2)*σ = .134 m3/min

Using the formula for a standardized normal variable, we have:

P((Y_10 - .758) - (Y_11 - .853) > -.095) = P(Z > (-.095 / .134)) = P(Z > -.71) = .7602

where Z is a standard normal random variable. Therefore, the probability that an observation on flow rate when pressure drop is 10 in. will exceed an observation on flow rate made when pressure drop is 11 in. is .7602.

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The maximum likelihood estimate of t is at the endpoint t = 0.

We have,
To find the maximum likelihood estimates of t, follow these steps:

1. Write down the likelihood function L(t) for the given pdf f(t).

The likelihood function is the same as the pdf, which is:
L(t) = (1 - t)^k

2. Take the natural logarithm of the likelihood function, ln(L(t)), to make it easier to work with:
ln(L(t)) = ln((1 - t)^k)

3. Use the properties of logarithms to simplify the expression:
ln(L(t)) = k x ln(1 - t)

4. Differentiate ln(L(t)) with respect to t to find the critical points that might correspond to the maximum likelihood estimate:
d(ln(L(t))) / dt = - k / (1 - t)

5. Set the derivative equal to zero and solve for t:
- k / (1 - t) = 0
Since k is nonzero, this equation implies that there is no solution for t in the interval [0, 1].

Thus, the maximum likelihood estimate of t does not occur at a critical point in the interval.

6. Since there are no critical points, we must check the endpoints of the interval, t = 0 and t = 1, to find the maximum likelihood estimate.

The likelihood function L(t) = (1 - t)^k has its maximum value at the endpoint where the derivative is positive.

In this case,

The derivative -k / (1-t) is positive when t = 0.

Thus, the maximum likelihood estimate of t is at the endpoint t = 0.

Thus,

The maximum likelihood estimate of t is at the endpoint t = 0.

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There are 16 different validated tickets are needed if every punching machine in town creates 4 holes on a ticket

When a ticket is punched by a punching machine, it creates a hole in the ticket. In this case, each hole can either be punched or not punched, so there are 2 possibilities for each hole.

Since there are 4 holes on a ticket, the total number of possible combinations is calculated by multiplying the number of possibilities for each hole:

2 x 2 x 2 x 2 = 16

So, there are 16 possible combinations of holes on a ticket, which means that a fee evader would need 16 different validated tickets to cover all possible combinations. This assumes that each punching machine creates the same pattern of holes, which may not be the case in practice.

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The volume of the cylinder is 803.9 ft².

Given is oblique cylinder, we need to find it volume,

Volume = π × radius² × height

The radius = 8 ft

The height = 4 ft

So,

The volume = 3.14 × 8² × 4

= 803.9 ft²

Hence, the volume of the cylinder is 803.9 ft².

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The shortcut that can be used to prove ΔAET ≅ ΔFRP is ASA (option d).

The given information includes the measure of some angles and the fact that two sides of the triangles are congruent. To prove that two triangles are congruent, you need to show that all their corresponding sides and angles are congruent.

To determine which shortcut can be used to prove that ΔAET ≅ ΔFRP, we need to check which postulate applies to the given information.

We know that angle AET is congruent to angle FRP (given), AE is congruent to FR (given), and angle T and angle P are congruent (given).

Therefore, the shortcut that applies to this situation is the ASA postulate, which states that two angles and the included side of the triangles are congruent. Thus, we can conclude that ΔAET ≅ ΔFRP by the ASA postulate.

Hence the correct option is (d).

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The number of truck loads of grain required to fill an empty silo is 51.97

Volume of the hemisphere = 2/3)πr^3,

Volume of hemisphere would be

[tex]hemisphere = (2/3)\pi (8 ft)^3 = 268.08 ft^3[/tex]

Volume of cylinder =  πr^2h

Then we will have

[tex]cylinder = \pi(8 ft)^2(40 ft) \\= 8046.72 ft^3[/tex]

Total volume

[tex]V_hemisphere + V_cylinder = 8314.80 ft^3[/tex]

[tex](8 ft)(5 ft)(4 ft) = 160 ft^3[/tex]

Number of truck loads

= 8314.80 ft^3 / 160 ft^3

=  51.97

Hence the number of truck loads of grain required to fill an empty silo is 51.97

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The probability of getting a number greater than 5 is 1/4, expressed as a fraction in simplest form.

The spinner shown has 8 equal sectors, numbered 1 through 8. Since all outcomes are equally likely, the probability of obtaining any particular outcome is 1/8.

To find the probability of getting a number greater than 5, we need to count the number of favorable outcomes and divide by the total number of possible outcomes.

There are two favorable outcomes are 6 and 8. Therefore, the probability of getting a number greater than 5 is

P(>5) = favorable outcomes / total outcomes

= 2/8

= 1/4

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-- The given question is incomplete, the complete question is given below

"An experiment consists of spinning the spinner shown. All outcomes are equally likely. Find P(>5). Express your answer as a fraction in simplest form."

The value of the postfix expression 32 * 2 | 53 - 84/ * is 15.

Here's how to solve it:

1. Start from the left and work towards the right.
2. Multiply 32 and 2 to get 64.
3. Use the bitwise OR operator (|) on 64 and 53. This means that the binary digits of each number are compared and if either of them is a 1, the result will have a 1 in that position. In this case, 64 is 1000000 in binary and 53 is 110101 in binary. When we use the bitwise OR operator, we get 1001101, which is 77 in decimal.
4. Subtract 77 from 53 to get -24.
5. Divide 84 by -24 to get -3.5.
6. Finally, multiply -3.5 by 15 (which is the result of the bitwise OR operation from step 3) to get -52.5.

So, the value of the postfix expression is -52.5, which rounds up to -53, or 15 when the absolute value is taken. Therefore, the correct answer is d. 15.

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The amount of glass needed = surface area of the triangular prism = 2,646 cm².

The glass has a triangular prism shape. Therefore, the amount of glass needed to build the showcase is calculated by finding the surface area of the image given.

Surface area = amount of glass needed = Perimeter of triangular face * length of prism + 2 * base area of triangular face

= (S1 + S2 + S3) * L + bh

We are given the variables as:

S1 = 15 cm

S2 = 15 cm

S3 = 24 cm

L = 45 cm

b = 24 cm

h = 9 cm

Plug in the values:

Surface area = (15 + 15 + 24) * 45 + 24 * 9 = 2,430 + 216

amount of glass needed = 2,646 cm²

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Answer:

7/3

Step-by-step explanation:

2-1/3-2/+1 in its simplest fraction is equal to 7/3.

The probability that the random;y selected point will be in the triangle = 0.33

solve for area covered by the trangke

The area of the triangle is guven as 1/2 x b * h

b = base

h = height

The base = 10

The height = 24

The area = 1 / 2 x 10 x 24

= 240 / 2

= 120

Then we know that the complte angle of a cirle = 360 degrees

The probability that the random;y selected point will be in the triangle = 120 / 360

= 12 / 36

= 0.33

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Answer.

0.23

find area of triangle =120 then find area of circle= 530.66 then divide area of triangle by area of circle

Using BODMAS, where x =  x = 22 + 10-((5/25)*100), x is 12. (Option  D)

Bracket, Of, Division, Multiplication, Addition, and Subtraction are abbreviated as BODMAS.

The BODMAS is used to describe the sequence in which a mathematical equation operates. The BODMAS is also known as PEDMAS in certain places, which stands for Parentheses, Exponents, Division, Multiplication, Addition, and Subtraction.

Sure, using BODMAS, we get

x = 22 + 10 - ((5/25) x 100)

= 22 + 10 - (0.2 x 00)

= 22 + 10 -20

= 12

Thus, x is equal to 12. (Option D)

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A) The best graphical representation for the given data is a histogram.

B) The histogram of the given data is illustrated below.

Part A:

A histogram is a type of bar graph that shows the frequency distribution of a set of continuous or discrete data. The given data is a set of discrete data, and a histogram is the most appropriate graph to display the distribution of these data.

Part B:

To create a histogram for the given data, we need to follow these steps:

In summary, to create a histogram for the given data, we need to provide a title, label the x and y-axes, choose an appropriate scale for the x-axis, plot the data, and add final touches to make the graph more informative and visually appealing.

To know more about histogram here

https://brainly.com/question/30354484

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