Find the value of x in the diagram. ​

We can observe from the graph that plan B is less expensive for the interval [0, 4], however plan A is better if you watch 4 or more movies each month.

We know that Plan A costs $14 per month + $2 each movie, thus the total cost if you see x movies is:

C₁(x) = $14 + $2*x

Plan B costs $10 a month + $3 per movie, therefore the total cost if you see x movies is:

C₂(x) = $10 + $3*x

The graph at the end may now be used to compare expenses. The blue line is C2(x), while the green line is C1(x).

The blue line in the graph may be seen between x = 0 and x = 4.

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Answer:

$300

Step-by-step explanation:

We know

4 headphones = $48

1 headphones = $12

25 headphones = $12 x 25 = $300

The expected value for a random variable is defined as the sum of the product of the values and probabilities of the variable.

In this case, the expected value for C is calculated by multiplying each possible value by its probability and adding the products. This is represented mathematically as:

Exp(C) = 0 x 0.60 + 1500 x 0.05 + 5000 x 0.13 + 15000 x 0.22 = 4025

The expected value of C can be interpreted as the average cost that the insurance company can expect to pay out for a randomly selected car of this model. This means that if the company insures a large number of these cars, they can expect the cost per car to average approximately $4025. The expected value does not represent the maximum cost for insuring this car model, or the cost for insuring a set number of cars. Instead, it provides an estimate of what the average cost per car will be.

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None of the transformations followed between the two given triangles. None of the options are correct.

Similar triangles are those triangles that have similar properties,i.e. angles and proportionality of sides.

Here,
The graph shows two triangles plotted on a coordinate plane. One triangle is at A (-4, 4), B (-8, 2), and C (-6, -6). Another triangle is at A '(4, -2), B '(2, 2), and C '(6, 0).

As we can see in the graph, the triangles are completely different, so the transformation of either of the ΔABC to ΔA'B'C' or ΔA'B'C' to ΔABC is not possible.

Thus, None of the transformations followed between the two given triangles. None of the options are correct.

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C. The warmer the temperature, the more often crickets chirp.

The graph shows a direct correlation between temperature and the rate of chirping. As the temperature increases, the rate of chirping increases as well. Using the formula y=mx+b, where m is the slope of the line and b is the y-intercept, the slope of the line in the graph is approximately 0.5, which means that for every 1 degree Celcius increase in temperature, the rate of chirping increases by 0.5 chirps per second. This proves that the warmer the temperature, the more often the crickets chirp.

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Answer:

18 cowry shells

Step-by-step explanation:

1 dozen which is 1(12) eggs = 3 shells

6 dozens will be 6(12) eggs = ?

Cross Multiply

(72 eggs × 3 shells) ÷ 12

= 216 ÷ 12

= 18

The domain and the range of the relation shown to the right are given as follows:

The relation is not a function, as there are inputs that are mapped to multiple outputs.

The domain of a function is the set that contains all the input values of the function, hence on the graph it is the values of x, thus the domain is of:

x >= -5.

The range of a function is the set that contains all the output values of the function, hence on the graph it is the values of y, thus the domain is of:

All real values.

Tracing a vertical line through the graph of the function, there are values of x for which the graph would be crossed two times, meaning that the relation is not a function.

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The mean is 45 and the variance is 18

The probability values are 0.951 and 0.937

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

n = 75

p = 0.6

The notation μ is the mean and it is calculated as

μ = n*p

So, we have

= 0.6*0.75

Evaluate

= 45.

The notation σ² is the variance and it is calculated as

σ² = n*p*(1-p)

So, we have

= 75*0.6*(1-0.6)

Evaluate

= 18.

Without continuity correction

The probability P(x>=52) without continuity correction is represented as

normalcdf(52, μ, σ)

So, we have

normalcdf(52, 45, √18)

Using a calculator, we get:

P(x>=52) = 0.951

With continuity correction.

The probability P(x>=52) with continuity correction is represented as

normalcdf(51.5, μ, σ)

So, we have

normalcdf(51.5, 45, √18)

Using a calculator, we get:

P(x>=52) = 0.937

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Complete question

Let X be a binomial random variable with n = 75 and p = 0.6

a) What is μ?

b) What is σ2?

c) What is the probability of P(x≥52), without continuity correction and with continuity correction.

Answer:

X ≥ -4

Step-by-step explanation:

Remember domain only focuses on the X values, so we can immediately rule out the answers containing Y. We can observe that the graph has a closed circle over x=-4. This tells us it will be included and so we must have "≥" in the answer choice. By simple process of elimination A, the first answer choice is the correct answer.

Also note that it would be read as "X is greater than or equal to negative 4," which is exactly what our graph depicts. The second choice, which looks like it could be correct at first would be read as "X is greater than negative 4," which excludes negative 4 in the answer. We need that negative 4 for it to be correct.

Answer:

Answer is Negative

Step-by-step explanation:

picture at the bottom provides proof.

When we select a random sample of 3 teachers and calculated the sample median from ages of 5 teachers in maths department, we get three samples of sample median 34, four samples of sample median 37 and three samples of sample median 42.

a. To list all 10 possible samples of size 3, we can use the combination formula C(5,3) = 5! / (3! * 2!) = 10.

The possible samples are:

Sample 1: (23, 34, 37) - Sample Median: 34

Sample 2: (23, 34, 42) - Sample Median: 34

Sample 3: (23, 34, 58) - Sample Median: 34

Sample 4: (23, 37, 42) - Sample Median: 37

Sample 5: (23, 37, 58) - Sample Median: 37

Sample 6: (23, 42, 58) - Sample Median: 42

Sample 7: (34, 37, 42) - Sample Median: 37

Sample 8: (34, 37, 58) - Sample Median: 37

Sample 9: (34, 42, 58) - Sample Median: 42

Sample 10: (37, 42, 58) - Sample Median: 42

b. To display the sampling distribution of the sample median, we can create a frequency table that shows how many times each sample median appears in the list of possible samples.

Sample Median Frequency

34                         3

37                         4

42                         3

This table shows that the sample median of 34 appears in 3 out of the 10 possible samples, the sample median of 37 appears in 4 out of the 10 possible samples, and the sample median of 42 appears in 3 out of the 10 possible samples.

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the probabilities are p(0)=0.027 p(5)=0.4 p(1)= 0.243 and p(2)= 0.9

and P(x) = nCx p^x (1 - p)^(n - x) = n! / (x!(n - x)!) * p^x * q^(n - x)

For n = 5, x = 1, p = 0.2:

P(1) = 5! / (1! (5 - 1)!) * 0.2^1 * (1 - 0.2)^(5 - 1) = 5 * 0.2 * 0.4096 = 0.4096

For n = 4, x = 2, q = 0.4:

P(2) = 4! / (2! (4 - 2)!) * (1 - 0.4)^2 * 0.4^(4 - 2) = 6 * 0.36 * 0.16 = 0.3456

For n = 3, x = 0, p = 0.7:

P(0) = 3! / (0! (3 - 0)!) * 0.7^0 * (1 - 0.7)^(3 - 0) = 1 * 1 * 0.027 = 0.027

For n = 5, x = 3, p = 0.1

P(3) = 5! / (3! (5 - 3)!) * 0.1^3 * (1 - 0.1)^(5 - 3) = 10 * 0.001 * 0.81 = 0.0081

For n = 4, x = 2, q = 0.6

P(2) = 4! / (2! (4 - 2)!) * (1 - 0.6)^2 * 0.6^(4 - 2) = 6 * 0.16 * 0.36 = 0.3456

For n = 3, x = 1, q = 0.9

P(1) = 3! / (1! (3 - 1)!) * (1 - 0.9)^1 * 0.9^(3 - 1) = 3 * 0.1 * 0.81 = 0.243

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The probability of an event in a binomial distribution is calculated using the formula Pr(X=x)=C(n,x)p^x(1-p)^(n-x)

(a) Pr(X=0)=C(3,0)p^0(1-p)^3=C(3,0)(0.7)^0(0.3)^3=0.027

(b) Pr(X=5)=C(5,5)p^5(1-p)^0=C(5,5)(0.4)^5(0.6)^0=0.01024

(c) Pr(X=1)=C(4,1)p^1(1-p)^3=C(4,1)(0.6)^1(0.4)^3=0.2304

(d) Pr(X=2)=C(5,2)p^2(1-p)^3=C(5,2)(0.4)^2(0.6)^3=0.3456

The probability of an event in a binomial distribution is calculated using the formula Pr(X=x)=C(n,x)p^x(1-p)^(n-x), where n is the number of trials, p is the probability of success, x is the number of successes, and C(n,x) is the number of combinations of n things taken x at a time. To calculate each of the probabilities given above, we first calculate the combinations C(n,x), followed by the probabilities for success and failure, and then multiply these values together to get the probability of the event. For example, in part (a), we calculate C(3,0)=1, p^0=1, and (1-p)^3=0.027, and the probability of the event Pr(X=0)=1*1*0.027=0.027.

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The Inequalities that represent the above situation are => x + y ≤ 50 and 2x + 3y ≥ 100 and the possible combinations are (20, 30) and (30, 20).
[ Graphs are given below ].

To form the required inequalities, represent the unknown quantities with variables like x, y,..etc.

In the given problem, we don't know the number of items that Jenny will sell so here we represent them with variables 'x' and 'y'.

And obtain expression according to the condition of the given problem. Since we need to represent them in inequalities use signs like less than, greater than, etc.  

Here we have

Jenny would like to make at least $100 in sales.

She estimates that she will sell at most 50 pieces of jewelry.

The bracelets that she is selling cost $2 and the necklaces cost $3  

Let Jenny sold 'x' bracelets and 'y' necklaces

Number of pieces that Jenny estimated = 50

=> x + y ≤ 50 ----- (1)

Jenny would like to make at least $100 in sales.

=> 2x + 3y ≥ 100 ---- (2)      

Let Jenny sell 20 bracelets and 30 necklaces  

=> 2(20) + 3(30) ≥ 100      

=> 40 + 90 ≥ 100  

=>  130 ≥ 100  [ which is true ]

Let Jenny sell 30 bracelets and 20 necklace

=> 2(30) + 3(20) ≥ 100  

=> 60 + 60 ≥ 100    

=> 120 ≥ 100    

∴ The possible combinations are (20, 30) and (30, 20)

Therefore,

The Inequalities that represent the above situation are => x + y ≤ 50 and 2x + 3y ≥ 100 and the possible combinations are (20, 30) and (30, 20).
[ Graphs are given below ].

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The area of R can be calculated using the definite integral of the function y = e^(2x-x^2) between x = 0 and x = 1.

The integral expression is ∫e^(2x-x^2)dx = ∫e^u du = e^u + C = e^(2x-x^2) + C. Evaluating the integral between x = 0 and x = 1 gives the area of R as e^2 - 1.

(b) The area of S can be calculated using the definite integral of the function y = e^(2x-x^2) between x = 0 and x = 1, minus the integral of the function y = 1 between x = 0 and x = 1. The integral expression for S is ∫e^(2x-x^2)dx - ∫1dx = ∫e^u du - x = e^u + C - x = e^(2x-x^2) + C - x. Evaluating the integral between x = 0 and x = 1 gives the area of S as e^2 - 1 - 1.

(c) The volume of the solid generated when R is rotated about the horizontal line y = 1 can be found using the integral expression ∫2πx(e^(2x-x^2) - 1)dx. This expression represents the area of the circular cross-sections of the solid multiplied by 2πx, which is the circumference of the circle. When evaluated between x = 0 and x = 1, this integral gives the volume of the solid generated when R is rotated about the horizontal line y = 1.

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On solving the equations y1 = 41 + 6x and y2 = 32 + 9x, it is obtained that -

For 3 books the cost will be the same.

The cost that will be same is $59.

What is an equation?

A mathematical definition of an equation is a claim that two expressions are equal when they are joined by the equals sign ("=").

The cost at Washington Printing is = $41 and $6 per book

The cost at Fairfax University is = $32 and $9 per book

Let the number of books be x.

Let y1 represent the total cost at Washington Printing.

Let y2 represent the total cost at Fairfax University.

So, the equations are -

y1 = 41 + 6x

y2 = 32 + 9x

According to the question -

y1 = y2

41 + 6x = 32 + 9x

Collect all the like terms together -

6x - 9x = 32 - 41

- 3x = - 9

x = 9/3

x = 3

Therefore, the number of books is 3.

The total cost can then be determined by the substituting the value of x in the equations -

41 + 6x = 41 + 6(3) = 41 + 18 = 59

32 + 9x = 32 + 9(3) = 32 + 27 = 59

Therefore, the total cost is $59.

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Of the 18 number of seniors not enrolled in an AP class or a club, 8 are enrolled in an AP class but not a club, and 10 are not enrolled in an AP class or a club.

First, we need to calculate the total number of seniors enrolled in AP classes and clubs. We know that 32 number of  seniors are enrolled in at least one AP class and 35 are in a club. That means that there are 67 seniors total enrolled in either an AP class or a club.

Next, we need to determine the number of seniors enrolled in both an AP class and a club. We can subtract the total number of seniors enrolled in either an AP class or a club from the total number of seniors at the school (80). This will give us the number of seniors not enrolled in either an AP class or a club, which is 18.

Finally, we can calculate the number of seniors enrolled in an AP class but not a club, and the number of seniors not enrolled in an AP class or a club. We know that there are 18 seniors not enrolled in either an AP class or a club, so we can subtract the number of seniors enrolled in an AP class (32) from 18 to get the number of seniors enrolled in an AP class but not a club, which is 8. We can also subtract the number of seniors enrolled in both an AP class and a club (67) from 18 to get the number of seniors not enrolled

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An equation that is perpendicular to line y = 1/2x - 8 and passes through (7, -6) in point-slope form is y + 6 = -2(x - 7).

Mathematically, the point-slope form of a straight line can be calculated by using this mathematical expression:

y - y₁ = m(x - x₁)

Where:

From the information provided, an equation of the first line is given by;

y = 1/2x - 8

Therefore, slope (m) = 1/2

In Mathematics, a condition that must be met for two lines to be perpendicular is given by:

m₁ × m₂ = -1

1/2 × m₂ = -1

m₂ = -2

Note: m represents the slope.

At point (7, -6), an equation of the line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - (-6) = -2(x - 7)

y + 6 = -2(x - 7)

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Complete Question:

Write an equation perpendicular to line y = 1/2x - 8 and passes through (7, -6) in point-slope form.

Answer:

Step-by-step explanation:

It is equivalent

4/12

1/3

1x4=4

         _

3x4=12

The small radius =5 cm; and the Long radius =5+4=9 cm.

Area (a + b) = 106 * pi cm^2 (given)

radius of a = x : radius of b = (x + 4)

Area a =[tex]pi * r^2 = pi * x^2\\[/tex]

Area b = [tex]pi * r^2 = pi * (x + 4)^2[/tex]

Area (a + b) = [tex]pi * x^2 + pi * (x + 4)^2[/tex] = 106 * pi cm^2

Area (a + b) =[tex]pi * x^2 + pi * (x^2 + 8x + 16)[/tex] = 106 * pi cm^2

Area (a + b) = pi * ([tex]2x^2[/tex]+ 8x + 16) = 106 * pi cm^2

Area (a + b) = [tex]2x^2 + 8x - 90 = 0[/tex]

2x^2 + 8x - 90 = x^2 + 4x - 45 = 0

x^2 + 4x - 45 = (x + 9)(x -5) = 0

x = 5 or x = -9

Therefore: radius a = 5 cm and radius b = 9 cm

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The question is incomplete

The sum of the areas of the two circles is 106pi cm^2 and the radius of the larger circle is 4cm longer than the radius Of the smaller circle. How could I find the length of the radii?

The percentage of the whole population voted. A pie chart is best used to show the percentage of the whole that different categories represent, in this case the percentage of the whole population that voted.

A pie chart is one of the most commonly used types of data visualization because it provides an easy way to understand the proportional distribution of the data that is being presented. In a presentation about voter turnout, a pie chart would be an excellent way to illustrate what percentage of the whole population voted. Pie charts can be used to show how a whole is divided into different categories, and how much of the whole each category represents. This type of chart is especially useful for presenting data in a visually appealing way and for making comparisons between different categories. In the case of voter turnout, a pie chart would be ideal for showing the percentage of the population that voted, and for comparing the proportions of those who did and did not vote.

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The value of 17% of 279,394 is 47496.98.

percentage, a relative value indicating hundredth parts of any quantity.

We have to find the value of 17% of 279,394

Convert 17% to decimal value by dividing with 100

17/100=0.17

Now multiply 0.17 with 279394

0.17×279394

47496.98

Hence, the value of 17% of 279,394 is 47496.98.

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Answer:

To solve for the time when the football hits the ground, we need to set h = 0 and solve for t. We can do this using the quadratic equation: 0 = -3.9t² + 15.6t. Solving for t results in t = 0 or t = 4. Therefore, the football hits the ground at t = 0 seconds or 4 seconds after it is kicked. The units of measurement for time are seconds.

The function f(x) will be : f(x) = 4x and f(2) = 8.

In mathematics, an expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Mathematical symbols can designate numbers (constants), variables, operations, functions, brackets, punctuation, and grouping to help determine the order of operations and other aspects of logical syntax.

Given that f(x) is a direct variation function.

Direct variation of {y} with {x} means that as {x} increases, {y} increases uniformly with it. Mathematically -

{K} = y/x = constant

Scale factor {K} is a dimensionless value that indicates the constant ratio value indicating direct variation.

It is given that -

f(3) = 12

We can write the slope of f(x) as -

m = (12 - 0)/(3 - 0)

m = 12/3

m = 4

So, f(x) will be -

f(x) = 4x

f(2) will be -

f(2) = 8

Therefore, the function f(x) will be : f(x) = 4x and f(2) = 8.

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A combination of loops and random numbers, a unique background and two unique shapes can be created.

The Applab Turtle project uses a combination of loops and random numbers to create a background and two unique shapes. The program starts by setting the background color of the canvas to a random number from 0 to 255. A for loop is then used to create a series of random circles. The loop is set to run 10 times, and each iteration uses the random number function to set the size of the circle and its position, resulting in a unique pattern of circles on the canvas. Following this, two for loops are used to create two unique shapes. The for loops are set to run a set number of times, and each iteration uses the random number function to set the size and position of the shape. By using a combination of loops and random numbers, a unique background and two unique shapes can be created.

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The equation of the plane is 2(x-2) + 4(y+5) - c(z-c/2) = 0

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

points (3,-7,c) and (1,-3,0)

The equation is then calculated as follows:

Start by calculating the midpoint M

M = ( (3+1)/2 , (-7-3)/2 , (c+0)/2 ) = (2,-5,c/2)

Then we need to find the vector parallel to the line, which is given by

v = (1-3, -3-(-7), 0-c) = (-2,4,-c)

Substitute the coordinates of the midpoint and the normal vector into the equation of a plane "ax + by + cz = d"

Where (a,b,c) is the normal vector and d is a constant.

-2(x-2) + 4(y+5) - c(z-c/2) = 0

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The area of the shaded region can be found by Area Decomposition Method.

Outer Area Minus Inner Area: Area of the shaded region = Area of the outer shape - Area of the unshaded inner shape

Area of Composite Figures: Area of the shaded region = Area of Region 1 + Area of Region 2 + ...... + Area of nth Region

The Area Decomposition Method:

Understanding the area decomposition method is the key to determining the area of a darkened zone.

For quick and efficient analysis, the area decomposition approach divides the fundamental forms present in a given complex figure.

To obtain the areas of the shaded zones, there are two scenarios.

They are of:

(1) Figures with holes and

(2) Composite figures.

Outer Area Minus Inner Area:

In figures with holes, take into account the area of the figure as a whole without the hole or hollow. By deducting the punch hole area from the whole figure, the remaining area can be found.

The area of the shaded region is equal to the Area of the outer shape minus the Area of the unshaded inner shape

Area of Composite Figures:

A composite plane figure's overall area is the same as the sum of its component areas.

The area of the shaded region is equal to the sum of the Area of Region 1, Area of Region 2, ......, Area of nth Region.

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A compound inequality to find the range of values for the width include the following:

x ≥ 8.

x ≤ 44.

Mathematically, the perimeter of a rectangle can be calculated by using this equation;

P = 2(L + x)

Where:

Note: The length of the dog run is 22 feet.

Since the perimeter of the dog run must be at least 52 feet, an inequality that models the situation is given by:

2(22) + x ≥ 52.

44 + x ≥ 52.

x ≥ 52 - 44

x ≥ 8.

Since the perimeter of the dog run must be no more than 88 feet, an inequality that models the situation is given by:

2(22) + x ≤ 88.

44 + x ≤ 88

x ≤ 88 - 44

x ≤ 44.

Therefore, a compound inequality to find the range of values for the width is 8 ≤ x ≤ 44.

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Answer:

$ 3916

Step-by-step explanation:

Amount= $5500

%decrease= 28.8% or .288

Amount decreased= 5500 x .288; $1584.

End of last year= 5500 - 1584; 3916