Please help !! I can’t figure this question out

For answering 5 questions wrong Hunter's score is 75 and for for answering x question wrong his score will be  f(x)=100-5*x.

A function is defined as the relationship between input and output, where each input has exactly one output. The inputs are the elements in the domain and the outputs are elements in the co-domain.

Hunter is taking a multiple choice test with a total of 100 points available.

Each question is worth exactly 5 points,

From that the number of questions in the test is ,[tex]\frac{100}{5}=20[/tex]questions

Hunter's test score (out of 100) if he got 5 questions wrong,

20-5=15 questions correct.

Each question carries 5 marks, then 15*5= 75 marks.

Hunter score will be 75 marks.

And again each question carries 5 marks, If hunter answers x questions wrong his score will be f(x)=100-5*x.

It is a linear equation.

Hence, for answering 5 questions wrong Hunter's score is 75 and for for answering x question wrong his score will be f(x)=100-5*x.

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Answer: The ratio $42/81$ can be simplified by dividing both the numerator and denominator by the greatest common factor of 42 and 81, which is 9.

$42/81 = (42 \div 9) / (81 \div 9) = 4/9$

So, the ratio $42/81$ can be expressed as the simplified fraction $\frac{4}{9}$.

Step-by-step explanation:

The conditions on a, b, and c for which the Linear equations has solutions are det(A) ≠ 0,  Rank[A|B] = Rank(A).

(a) 3x + 2y - z = a

x + y + 2z = b

5x + 4y + 32 = c

For this system to have solutions, the augmented matrix [A|B]  must have a unique solution. This is equivalent to determinant of the coefficient matrix A is non-zero, and the system is consistent (the row rank of the augmented matrix is equal to the row rank of the coefficient matrix). Therefore, the conditions on a, b, and c for this system to have solutions are:

det(A) ≠ 0

Rank[A|B] = Rank(A)

(b) -3x + 2y + 4z = a

-x - 2y + 3z = b

-x - 6y + 232 = 0

For this system to have solutions, the conditions are the same as in (a):

det(A) ≠ 0

Rank[A|B] = Rank(A)

(c) 4x - 2y + 3z = a

2x - 3y – 2z = b

4x - 2y + 32 = c

For this system to have solutions, the conditions are the same as in (a) and (b):

det(A) ≠ 0

Rank[A|B] = Rank(A)

For each system of linear equations to have solutions, the number of equations must be equal to the number of variables, and the determinant of the coefficient matrix must be non-zero.

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Since there are 20 professors and each professor is on exactly 3 committees, there must be a total of 60 committee positions.

Let's assume that it is not possible to select a distinct representative from each committee. This means that there must be some committee that does not have a distinct representative.

Let's consider the professors on this committee. Since each professor is on exactly 3 committees, each of the professors on this committee must also be on 2 other committees.

Let's consider the 2 other committees for each of the professors on this committee. Since there are 6 professors on the committee, there are a total of 12 other committees to consider.

Each of these 12 committees must have a representative chosen from the remaining 14 professors, since we have assumed that it is not possible to select a distinct representative from the original 10 committees.

However, since there are only 14 professors remaining, and each professor can only be on 3 committees, it is not possible to choose a representative from all 12 of these committees without choosing at least one professor to be on 4 committees. This is a contradiction, since we have assumed that each professor is on exactly 3 committees.

Therefore, our assumption that it is not possible to select a distinct representative from each committee must be false, and it is indeed possible to do so.

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The probability of all three of the selected homes having a security system is 0.3277, the probability of none of the three selected homes having a security system is 0.0414, and the probability of at least one of the selected homes having a security system is 0.9709. The events are dependent.

The probability of all three of the selected homes having a security system is 0.3277, which can be calculated using the formula P(A and B and C) = P(A) x P(B) x P(C). In this case, P(A) is the probability of the first home having a security system, which is 8/11 (since 8 out of 11 homes have a security system). P(B) is the probability of the second home having a security system, which is 7/10 (since 7 out of 10 remaining homes have a security system). P(C) is the probability of the third home having a security system, which is 6/9 (since 6 out of 9 remaining homes have a security system). Therefore, P(A and B and C) = (8/11) x (7/10) x (6/9) = 0.3277.

The probability of none of the three selected homes having a security system is 0.0414, which can be calculated using the formula P(A and B and C) = P(A') x P(B') x P(C'). In this case, P(A') is the probability of the first home not having a security system, which is 3/11 (since 3 out of 11 homes do not have a security system). P(B') is the probability of the second home not having a security system, which is 3/10 (since 3 out of 10 remaining homes do not have a security system). P(C') is the probability of the third home not having a security system, which is 3/9 (since 3 out of 9 remaining homes do not have a security system). Therefore, P(A and B and C) = [tex](3/11) x (3/10) x (3/9) = 0.0414[/tex].

The probability of at least one of the selected homes having a security system is 0.9709, which can be calculated using the formula P(A or B or C) = 1 - P(A' and B' and C'). In this case, P(A' and B' and C') is the probability of none of the three selected homes having a security system, which is 0.0414 (as calculated above). Therefore, P(A or B or C) = 1 - 0.0414 = 0.9709.

The events are dependent because each selection is affected by the previous selections. For example, the probability of the second home having a security system is affected by whether or not the first home had a security system. If the first home had a security system, then there are 7 out of 10 remaining homes with a security system; if the first home did not have a security system, then there are 8 out of 10 remaining homes with a security system.

The probability of all three of the selected homes having a security system is 0.3277, the probability of none of the three selected homes having a security system is 0.0414, and the probability of at least one of the selected homes having a security system is 0.9709. The events are dependent.

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The system of equations has one solution, which is (2, 3).

To find the solution to a graphed system of equations, you need to find the points where the graphs of the two equations intersect. These points represent the coordinates of the solution, which are the values of x and y that make both equations true simultaneously.

The system of equations as shown in the graph has one solution, which is: (2, 3). This is because the liens intersect at the point (2, 3).

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a) the population variance and sd use n in the denominator instead of n-1.  - false

b) the sample variance and sd inherit the strengths and weaknesses of the sample mean. - True

c) if we did not square the distances in the numerator, we would end up with 0. -false

d) the sample variance and sd are robust to outliers and/or extreme points.-false

a) False. The population variance and standard deviation use N in the denominator, where N is the size of the population. The sample variance and standard deviation use n-1 in the denominator, where n is the size of the sample.

b) True. The sample variance and standard deviation inherit the strengths and weaknesses of the sample mean, since they are based on the deviations of the sample values from the sample mean.

c) False. If we did not square the distances in the numerator, we would end up with the mean deviation, which is a valid measure of dispersion but not the variance.

d) False. The sample variance and standard deviation are not robust to outliers and/or extreme points, as they are sensitive to the values of the data points and can be significantly influenced by the presence of outliers.

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Answer:

35 + 83) − (72 − 63) = 109

Step-by-step explanation:

To subtract the polynomials (35 + 83) − (72 − 63), we'll simplify each term in the polynomials first and then subtract the corresponding terms:

(35 + 83) = 118

(72 - 63) = 9

Now we can subtract the corresponding terms:

118 - 9 = 109

So the result of (35 + 83) − (72 − 63) is 109.

Aɳʂɯҽɾҽԃ Ⴆყ ɠσԃKEY ꦿ

Answer:

Okay first do what is in the parenthesis:

35 + 83 = 118

72 - 63 = 9

Now subtract:

118 - 9 = 109

The answer is: 109

Step-by-step explanation:

Hope it helps!

The interest rate that can explain how $4081 grew to $8404.54 in 21 years with annual compounding is option (a) 3.5%, rounded to the nearest tenth.

The interest rate and the compounding frequency determine the growth of the initial amount over time. In this problem, we need to find the interest rate that can explain how an initial amount grew to a specific amount over a certain number of years.

The problem asks us to find the interest rate that can explain how an initial amount of $4081 grew to $8404.54 in 21 years with annual compounding. We can use the formula for compound interest to solve this problem:

[tex]A = P(1 + r/n)^{nt}[/tex]

where A is the future value of the investment, P is the initial amount, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.

In this case, we know that P = $4081, A = $8404.54, n = 1 (annual compounding), and t = 21. We need to find the value of r that makes the equation true.

We can rearrange the formula to solve for r:

[tex]r = n[(A/P)^{1/nt} - 1][/tex]

Substituting the known values, we get:

[tex]r = 1[(8404.54/4081)^{1/(1*21)} - 1][/tex]

r = 0.0353 or 3.53%

This means that if the initial amount was invested with an annual interest rate of 3.5%, it would have grown to the given future value after 21 years.

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Answer:

= 4y + x - 16 = 0.

Step-by-step explanation:

Point are (-8,6)

where

x1 = -8 and y1 = 6

the equation o the line is given by the formula

[tex] = \frac{y - y1}{x - x1} = m[/tex]

where (m) is the gradient

the gradient (m) = -1/4

therefore the equation of the line

[tex] \frac{y - y1}{x - x1} = m \\ = \frac{y - 6}{x - ( - 8)} = \frac{ - 1}{4} \\ = \frac{y - 6}{x + 8} = \frac{ - 1}{4} \\ = 4(y - 6) = - 1(x + 8) \\ = 4y - 24 = - x - 8 \\ = 4y = - x - 8 + 24 \\ = 4y = - x + 16 \\ = 4y + x - 16 = 0.[/tex]

therefore the equation of the line is = 4y + x - 16 = 0.

note// y and x where not given any value.

Option A. No, this is not possible. Since A and B are not mutually exclusive events, we have P(A ∩ B) = P(A) + P(B) - P(A ∪ B), where P(A ∪ B) is the probability that the selected student has either a Visa or a MasterCard.

Since 0.5 > 1 - P(A ∪ B) = P(A') ∩ P(B'), we must have P(A') ∩ P(B') < 0, which is impossible.

B. The probability that the selected student has at least one of these two types of cards is given by P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.7 + 0.4 - 0.3 = 0.8.

C. The probability that the selected student has neither type of card is given by P(A' ∩ B') = 1 - P(A ∪ B) = 1 - 0.8 = 0.2.

D. The probability of the event A ∩ B' is given by P(A ∩ B') = P(A) - P(A ∩ B) = 0.7 - 0.3 = 0.4.

E. The probability that the selected student has exactly one of the two types of cards is given by P((A ∩ B') ∪ (A' ∩ B)) = P(A ∩ B') + P(A' ∩ B) = 0.4 + 0.1 = 0.5.

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The measure of the angle B is found to be 184 degrees.

The angles given in question:

line AB || DC

∠B = 9x + 2

∠C = 5x - 4

So,

∠B + ∠C = 180 ( co interior angles )

9x + 2 + 5x - 4 = 180

9x  - 2 = 180

9x = 180 + 2

x = 182/9

Then,

∠B = 9x + 2

∠B = 9*182/9 + 2

∠B = 182 + 2

∠B = 184

Thus, the measure of the angle B is found to be 184 degrees.

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The diagram of the correct question is attached.

The value of covariance cov(y1, y2) is -0.222

                      y1      

y2         0         1         2          Total

0          1/9      2/9     1/9          4/9

1          2/9      2/9      0            4/9

2          1/9        0       0            1/9

Total   4/9       4/9     1/9          1      

  marginal distribution of y2:

y2      P(y2)      y2P(y2)      y2^2P(y2)

0         4/9        0.0000        0.0000

1          4/9         0.4444        0.4444

2          1/9         0.2222        0.4444

total      1          0.66667       0.88889

E(y2) = 0.6667

E(y2^2) = 0.8889

Var(y2) = E(y2^2)-(E(y2))^2=0.4444

marginal distribution of y1  

y1       P(y1)       y1P(y1)      y1^2P(y1)

0         4/9       0.0000      0.0000

1          4/9        0.4444       0.4444

2          1/9         0.2222      0.4444

total  1.00          0.6667      0.8889

E(y1) = 0.6667

E(y1^2) = 0.8889

Var(y1)= σy = E(y1^2)-(E(y1))^2 = 0.4444

E(XY) =∑xyP(x,y)=0.2222

Cov(y1,y2) = -0.222

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(a) The diver hit the water after 2.5 seconds

(b) The diver hit the water with a velocity of 80 ft/s

Since the cliff diver plunges from a height of 100 ft above the water surface and the distance the diver falls in seconds is given by the function d(t) = 16t². We can say d(t) = 100ft. So we have:

d(t) = 16t² = 100

t² = 100/16

t² = √(100/16)

t = 10/4

t = 2.5 seconds

The derivative of the distance will give us the velocity. That is:

v(t) = d'(t) = 32t

Substitute t =2.5  into v(t):

v(2.5) = 32(2.5)

v(2.5) = 80 ft/s

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A)  We can draw a line perpendicular to side ED from the center of the hexagon, which will bisect side ED and create two additional triangles.

B) Area of each triangle = 31.16 square inches (rounded to two decimal places)

C) Total surface area = 375.87 square inches (rounded to two decimal places)

Any geometric shape with three dimensions can have its surface area determined. The area or region that an object's surface occupies is known as its surface area.

Now in the given question,

Part A: To decompose this shape into triangles, we can draw lines connecting the center of the hexagon to each of its vertices, creating six triangles. We can also draw a line perpendicular to side ED from the center of the hexagon, which will bisect side ED and create two additional triangles.

Part B: Each of the six triangles created by connecting the center of the hexagon to its vertices is an equilateral triangle, because all sides of the hexagon are equal in length. The area of an equilateral triangle can be calculated using the formula:

Area = (√3 / 4) x side²

where side is the length of one side of the equilateral triangle. In this case, the length of each side of the equilateral triangle is also 12 inches, so we have:

Area = (√3 / 4) x 12²

Area = (√3 / 4) x 144

Area = 36√3 square inches

Each of the two triangles created by drawing a perpendicular from the center of the hexagon to side ED is a right triangle, because one of its angles is 90 degrees. We can use the Pythagorean theorem to calculate the length of the other two sides of the right triangle. We know that one side has length 10.4 inches and the hypotenuse (which is also a side of the hexagon) has length 12 inches. Let x be the length of the other side of the right triangle. Then we have:

x² + 10.4² = 12²

x²+ 108.16 = 144

x² = 35.84

x = √35.84

x = 5.99 inches (rounded to two decimal places)

The area of each right triangle can be calculated using the formula:

Area = (1/2) x base x height

where the base is 5.99 inches (the length of the side opposite the 90 degree angle) and the height is 10.4 inches (the length of the side adjacent to the 90 degree angle). We have:

Area = (1/2) x 5.99 x 10.4

Area = 31.16 square inches (rounded to two decimal places)

Part C: To determine the area of the table's surface, we need to add up the areas of all eight triangles. There are six equilateral triangles, each with an area of 36sqrt(3) square inches, and two right triangles, each with an area of 31.16 square inches. Therefore, the total area of the table's surface is:

Total area = 6 x 36√3 + 2 x 31.16

Total area = 216√3 + 62.32

Total area = 375.87 square inches (rounded to two decimal places)

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The value of x is given as follows:

x = 34.

The angle measures are given as follows:

In a parallelogram, consecutive angles are supplementary, meaning that the sum of their measures is of 180º.

Angles A and B are consecutive, hence the value of x is obtained as follows:

2x + 10 + 3x = 180

5x = 170

x = 34.

Then the angle measures are given as follows:

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Answer:

it could be D but D gives us 661.63 C gives us 613.07 B gives us 892.165 and A gives us 883.98 so your closes answer is D

Step-by-step explanation:

8.42x109=917.78-253.8=663.98

Answer: We can calculate the size of the lot in acres and then compare it to the minimum size of 3/4 of an acre that the family is looking for.

1 acre = 43,560 square feet

So, a 30,000 sq-ft lot is equivalent to:

30,000 sq-ft / 43,560 sq-ft/acre = 0.6864 acres

Since 0.6864 acres is less than 3/4 of an acre, the family's agent will be wasting her time by showing them the house with a 30,000 sq-ft lot. The lot is not big enough for the family's needs.

Step-by-step explanation:

A sequence of transformations is applied to a polygon. [ The following statements represent a sequence of transformations where the resulting polygon is similar to the original polygon but have a smaller area than the original polygon is that: a dilation about the origin by a scale factor of 2 3 followed by a rotation of 90° counterclockwise about the origin and a translation 5 units left.

To measure the magnitude of the orbital angular momentum L and the projection of the angular momentum along a particular axis (say, the z-axis), we need to use the mathematical formalism of quantum mechanics.

In quantum mechanics, the orbital angular momentum L of a particle is an operator that acts on the wave function describing the particle's motion. Similarly, the z-component of the angular momentum Lz is also an operator.

The magnitude of the orbital angular momentum L can be calculated from the components of the angular momentum operator using the expression:

L^2 = L₁^2 + L₂^2 + L₃^2

where L₁, L₂, and L₃ are the x, y, and z components of the angular momentum operator, respectively. To measure the magnitude of the orbital angular momentum, we would need to measure the values of L₁ L₂, and L₃ and use them to calculate L^2.

The projection of the angular momentum along a particular axis (say, the z-axis) is given by the operator L₃. To measure the z-component of the angular momentum, we would need to measure the value of L₃ for a particular state of the system.

In practice, these measurements are often carried out using experiments involving the interaction of particles with magnetic fields. The behavior of the particles in the magnetic field allows us to infer information about the angular momentum of the particles.

The measurement of the magnitude and projection of angular momentum is an important part of many areas of physics, including quantum mechanics and solid-state physics.

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The distance traveled in the time interval 0 ≤ t ≤ 6, given the velocity function v(t) = t^2 - 6t - 16, can be determined by calculating the definite integral of the absolute value of the velocity function over the given time interval. The result is 128 units of distance.

To find the distance traveled in the given time interval, we need to integrate the absolute value of the velocity function v(t) = t^2 - 6t - 16 over the interval 0 ≤ t ≤ 6. The reason for taking the absolute value is that distance is a scalar quantity and does not depend on the direction of motion.

Taking the integral of the absolute value of the velocity function, we have:

∫|v(t)| dt = ∫|t^2 - 6t - 16| dt

To evaluate this integral, we need to split it into intervals where the velocity function is positive and negative. The absolute value function essentially removes the negative sign from the expression inside the absolute value brackets.

Next, we find the points where the velocity function changes sign by setting v(t) = 0:

t^2 - 6t - 16 = 0

Solving this quadratic equation, we find t = -2 and t = 8 as the points where the velocity changes sign.

Now, we evaluate the integral over the intervals [0, 2] and [2, 6] separately, considering the absolute value of the velocity function within each interval.

∫|v(t)| dt = ∫(t^2 - 6t - 16) dt over [0, 2] + ∫(-(t^2 - 6t - 16)) dt over [2, 6]

Evaluating the definite integrals, we obtain:

(128/3) + (128/3) = 256/3

Therefore, the distance traveled in the time interval 0 ≤ t ≤ 6 is 256/3 units of distance, which is approximately 85.33 units of distance.

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Hi! The answer would be 445.272727273

Glad I can help :)

The value of x and y is 31° and 101°.  Value is the measure of what something is worth, either financially or in terms of importance.

[tex]$\angle \mathrm{PRT}+\angle \mathrm{RTP}+\angle \mathrm{TPR}=180^{\circ} \text { (angle sum property of triangle) } \\[/tex]

[tex]& \Rightarrow \mathrm{x}+\left(180^{\circ}-\angle \mathrm{RTQ}\right)+60^{\circ}=180^{\circ} \text { (linear pair) } \\[/tex]

[tex]& \Rightarrow \mathrm{x}+\left(180^{\circ}-97^{\circ}\right)+60^{\circ}=180^{\circ} \\[/tex]

We get,

[tex]& \Rightarrow \mathrm{x}=31^{\circ} \\[/tex]

[tex]& \text { Now } \angle \mathrm{PRT}+\angle \mathrm{TRQ}+\angle \mathrm{QRS}=180^{\circ} \text { (angle of straight line) } \\[/tex]

[tex]& \Rightarrow \mathrm{x}+48^{\circ}+\mathrm{y}=180^{\circ} \\[/tex]

Simplifying,

[tex]& \Rightarrow 31^{\circ}+48^{\circ}+\mathrm{y}=180^{\circ} \\[/tex]

Then we get,

[tex]& \Rightarrow \mathrm{y}=101^{\circ}[/tex]

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The average rate of change over the interval (-2,0) is -2.

The average rate of change describes how quickly one quantity changes in comparison to another. It indicates how much the function changed per unit during the specified interval.

Given that the interval is (-2,0). The coordinate of the point from the interval -2,0 is ( -2 , 4 ).

The average rate of change will be calculated as:-

Rate of change = ΔY / ΔX

Rate of change = ( 4 - 0 ) ( -2 - 0 )

Rate of change = -2

Therefore, the average rate of change will be -2.

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The question is geometric distribution. We are concerned with just the first success. That means there is no fixed number of trials

Before a situation can model binomial distribution.

For question a.

The probability of success is not the same for each trial. Since we are not putting any back. So it cannot be modeled as binomial distribution

For question b.

This has many outcomes not just success and failure.

Lan needs to record Pluto bar, jolly farm, husky kisses, finish fish, and hair heads.

For question c.

The question is geometric distribution. We are concerned with just the first success. That means there is no fixed number of trials

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The value of C is 108°

Given [tex]a=33,\ \ b=26, \ \ \angle B=31^0 .[/tex]

We have to find [tex]\angle C, \ \angle A \ and\ c[/tex]

We can use the cosine formula to find these:

[tex]cosB=\frac{a^2+c^2-b^2}{2ac}\\cos31^o=\frac{33^2+c^2-26^2}{2(33)(c)}\\0.8572=\frac{413+c^2}{66c}\\56.5752c=413+c^2\\i.e. c^2-56.5752c+413=0\\\Rightarrow c=47.9647, \ 8.6104\\So, \mathbf{c=48 \ or \ c=9}\\Now if c=48:\\cosA=\frac{b^2+c^2-a^2}{2bc}\\=\frac{26^2+48^2-33^2}{2(26)(48)}\\[/tex]

[tex]=\frac{1891}{2496}\\=0.75761\\\therefore A=cos^{-1}(0.75761)=40.746^0\approx 41^0\\i.e. \mathbf{\angle A= 41^0}\\cosC=\frac{a^2+b^2-c^2}{2ab}\\=\frac{33^2+26^2-48^2}{2(33)(26)}\\=\frac{-539}{1716}\\=-0.3141\\\therefore C=cos^{-1}(-0.3141)=108.306^0\approx 108^0[/tex]

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