Calculate the fuel economy of your family or personal car in terms of gallons per mile and the total cost for one fill-up git hub.

Answer:

A relationship is said to have direct variation when one variable changes and the second variable changes proportionally; the ratio of the second variable to the first variable remains constant. For example, when y varies directly as x, there is a constant, k, that is the ratio of y:x.

Answer:

10-[6-4+(5)]×2

10-[2+5]×2

10-(7)×2

10-14= -4

Answer: a. 0.40   b. 0.23  c . 0.435   d . 0.25

Step-by-step explanation:

                                   melanin      content    Total

                                            high   low

moisture   high                     13      10                23

content    low                       47      30                77

 Total                                   60      40               100

Let A denote the event that a sample has low melanin content, and let B denote the event that a sample has high moisture content.

a) Total skin samples has low melanin content = 10+30=40

P(A)=[tex]\dfrac{40}{100}=0.40[/tex]

b) Total skin samples has high moisture content = 13+10=23

P(B) =[tex]\dfrac{23}{100}=0.23[/tex]

c) A ∩ B =  Total skin samples has both low melanin content and high moisture content =10

P(A ∩ B) =[tex]\dfrac{10}{100}=0.10[/tex]

Using conditional probability formula , [tex]P (A|B)=\dfrac{P(A\cap B)}{P(B)}[/tex]

[tex]P (A|B)=\dfrac{0.10}{0.23}=0.434782608696\approx0.435[/tex]

d)  [tex]P (B|A)=\dfrac{P(A\cap B)}{P(A)}[/tex]

[tex]P (B|A)=\dfrac{0.10}{0.40}=0.25[/tex]

Answer:

1. 3, and 2. m x n

Step-by-step explanation:

1. for an upper triangular 2x2 matrix i.e. (a,0,c,d), three (03) unknown elements a, c, and d are needed to be specified.

2. for m x n matrix, m*n elements are needed to be specified.

Final answer:

To specify an upper triangular 2x2 matrix, 3 unknown real numbers are needed. For an m × n matrix, m × n unknown real numbers are required.

Explanation:

The question asks how many unknown real numbers are needed to specify each of the given matrices: an upper triangular 2x2 matrix, and an m × n matrix.

1. An Upper Triangular 2x2 Matrix

An upper triangular matrix has the form:

(a, b)
(0, c)

Thus, to specify an upper triangular 2x2 matrix, 3 unknown real numbers are needed: a, b, and c.

2. An m × n Matrix

An m × n matrix has m rows and n columns. To specify such a matrix, one needs m × n unknown real numbers, representing each element in the matrix.

Answer:

dont see much information here but as far as lawsuits go id aim for the highest answer

Step-by-step explanation:

_____________________________________

Answer:

True

Step-by-step explanation:

The time between customer arrivals is called inter-arrival time. According to Queueing Notation, the inter-arrival time can be model based on difference probability distribution. The probability distribution by which the inter-arrival time can be modeled include:

Answer:

For x=45

sample proportion=45/140=0.321

z=(0.321-0.30)/sqrt(0.3*(1-0.3)/140)

z=0.54

For x=47

sample proportion=47/140=0.336

z=(0.336-0.30)/sqrt(0.3*(1-0.3)/140)

z=0.93

Now,

P(0.54<z<0.93)=P(z<0.93)-P(z<0.54)

=0.8238-0.7054

=0.118

So,correct option is 0.119

Answer:

(x-2)/5(x-2)

cancel x-2 from the numerator and the denominator and the answer is 1/5

Answer:

a) No it doesn't mean that linear model is inappropriate

b) No. The prediction using this model will not be accurate.

Step-by-step explanation:

a)

For answering this part, firstly consider the concept of [tex]R^{2}[/tex]

The [tex]R^{2}[/tex] also known as coefficient of determination is used to determine the amount of variability in dependent variable is explained by the linear model. Lower [tex]R^{2}[/tex] depicts that less variation of dependent is explained by the independent variable using the linear model. The linearity of model is determined by scatter plot. Thus, if the [tex]R^{2}[/tex] is lower, it doesn't mean that linear model is inappropriate.

b)

The predictions made by the model having lower [tex]R^{2}[/tex] value are erroneous. The model is used for prediction if the linear model explains the larger portion of variability in dependent variation. If the predictions made from the model that have lower [tex]R^{2}[/tex] value then the predicted values will not be close to the actual value and thus residuals will not be minimum as residuals are the difference of actual and predicted values.

Answer:

a.

lower Quartile= 57.5

Upper Quartile=81

b.

23.5

c.

box-plot is attached in excel file

Step-by-step explanation:

The data is arranged in ascending order so, the lower quartile denoted as Q1 can be calculated as under

Q1=((n+1)/4)th score=(41/4)th score=(10.25)th score

Q1=10th score+0.25(11th-10th)score

Q1=57+0.25(59-57)=57+0.5=57.5

Q1=57.5

The data is arranged in ascending order so, the third quartile denoted as Q3 can be calculated as under

Q3=(3(n+1)/4)th score=(3*41/4)th score=(30.75)th score

Q3=30th score+0.75(31th-30th)score

Q3=81+0.75(81-81)=81+0=81

Q3=81

b)

Interquartile range=IQR=Q3-Q1=81-57.5=23.5

IQR=23.5

c)

The box-plot is made in excel and it shows no outlier. The box-plot shows the 5-number summary(minimum-Q1-median-Q3-maximum) as 25-57.5-72-81-98.

Answer: 2520 ways

Step-by-step explanation:

7!/2!

Answer:

no

Step-by-step explanation:

The probability that the project will be finished in 62 weeks or less is 0.6915.

The probability that the project will be finished in 66 weeks or less is 0.9332.

The probability that the project will take longer than 65 weeks is 0.1056.

Given that:

The time to complete a construction project is normally distributed.

The mean is :

μ = 60 weeks

The standard deviation is:

σ = 4 weeks

The z-score is:

[tex]z=\frac{x-\mu }{\sigma }[/tex]

When x = 62,

[tex]z=\frac{62-60}{4}[/tex]

   [tex]=0.5[/tex]

So, P(x ≤ 62) = P(z ≤ 0.5).

From the standard table, P(z ≤ 0.5) = 0.6915

When x = 66,

[tex]z=\frac{66-60}{4}[/tex]

   [tex]=1.5[/tex]

So, P(x ≤ 66) = P(z ≤ 1.5).

From the standard table, P(z ≤ 1.5) = 0.9332

When x = 65,

[tex]z=\frac{65-60}{4}[/tex]

   [tex]=1.25[/tex]

So, P(x > 65) = P(z > 1.25).

                     = 1 - P(z ≤ 1.25).

From the standard table, P(z ≤ 1.25) = 0.8944

So, P(x > 65) = P(z > 1.25)

                     = 1 - 0.8944

                     = 0.1056

Hence, the probabilities are 0.6915, 0.9332, and 0.1056 respectively.

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The probability the project will be finished in 62 weeks or less is approximately 69.15%. The probability the project will be finished in 66 weeks or less is approximately 93.32%. The probability the project will take longer than 65 weeks is approximately 10.56%.

To find the probability that the project will be finished in 62 weeks or less, we need to calculate the z-score. The z-score formula is (x - μ) / σ where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Plugging in the values, we get (62 - 60) / 4 = 0.5. Using a z-score table, we can look up the probability corresponding to a z-score of 0.5, which is approximately 0.6915 or 69.15%.

Similarly, to find the probability that the project will be finished in 66 weeks or less, we calculate the z-score: (66 - 60) / 4 = 1.5. Looking up a z-score of 1.5 in the table, we find the probability is approximately 0.9332 or 93.32%.

To find the probability that the project will take longer than 65 weeks, we can subtract the probability of it being finished in 65 weeks or less from 1. Using the z-score formula, we get (65 - 60) / 4 = 1.25. The probability of finishing in 65 weeks or less is the area to the left of this z-score, which is approximately 0.8944 or 89.44%. Subtracting from 1, we get the probability of taking longer than 65 weeks is approximately 0.1056 or 10.56%.

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Answer:

1) a.) Your confidence interval estimate is narrower

2) c.) The width of a confidence interval estimate for a proportion will be narrower for 90% confidence level than for a 95% confidence level

Step-by-step explanation:

Confidence Interval can be stated as  M±ME where

Margin of Error determines the range of the confidence interval around the mean.

Margin of error (ME) of the mean can be calculated using the formula

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

From the formula we can reach the following conclusions:

Since your sample size (49) is bigger than your friend's (36), your confidence interval is narrower, because margin of error is narrower.

Since the confidence level 90% has smaller statistic than the confidence level 95%, its confidence interval is narrower.

That is, we can estimate narrower confidence intervals with less confidence.

Answer:

2.95 cent

Step-by-step explanation:

1 gallon = 231 cubic inches

1 litre = 1000ml = 61.0237 cubic inches

1 galloon = 231 / 61.0237 = 3.7854118 liters

if Carol paid $0.78 per litre

1 galloon = 0.78 x 3.7854118 = 2.952621204 ≅ 2.95 cent

Answer:

236.68 feet needed for fencing

Explanation:

You are designing a rectangular enclosure with 2 rectangular sections separated by parallel walls

Let L be the length and W be the width

Perimeter of rectangle = 3(length )+ 4 (width)

[tex]P=3L+4W[/tex]

Area of the rectangle = length * width

 [tex]A=2LW[/tex]

[tex]W= \frac{A}{2L}[/tex]

[tex]W= \frac{1700}{2(60)}=14.17[/tex]

Replace the values in perimeter

[tex]P=3L+4W[/tex]

[tex]P=3(60)+4(14.17)=236.68[/tex]

So 236.68 feet needed for fencing

If the coefficient matrix has a pivot in each column, it means that it is shaped like this:

[tex]A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right][/tex]

So, the correspondant system

[tex]Ax = b[/tex]

will look like this:

[tex]\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right][/tex]

This turn into the following system of equations:

[tex]\begin{cases}a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1\\a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2\\a_{3,3}x_3+a_{3,4}x_4=b_3\\a_{4,4}x_4=b_4\end{cases}[/tex]

The last equation is solvable for [tex]x_4[/tex]: we easily have

[tex]x_4=\dfrac{b_4}{a_{4,4}}[/tex]

Once the value for [tex]x_4[/tex] is known, we can solve the third equation for [tex]x_3[/tex]:

[tex]x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}[/tex]

(recall that [tex]x_4[/tex] is now known)

The pattern should be clear: you can use the last equation to solve for [tex]x_4[/tex]. Once it is known, the third equation involves the only variable [tex]x_3[/tex]. Once

Answer:

Step-by-step explanation:

Hello!

You have two variables of interest.

X: Setup size (inches)

Y: Type the number of channels available.

Qualitative variables are those who describe characteristics of the subject of study, for example, the eye color of a person.

Quantitative variables are those that count quantities, for example, the shoe size of a person.

Continuous and discrete variables are quantitative. The difference is that the continuous variables are those who count in a determined range of valours, but between two observed values, there are infinite possible outcomes, for example, the body temperature of a cat. The normal temperature of a cat is around 38ºC, using a normal thermometer you measure the body temperature of two cats and obtain the following values 37.8 and 37.9 if you change the thermometer to one designed to take more precise measurements, it is possible that you obtain more values, for example, 37.81 and 39.94 and with a more precise tool you may become temperatures with more digits, this means that within this two temperatures there are infinite values of temperature, only limited by the equipment available.

A discrete variable is a quantitative variable but between the values, these variables take there are no other possible observations, regardless of the method of equipment used. An example of a discrete variable is the amount of money in a pocket. If you have two bills in one pocket, one is a 10 dollar bill and the other is a 20 dollar bill, there are no possible values in between, you either have ten or twenty, there is not possible, in this example, to count 15 dollars.

Then the variable "Y: Type number of channels available." is quantitative discrete, it counts the number of channels and between each channel there is nothing.

The variable "X: Setup size (inches)", the "inch" is a unit of length, and these variables are usually continuos, but in this example, your variable describes the screen width of the televisions and the type of image definition. Both are characteristics of the TVs so the variable is a qualitative one.

I hope it helps!

Answer:

[tex] [tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)[/tex]=[tex]1-384 +512+3\sqrt{8} -18(8)^{5/2} +3 (8)^{7/2} =1223.601[/tex]

And the limit on this case exists.

Step-by-step explanation:

We want to find the following limit:

[tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)[/tex]

First we can distribute the polynomials like this:

[tex] lim_{x \to 8} (1-6x^2 +x^3+3\sqrt{x} -18 x^{5/2} +3x^{7/2})[/tex]

And Now we can use the distributive property for the limit and we got:

[tex] lim_{x \to 8} 1 - 6 lim_{x \to 8} x^2 + lim_{x \to 8} x^3 +3 lim_{x \to 8} \sqrt{x} -18 lim_{x \to 8} x^{5/2} + 3 lim_{x \to 8} x^{7/2}[/tex]

And now we can evaluate the limit and we got:

[tex] [tex] lim_{x \to 8} (1+3\sqrt{x})(1-6x^2 +x^3)[/tex]=[tex]1-384 +512+3\sqrt{8} -18(8)^{5/2} +3 (8)^{7/2} =1223.601[/tex]

And the limit on this case exists.

To solve limit problems in mathematics, limit laws are often very useful. In this specific case, as the function is a polynomial and defined for all real number values, a direct substitution of x=8 into the function is sufficient. Therefore, the limit as x approaches 8 for function 1 + 3x5 - 6x2 + x3 is calculable.

In the field of mathematics, limit laws are used quite frequently for evaluating limits. In this case, we want to calculate the limit as x approaches 8 for the function 1 + 3x5 - 6x2 + x3.

For a given polynomial function like this one, an easy and very straightforward approach is to substitute the value x is approaching (in this scenario, x = 8) directly into the polynomial function.

So, after substitution, our function becomes: 1 + 3*(8)^5 - 6*(8)^2 + (8)^3. Simplifying it further, the limit as x approaches 8 of this function gives us a definite numeric value.

Always remember while applying limit laws, you might at times need the limit laws to evaluate complex limit problems but in this given scenario, direct substitution works perfectly fine because this polynomial function is defined for all real number values of X.

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Answer:

y=-2· e^(0.01)/ x²

Step-by-step explanation:

We calculate the given differential equation, we get

dy/dx = e^(−0.01) xy²

dy/y² = e^(−0.01) x dx

∫ y^(-2) dy= e^(−0.01) ∫ x dx

- y^(-1) = e^(−0.01) x²/2

-1/y=  e^(−0.01) x²/2

y=-2/ e^(−0.01) x²

y=-2· e^(0.01)/ x²

We use the site desmos.com, to plot graph for the solution of the the given differential equation. We get a graph.

Answer:

Step-by-step explanation:

Check the attachment for the solution

Answer:

Step-by-step explanation:

The objective is to determine which of the following matrices are in reduced echelon form and which others are only in echelon form. The given matrices are

                       [tex]\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0& 2 & 0 & 0 \\ 0& 0 & 1 & 1 \end{bmatrix}[/tex],  [tex]\begin{bmatrix} 1 & 0 & 1 & 1 \\ 0& 1& 1 & 1 \\ 0& 0 & 0 & 0 \end{bmatrix}[/tex]  and   [tex]\begin{bmatrix} 0& 0 & 0 & 0 \\ 1& 3 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0& 0 & 0 & 1 \end{bmatrix}[/tex].

First, recall what is an echelon and reduced echelon form of a matrix.

A matrix is said to be in a Echelon form if

A matrix is said to be in  a Reduced Echelon form if

A column that contains a leading 1 which is the only non-zero element is called a pivot column.

Now, let's have a look at the first matrix

                                 [tex]\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0& 2 & 0 & 0 \\ 0& 0 & 1 & 1 \end{bmatrix}[/tex]

As we can see, it doesn't have any zero rows. Each leading entry in a row is placed to the right of the leading entry from the row above and all elements below the leading entries in all columns are equal to zero. Therefore, this matrix is in an Echelon form.

In the second row, the leading entry is 2, not 1, so because of the first property of the Reduced Echelon form, it is not in a Reduced Echelon form.

Notice that it can be transformed to the Reduced Echelon form by multiplying the second row by [tex]\frac{1}{2}.[/tex]

The second matrix is

                                         [tex]\begin{bmatrix} 1 & 0 & 1 & 1 \\ 0& 1& 1 & 1 \\ 0& 0 & 0 & 0 \end{bmatrix}[/tex]

There is a zero row, and all non-zero rows are placed above it. Each leading entry in a row, which is the first non-zero entry, is placed to the right of the entry of the row above it and all elements below the leading entry are equal to zero in each column, so it is in the Echelon form.

It is also in the Reduced Echelon form, since all non-zero rows the leading entry is 1 and it is the only non zero element in each column.

The least given matrix is

                                        [tex]\begin{bmatrix} 0& 0 & 0 & 0 \\ 1& 3 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0& 0 & 0 & 1 \end{bmatrix}[/tex]

This matrix doesn't satisfy the condition that if there is any zero-row, it must be below all other non-zero rows, so it is not in Echelon form.

A matrix that is not in an Echelon form, it is not in an Reduced Echelon form either.

Therefore, this matrix is not in an Reduced Echelon form.

The question pertains to calculating the probability of drawing exactly 3 jacks from 51 randomly drawn cards from a 52-card deck. While the probability is very high, it's not an absolute certainty. The exact calculations involve complex combinatorial mathematics.

The subject of this question pertains to probability in mathematics, specifically to calculate the likelihood of drawing 3 jacks when selecting 51 cards from a shuffled deck of 52 cards.

First off, we need to understand that in a well-shuffled 52-card deck, there are 4 Jacks. Even if you select 51 out of 52 cards, there isn't a guarantee that you will select 3 jacks because the selection is random. The scenario you provided indicates a nearly certain event (since you're pulling nearly all the cards), but it still isn't an absolute certainty.

The exact probability computation for this kind of problem are more complex as they would involve combinatorial calculations. For simplicity, let's consider a similar but simpler scenario. Let's assume you are drawing just 4 cards instead. The probability of getting exactly 3 Jacks would be a combination of the probability of picking a Jack, and the probability of picking a non-Jack card. This would be calculated as (C(4,3) * C(48,1)) / C(52,4), with C representing the combination formula. This gives us how many ways we can draw 3 Jacks and a non-Jack divided by how many ways we can draw any 4 cards.

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Final answer:

The probability of drawing 3 jacks from a standard deck of 52 cards when selecting 51 is 1 (certainty), as it is a guaranteed event given the conditions.

Explanation:

The question asks about the probability of a certain event occurring when dealing with a standard deck of 52 cards. In this case, the event is being certain to get 3 jacks when selecting 51 cards out of 52. Since there are 4 jacks in the deck, and upon drawing 51 cards you're left with only 1 card that is not drawn, it is guaranteed that you'll have the 3 jacks among the drawn cards.

Hence, the probability is 1 (certainty), as there is only one card you're not drawing and 4 chances to have drawn a jack, which means you will always end up with all 3 jacks among the chosen 51 cards.

Answer:its A

Step-by-step explanation:it was

The formula is used to find the diameter of circle is: [tex]d = \frac{C}{3.14}[/tex]

The diameter of circle is 5.1 cm

Solution:

Given that,

Circumference (C) of a circle is 16 cm

The formula for circumference of circle when diameter is given is:

[tex]C = \pi d\\\\\pi \text{ is a constant equal to 3.14}\\\\C = 3.14d[/tex]

Rearrange the formula to get "d"

Divide both sides by 3.14

[tex]d = \frac{C}{3.14}[/tex]

The above formula is used to find the diameter of circle

Given that, circumeference = C = 16 cm

Substituting we get,

[tex]d = \frac{16}{3.14}\\\\d = 5.095 \approx 5.1[/tex]

Thus diameter of circle is 5.1 cm

No, it is not advisable to predict the stopping time for your large SUV using model trained for compact cars.

Prediction means generating the values of the dependent variable using some specific models in machine learning.

Given that, the model is trained on 20 compact cars and the model is developed such that it predicts the stopping time for a vehicle by using its weight.

Here the dependent variable is stopping time which is required to be predicted. As the model is trained on compact cars that is medium size cars and if we expect the same model to predict stopping time for large SUV, then model is going to predict false stopping time as the weights for large SUV is quiet higher than the compact cars. So, model may consider it as an outlies and will lead to incorrect prediction.

Therefore, it is not advisable to use the same model for predicting the stopping time for your large SUV.

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Final answer:

Using a stopping time model developed from compact car data to predict the stopping time for a large SUV is not advisable due to differences in vehicle dynamics, which may lead to inaccurate results.

Explanation:

When a consumer group develops a model to predict the stopping time of a vehicle based on its weight, the model must be used within the context of the data from which it was derived. Using the model, which was built on data from 20 compact cars, to predict the stopping time of a larger SUV is not advisable due to differences in vehicle dynamics, size, weight distribution, and potentially different braking systems. Models are designed to be predictive within the range of data they are based on, and extrapolating them beyond that range can lead to inaccurate predictions. Specifically, the heavier mass of an SUV compared to compact cars means that it would likely have a longer stopping distance due to greater momentum, and this may not be represented in a model calibrated to lighter vehicles.

Answer:

The x-coordinate of the solution to this system of equations is 1.

Step-by-step explanation:

Given,

[tex]6x - 5y = 5\\\\3x + 5y = 4[/tex]

We have to find out the x-coordinate of the equation.

Solution,

Let [tex]6x-5y=5\ \ \ \ equation\ 1[/tex]

Again let [tex]3x+5y=4\ \ \ \ \ equation \ 2[/tex]

Now using elimination method we will solve the equations.

For this we will add equation 1 and equation 2 and get;

[tex](6x-5y)+(3x+5y)=5+4\\\\6x-5y+3x+5y=9\\\\9x=9[/tex]

Now on dividing both side by '9' we get;

[tex]\frac{9x}{9}=\frac{9}{9}\\\\x=1[/tex]

Hence The x-coordinate of the solution to this system of equations is 1.

1

ur welcome homie

poggers

Answer:

I'm pretty sure it would be 345, just add the two 178 and 167

Answer:

Central angle= 1.15 radians

Step-by-step explanation:

[tex]Arc\,\,length=s= 1.22\,m\\Radius=r=1.06\,m\\\\Central\,\, angle=\theta=?\\\\Using\\\\ s=r\theta\\\\\theta=\frac{s}{r}\\\\\theta= \frac{1.22}{1.06}\\\\\theta=1.15 \,rad[/tex]

Answer:

2(x+6)(x-3)

Step-by-step explanation:

Factor the GCF out of the trinomial on the left side of the equation.

[tex]2x^2 + 6x - 36 =2(x^2 + 3x - 18)[/tex]

Greatest common factor of 2, 6, 18 is 2

so GCF is 2

divide each term when we take out GCF 2

so [tex]2(x^2 + 3x - 18)[/tex]

now factor the trinomial

product is -18 and sum is +3

6 times -3 is -18  and 6-3=3

[tex]2(x^2+3x-18)\\2(x+6)(x-3)[/tex]

Answer:

The answer to your question is the second option

Step-by-step explanation:

A Quadratic function is a polynomial of degree two. That means that the higher exponent is 2.

a) This option is incorrect because the highest power is 3 not two.

b) This option is the right answer, the highest power is 2, so, it is a quadratic function.

c) This option is incorrect, the highest power is -2.

d) This option is incorrect, the highest option is 1.

Answer:

Option 2 is the correct answer

Step-by-step explanation:

A quadratic function is a function in which the highest power to which the variable is raised is 2

1) f(x) = −8x3 − 16x2 − 4x

The given function is a cubic function because the highest power

to which the variable,x is raised is 3

2) f(x) = 3x²/4 + 2x - 5

The given function is a quadratic function because the highest power

to which the variable,x is raised is 2

3) f(x) = 4/x² - 2/x + 1

It can be rewritten as

f(x) = 4x^-2 - 2x^-1 + 1

The given function is not a quadratic function because the highest power to which the variable,x is raised is - 2

4) f(x) = 0x2 − 9x + 7

It can be rewritten as

f(x) = - 9x + 7

The given function is not a quadratic function because the highest power to which the variable,x is raised is 1

Answer:

For the set X = {a, b, c}, the following three relations satisfy the required conditions in (a), (b) and (c) respectively.

(a) R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)} is reflexive and symmetric but not transitive .

(b) R = {(a, a), (b, b), (c, c), (a, b)} is reflexive and transitive but not symmetric .

(c) R = {(a,a), (a, b), (b, a)} is symmetric and transitive but not reflexive .

Step-by-step explanation:

Before, we go on to check these relations for the desired properties, let us define what it means for a relation to be reflexive, symmetric or transitive.

Given a relation R on a set X,

R is said to be reflexive if for every [tex]a \in X, (a,a) \in R[/tex].

R is said to be symmetric if for every [tex](a, b) \in R, (b, a) \in R[/tex].

R is said to be transitive if [tex](a, b) \in R[/tex] and [tex](b, c) \in R[/tex], then [tex](a, c) \in R[/tex].

(a) Let R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)}.

Reflexive: [tex](a, a), (b, b), (c, c) \in R[/tex]

Therefore, R is reflexive.

Symmetric: [tex](a, b) \in R \implies (b, a) \in R[/tex]

Therefore R is symmetric.

Transitive: [tex](a, b) \in R \ and \ (b, c) \in R[/tex] but but (a,c) is not in  R.

Therefore, R is not transitive.

Therefore, R is reflexive and symmetric but not transitive .

(b) R = {(a, a), (b, b), (c, c), (a, b)}

Reflexive: [tex](a, a), (b, b) \ and \ (c, c) \in R[/tex]

Therefore, R is reflexive.

Symmetric: [tex](a, b) \in R \ but \ (b, a) \not \in R[/tex]

Therefore R is not symmetric.

Transitive: [tex](a, a), (a, b) \in R[/tex] and [tex](a, b) \in R[/tex].

Therefore, R is transitive.

Therefore, R is reflexive and transitive but not symmetric .

(c) R = {(a,a), (a, b), (b, a)}

Reflexive: [tex](a, a) \in R[/tex] but (b, b) and (c, c) are not in R

R must contain all ordered pairs of the form (x, x) for all x in R to be considered reflexive.

Therefore, R is not reflexive.

Symmetric: [tex](a, b) \in R[/tex] and [tex](b, a) \in R[/tex]

Therefore R is symmetric.

Transitive: [tex](a, a), (a, b) \in R[/tex] and [tex](a, b) \in R[/tex].

Therefore, R is transitive.

Therefore, R is symmetric and transitive but not reflexive .

Relation from the set of two variables is subset of certain product. The relation for the condition are,

[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]

[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]

[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]

Relation from the set of two variables is subset of certain product. Relation are of three types-

1) Reflexive and symmetric but not transitive -

Let a data set as,

[tex]X=1,2,3[/tex]

For the data set the relation can be given as,

[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]

[tex]R_1[/tex] is reflexive as it can be represent as [tex]R_1(a,a)[/tex] for,

[tex]a=1,2,3, \;\;\;\;\; [/tex]

[tex]a[/tex] ∈ [tex]X[/tex]

[tex]R_1[/tex] is symmetric as it can be represent as [tex]R_1(a,b)[/tex] for,

[tex]a,b \;\;\;\;(1,2) (2,1)[/tex]

[tex]a,b[/tex] ∈ [tex]X[/tex]

[tex]R_1[/tex] is not transitive as it can be represent as [tex]R_1\neq (a,c)[/tex] .

[tex]a,c\neq \;\;\;\;(1,3) (3,1)[/tex]

2)  Reflexive and transitive but not symmetric

Let a data set as,

[tex]X=1,2,3[/tex]

For the data set the relation can be given as,

[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]

[tex]R_2[/tex] is reflexive as it can be represent as [tex]R_2(a,a)[/tex] for,

[tex]a=1,2,3, \;\;\;\;\; [/tex]

[tex]a[/tex] ∈ [tex]X[/tex]

[tex]R_1[/tex] is transitive as it can be represent as [tex]R_1(a,c)[/tex] for,

[tex]a,c \;\;\;\;(1,3) (3,1)[/tex]

[tex]a,c[/tex] ∈ [tex]X[/tex]

[tex]R_1[/tex] is not symmetric as it can be represent as [tex]R_1\neq (a,b)[/tex] .

[tex]a,b\neq \;\;\;\;(1,2) (2,1)[/tex]

3) Symmetric and transitive but not reflexive

Let a data set as,

[tex]X=1,2,3[/tex]

For the data set the relation can be given as,

[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]

[tex]R_1[/tex] is symmetric as it can be represent as [tex]R_3(a,b)[/tex] for,

[tex]a,b=(1,2),(2,1) \;\;\;\;\; [/tex]

[tex]a,b[/tex] ∈ [tex]X[/tex]

[tex]R_3[/tex] is transitive as it can be represent as [tex]R_3(a,c)[/tex] for,

[tex]a,c \;\;\;\;(1,3) (3,1)[/tex]

[tex]a,c[/tex] ∈ [tex]X[/tex]

[tex]R_1[/tex] is not reflexive as it can be represent as [tex]R_3\neq (a,a)[/tex] .

[tex]a,a\neq \;\;\;\;(1,1) [/tex]

Thus the relation for the condition are,

[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]

[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]

[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]

Learn more about the Reflexive, symmetric and transitive relation here;

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