(b) Suppose the 4 × 6 coefficient matrix for a system of linear equations has 4 pivot columns. Say as much as you can about the solutions to the corresponding system of equations, with explanation.

Answer:

You can use the given hint as follows:

Step-by-step explanation:

Let [tex]A[/tex] be a square matrix that is a skew-symmetric matrix. Since the matrix [tex]R={\bf x}^{T}A{\bf x}[/tex] is matrix of size [tex]1\times 1[/tex] then it can be identified with an scalar. It is clear that [tex]R=R^{T}[/tex]. Then applying the properties of transposition we have

[tex]({\bf x}^{T}A{\bf x})^{T}=({\bf x}^{T})A^{T}({\bf x}^{T})^{T}={\bf x}^{T}(-A){\bf x}=-{\bf x}^{T}A{\bf x}[/tex]

Then,

[tex]{\bf x}^{T}A{\bf x}+{\bf x}^{T}A{\bf x}=0[/tex]

[tex]2{\bf x}^{T}A{\bf x}=0[/tex]

Then,

[tex]{\bf x}^{T}A{\bf x}=0[/tex]

For all column vector [tex]{\bf x}[/tex] of size [tex]n\times 1[/tex] .

As per the question,

Note:

''A positive integer is any integer that is greater than zero (0)''.  Also, a positive integer is any integer that is a member of the set of natural numbers, i.e., counting numbers; therefore, the positive integers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 34, 25, 26, 27, 28, 29, 30, 31, 32, 33, ...

“Relatively Prime” (also called “co-prime”) numbers are numbers whose HCF is 1.  Any consecutive positive integers are co-prime (e.g.: 42 and 43)

Now,

First case: If n = 15

Positive integers (less than n = 15) are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 and 14.

Relatively prime are : (1 and 15), (2 and 15), (4 and 15), (7 and 15), (8 and 15),  (11 and 15), (13 and 15) and (14 and 15).

Second case: If n = 18

Positive integers (less than n = 18) are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 and 17.

Relatively prime are : (1 and 18), (5 and 18), (7 and 18), (11 and 18), (13 and 18) and (17 and 18).

Third case: If n = 22

Positive integers (less than n = 22) are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 and 21.

Relatively prime are : (1 and 22), (3 and 22), (5 and 22), (7 and 22), (9 and 22),  (13 and 22), (15 and 22),  (17 and 22),  (19 and 22) and (21 and 22).

Fourth case: If n = 30

Positive integers (less than n = 30) are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28 and 29.

Relatively prime are : (1 and 30), (7 and 30), (11 and 30), (13 and 30),

(17 and 30), (19 and 30), (23 and 30) and (29 and 30).

Fifth case: If n = 35

Positive integers (less than n = 35) are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33 and 34.

Relatively prime are : (1 and 35), (2 and 35), (3 and 35), (4 and 35), (6 and 35), (8 and 35), (9 and 35), (11 and 35), (12 and 35), (13 and 35), (16 and 35), (17 and 35), (18 and 35), (19 and 35), (22 and 35), (23 and 35), (24 and 35), (26 and 35), (27 and 35), (29 and 35), (31 and 35), (32 and 35), (33 and 35) and (34 and 35).

Answer:

Step-by-step explanation:

a) We want to prove that [tex]f^{-1}(A\cap B)\subset f^{-1}(A)\cap f^{-1}(B)[/tex]. Then, we can do that proving that every element of  [tex]f^{-1}(A\cap B)[/tex] is an element of [tex]f^{-1}(A)\cap f^{-1}(B)[/tex] too.

Then, suppose that [tex]x\in f^{-1}(A\cap B)[/tex]. From the definition of inverse image we know that [tex]f(x)\in A\cap B[/tex], which is equivalent to [tex]f(x)\in A[/tex] and [tex]f(x)\in B[/tex]. But, as [tex] f(x) \in A [/tex] we can affirm that [tex]x\in f^{-1}(A)[/tex] and, because  [tex]f(x)\in B[/tex] we have [tex]x\in f^{-1}(B)[/tex].

Therefore, [tex]x\inf^{-1}(A)\cap f^{-1}(B)[/tex].

b) We want to prove that [tex]f(A\cap B) \subset f(A)\cap f(B)[/tex]. Here we will follow the same strategy of the above exercise.

Assume that [tex]y\in f(A\cap B)[/tex]. Then, there exists [tex]x\in A\cap B[/tex] such that [tex]y=f(x)[/tex]. But, as [tex]x\in A\cap B[/tex] we know that [tex]x\in A[/tex] and [tex]x\in B[/tex]. From this, we deduce [tex]f(x)=y\in f(A)[/tex] and [tex]f(x)=y\in f(B)[/tex]. Therefore, [tex]y\in f(A)\cap f(B)[/tex].

c) Consider the constant function [tex]f(x)=1[/tex] for every real number [tex]x[/tex]. Take the sets [tex]A=(0,1)[/tex] and [tex]B=(1,2)[/tex].

Notice that [tex]A\cap B = (0,1)\cap (1,2)[/tex]=Ø, so [tex]f(A\cap B)[/tex]=Ø. But [tex]f(A) = \{1\}[/tex] and [tex]f(B) = \{1\}[/tex], so [tex]f(A)\cap f(B) =\{1\}[/tex].

Answer:

The correct answers are marked.

Step-by-step explanation:

Line k is indicated as perpendicular to RX by the little square at their point of intersection.

Congruence of different line segments will be indicated by marks on them, or by the nature of the geometry containing them (a parallelogram, for instance). There is nothing in this diagram indicating RZ is congruent to GR.

The named points, X, B, V, N, are all shown as being in plane F, so are coplanar.

A plane contains an infinite number of points. In the diagram, there are 5 named points in plane F.

The endpoint of ray RH is point R, which is on line K. However, that is the extent of their intersection. Ray RH heads off in a different direction than line k, so is not part of it.

Segment VN is in plane F; ray RH is not in the plane, but is skew to segment VN. They are not headed in the same direction.

Points R and G are both on line k, so line RG is the same as line k.

In the given figure, line K is perpendicular to RX, points X, B, V, and N are coplanar in plane F, and line RG is the same as line K.

1. Line K is perpendicular to RX:

  - This statement is true because there is a right-angle symbol (∟) at the intersection point of lines K and RX, indicating that line K is perpendicular to line RX.

2. RZ is congruent to GR:

  - There is no information or markings in the given description that suggest RZ is congruent to GR. Congruence typically requires specific markings or information about the lengths or angles of the line segments, which are not provided in this description.

3. Points X, B, V, and N are coplanar:

  - This statement is true. Coplanar points are points that lie in the same plane. In the description, it's mentioned that all these points are in plane F, so they are indeed coplanar.

4. Plane F contains six points:

  - This statement is not true. Plane F contains only five named points (R, X, B, V, N) as indicated in the description.

5. Ray RH is part of line K:

  - This statement is not true. While the endpoint of ray RH is point R, which is on line K, ray RH heads off in a different direction than line K and does not extend along line K, so it's not considered part of line K.

6. VN is headed the same direction as RH:

  - This statement is not true. Segment VN and ray RH are not headed in the same direction. VN is a line segment in plane F, while RH is a ray that extends from point R but does not align with VN.

7. Line RG is the same as line K:

  - This statement is true. Both line RG and line K coincide and are represented by the same line in the figure, indicating that they are the same line.

So, statements 1, 3, and 7 are the accurate descriptions of the figure based on the information provided.

To know more about coplanar, refer here:

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Answer:

3x^2-2x+2-x^2-5x+5

2x^2-7x+7

Answer:

4x^2+3x-3

Step-by-step explanation:

(3x^2-2x+2)-(x^2+5x-5)

3x^2+1x^2=4x^2

2x-5x=3x

2-5=3

4x^2+3x-3

Answer:

3,500,000,000

Step-by-step explanation:

The total number of human genomic characters is 3.5 billion.

Expressing this quantity numerically without using a decimal we can write it as ;

3,500,000,000

Answer: There are 9 T.V. and 29 radio ads.

Step-by-step explanation:

Since we have given that

Total amount spend on TV and radio = $41,500

Total number of voters using the allocated funds = 148,000

Let the number of TV be 'x'.

Let the number of radio ads be 'y'.

Cost of each TV = $3000

Cost of each radio ads = $500

Number of voters see T.V. = 10,000

Number of voters use radio = 2000

So, According to question, it becomes,

[tex]3000x+500y=\$41500\implies\ 30x+5y=415\\\\10000x+2000y=148000\implies 10x+2y=148[/tex]

Using the graphing method, we get that

These two lines are intersect at (9,29).

Hence, there are 9 T.V. and 29 radio ads.

Answer:  9 TV ads and 29 radio ads will contact 148,000 voters using the allocated funds .

Step-by-step explanation:

Let x denotes the number of users of TV ads and y denotes the number of radio ads.

Then by considering the given information, we have the foolowing system of equation:-

[tex]\text{Number of voters}\ :10000x+2000y=148000----(1)\\\\\text{Total costs}\ :3000x+500y=41500------(2)[/tex]

Multiply 4 on both sides of equation (2) , we get

[tex]12000x+2000y=166000---------(3)[/tex]

Subtract (1) from (3) , we get

[tex]2000x=18000\\\\\Rightarrow\ x=\dfrac{18000}{2000}=9[/tex]

Put x= 9 , in (2), we get;

[tex]3000(9)+500y=41500\\\\\Rightarrow\ 27000+500y=41500\\\\\Rightarrow\ 500=14500\\\\\Rightarrow\ y=\dfrac{14500}{500}=29[/tex]

Hence, the number of TV ads will be 9 and the number of radio ads will will be 29.

Answer:

Cos x = 1 - [tex]\frac{x^2}{2!}[/tex] + [tex]\frac{x^4}{4!}[/tex] - [tex]\frac{x^6}{1!}[/tex] + ...

Step-by-step explanation:

We use Taylor series expansion to answer this question.

We have to find the expansion of cos x at x = 0

f(x) = cos x, f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, f''''(x) = cos x

Now we evaluate them at x = 0.

f(0) = 1, f'(0) = 0, f''(0) = -1, f'''(0) = 0, f''''(0) = 1

Now, by Taylor series expansion we have

f(x) = f(a) + f'(a)(x-a) + [tex]\frac{f''(a)(x-a)^2}{2!}[/tex] + [tex]\frac{f'''(a)(x-a)^3}{3!}[/tex] + [tex]\frac{f''''(a)(x-a)^4}{4!}[/tex] + ...

Putting a = 0 and all the values from above in the expansion, we get,

Cos x = 1 - [tex]\frac{x^2}{2!}[/tex] + [tex]\frac{x^4}{4!}[/tex] - [tex]\frac{x^6}{1!}[/tex] + ...

Final answer:

The expansion of cos x about the point x=0 is given by the Maclaurin series of cos x, which is 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...

Explanation:

The expansion of cos x about the point x=0 is given by the Maclaurin series of cos x. The Maclaurin series is a special case of the Taylor series, which is a way to approximate a function using a sum of terms.

The Maclaurin series of cos x is:

cos x = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...

This series can be derived by expanding the cosine function using its power series representation and evaluating it at x = 0.

Answer:

168 of these adults only watched TV last Friday night

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the adults that watched TV

-The set B represents the adults that hung out with friends.

-The set C represents the adults that ate pizza

-The set D represents the adults that did not do any of these three activities.

We have that:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

In which a is the number of adults that only watched TV, [tex]A \cap B[/tex] is the number of adults that both watched TV and hung out with friends, [tex]A \cap C[/tex] is the number of adults that both watched TV and ate pizza, and [tex]A \cap B \cap C[/tex] is the number of adults that did all these three activies.

By the same logic, we have:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

This diagram has the following subsets:

[tex]a,b,c,D,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)[/tex]

There were 525 adults suveyed. This means that:

[tex]a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 525[/tex].

Solution

We build the sets from what the problem states:

96 did not do any of these three activities:

[tex]D = 96[/tex]

25 watched TV, hung out with friends, and ate pizza:

[/tex]A \cap B \cap C = 25[/tex]

32 hung out with friends and ate pizza, but did not watch TV:

[tex]B \cap C = 32[/tex]

50 watched TV and hung out with friends, but did not eat pizza:

[tex]A \cap B = 50[/tex]

26 watched TV and ate pizza, but did not hang out with friends:

[tex]A \cap C = 26[/tex]

152 ate pizza:

[tex]C = 152[/tex]

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

[tex]152 = c + 26 + 32 + 25[/tex]

[tex]c = 69[/tex]

166 hung out with friends

[tex]B = 166[/tex]

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]166 = b + 32 + 50 + 25[/tex]

[tex]b = 59[/tex]

How may 18-24 year olds (of these three activities) only watched TV last Friday night?

We can find the value of a from the following equation:

[tex]a + 59 + 69 + 96 + 50 + 26 + 32 + 25 = 525[/tex]

[tex]a = 525 - 357[/tex]

[tex]a = 168[/tex]

168 of these adults only watched TV last Friday night

Answer:

The correct option is b.

Step-by-step explanation:

The given function is

[tex]f(x)=-5x+1[/tex]

We need to find the value of f(-2). It means we have to find the value of given function at x=-2.

Substitute x=-2 in the given function to find the value of f(-2).

[tex]f(-2)=-5(-2)+1[/tex]

On simplification we get

[tex]f(-2)=10+1[/tex]

[tex]f(-2)=11[/tex]

The value of f(-2) is 11. Therefore the correct option is b.

Answer:

It takes the frog 7 hours to get out of the well.

Step-by-step explanation:

We know that the well is 11 feet deep and the frog can climb 3 feet per hour.

Each time it climbs, it rests for an hour, and decreases its height by 1 foot.

So, if the frog reaches 3 feet in the first hour, then in the next hour it is 2 feet.

Lets calculate with the same pattern:

1st hour: 3 feet

2nd hour: [tex]3-1=2[/tex] feet

3rd hour: [tex]2+3=5[/tex] feet

4th hour: [tex]5-1=4[/tex] feet

5th hour: [tex]4 +3= 7[/tex] feet

6th hour: [tex]7-1=6[/tex] feet

5th hour: [tex]6+3=9[/tex] feet

6th hour: [tex]9-1=8[/tex] feet

7th hour:  [tex]8+3= 11[/tex] feet

Therefore, it takes the frog 7 hours to get out of the well.

Final answer:

To find out how long it takes for a frog to climb out of an 11-foot deep well, we calculate the net gain of height over time considering its climbing rate and slipping back. It will take the frog a total of 11 hours to escape the well.

Explanation:

The question involves a frog climbing out of a well and deals with a sequence of movements that include climbing and slipping back. Each hour, the frog climbs 3 feet but then slips back 1 foot during the rest hour.

To solve this, we perform a step-by-step calculation to determine the total time required for the frog to climb out of an 11-foot deep well. The frog makes a net gain of 2 feet for every 2 hours (3 feet up in the first hour and slips back 1 foot in the next hour).

However, on the final climb, the frog does not slip back since it will climb out of the well. Therefore, in the 11th hour, the frog climbs the remaining 1 foot and escapes the well.

So, the total time taken is 11 hours.

Answer:

11.78 hours.

Step-by-step explanation:

We are asked to find the time that an "H" cylinder will last having 1350 PSI of gas in it at a flow of 6 LPM.

[tex]\text{Duration of tank (in minutes)}=\frac{\text{Tank pressure (in PSI)}\times\text{Conversion factor}}{\text{Flow (LPM)}}[/tex]

[tex]\text{Duration of H tank (in minutes)}=\frac{1350\times 3.14}{6}[/tex]

[tex]\text{Duration of H tank (in minutes)}=\frac{4239}{6}[/tex]

[tex]\text{Duration of H tank (in minutes)}=706.5[/tex]

To convert the time in hours, we will divide 706.5 by 60 as i hour equals 60 minutes.

[tex]\text{Duration of H tank (in hours)}=\frac{706.5}{60}[/tex]

[tex]\text{Duration of H tank (in hours)}=11.775[/tex]

Therefore, it will take 11.78 hours to last the given cylinder.

Answer:

(a) 1:2

(b) 6 cups

(c) [tex]\dfrac{2}{3}[/tex]

Step-by-step explanation:

(a) Given,

amount of rice mixture contains= 1 cup

amount of water mixture contains= 2 cups

[tex]\textrm{So, the ratio of rice to water}\ =\ \dfrac{\textrm{amount of rice in mixturte}}{\textrm{amount of water in mixture}}[/tex]

                                                              [tex]=\ \dfrac{1}{2}[/tex]

So, the ratio of rice to water is 1:2.

(b) Amount of rice in mixture = 3 cups

[tex]\textrm{So, the ratio of rice to water}\ =\ \dfrac{\textrm{amount of rice in mixturte}}{\textrm{amount of water in mixture}}[/tex]

                      [tex]=>\ \dfrac{1}{2}\ =\ \dfrac{3}{\textrm{amount of water in mixture}}[/tex]

                       => amount of water in mixture = 3 x 2

                                                                          = 6 cups

(c) [tex]\textrm{amount of rice in mixture}\ =\dfrac{1}{3}[/tex]

 [tex]\textrm{So, the ratio of rice to water}\ =\ \dfrac{\textrm{amount of rice in mixturte}}{\textrm{amount of water in mixture}}[/tex]

   [tex]=>\ \dfrac{1}{2}\ =\ \dfrac{\dfrac{1}{3}}{\textrm{amount of water in mixture}}[/tex]

  [tex]=>\textrm{amount of water in mixture}\ =\ \dfrac{2}{3}[/tex]

So, the amount of water in the mixture will be [tex]\dfrac{2}{3}[/tex] cup.

Answer:

There are 4,148,350,734,528 ways

Step-by-step explanation:

We have

(a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators?

We want to choose 5 Demonstrators and 5 Repudiators. The number of ways to do this is [tex]{44} \choose {5}[/tex] and [tex]56 \choose 5[/tex] respectively. Therefore, the number of ways to select the committee is given by:

[tex]{{44}\choose {5}} \times {{56}\choose{5}}=\frac{44!}{39!5!}\times\frac{56!}{51!5!}=\frac{44!56!}{51!39!5!5!}=\frac{44\times43\times42\times41\times40\times56\times55\times54\times53\times52}{5!5!}=\\\\=\frac{44\times43\times42\times41\times8\times56\times11\times54\times53\times52}{4!4!}= \frac{11\times43\times42\times41\times2\times56\times11\times54\times53\times52}{3!3!}=\\\\\frac{11\times43\times14\times41\times2\times56\times11\times18\times53\times52}{2!2!}=[/tex]

[tex]11\times43\times14\times41\times28\times11\times18\times53\times52=4,148,350,734,528[/tex]

(b) Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?

[tex]44\times43\times56\times55=5,827,360.[/tex]

This is because there are (in the case of Demonstrators) 44 possibilities to choose an speaker and after choosing one, there would be 43 possibilities to choose a vice speaker. The same situation happens in the case of Repudiators.

[tex]5\times4\times5\times4=400[/tex].

Final answer:

To select a committee with an equal number of members from both parties, we use the combination formula to calculate the number of ways to choose 5 out of 44 Demonstrators and 5 out of 56 Repudiators, and then multiply these numbers together. For selecting speakers and vice speakers, we multiply the number of possible choices for each position within each party. The committee selection results in 4,149,395,102,528 ways, and selecting speakers and vice speakers yields 5,825,760 ways.

Explanation:

For part (a), if the committee requires an equal number of members from both parties, and we have 44 Demonstrators and 56 Repudiators, we need to select 5 members from each party to have a committee of 10 with an equal number of senators from both parties.

The number of ways to select 5 Demonstrators from 44 is given by the combination formula C(44, 5), which represents the number of ways to choose 5 members out of 44 without regard to order. Similarly, the number of ways to select 5 Repudiators from 56 is C(56, 5).

To find the total number of ways to form the committee, we multiply the two results:

The total number of ways is 1,086,008 * 3,819,416 = 4,149,395,102,528 ways.

For the selection of speakers and vice speakers, for each party, we can choose 1 speaker and 1 vice speaker. There are 44 Demonstrators, so there are 44 choices for the speaker and 43 choices for the vice speaker since the same person cannot hold both positions. Similarly, there are 56 Repudiators, so there are 56 choices for the speaker and 55 for the vice speaker.

The total number of ways to select the speakers and vice speakers for both parties is:

Final calculation: (44 * 43) * (56 * 55) = 1,892 * 3,080 = 5,825,760 ways.

Answer:

Take 5 ml of dye and add 3 ml of water

Thus,

The total volume of solution becomes = 8 mL

now,

This solution of 8 mL contains [tex]\frac{\textup{5}}{\textup{8}}[/tex] part of dye and [tex]\frac{\textup{3}}{\textup{8}}[/tex] part of water.

Next step is to take out 2 mL of solution

thus,

Volume of dye in 2 mL solution = [tex]\frac{\textup{5}}{\textup{8}}\times2\ mL[/tex]

or

Volume of dye in 2 mL solution = 1.25 mL

hence,

the 1.25 mL dye is measured.

Step-by-step explanation:

Given:

10-mL graduate calibrated in 1-mL units

dye solution to be measured = 1.25 mL

Now,

take 5 ml of dye and add 3 ml of water

Thus,

The total volume of solution becomes = 8 mL

now,

This solution of 8 mL contains [tex]\frac{\textup{5}}{\textup{8}}[/tex] part of dye and [tex]\frac{\textup{3}}{\textup{8}}[/tex] part of water.

Next step is to take out 2 mL of solution

thus,

Volume of dye in 2 mL solution = [tex]\frac{\textup{5}}{\textup{8}}\times2\ mL[/tex]

or

Volume of dye in 2 mL solution = 1.25 mL

hence,

the 1.25 mL dye is measured.

To measure 1.25 mL of a dye solution using a 10-mL graduate, fill it with water up to the 1 mL mark. Add the dye solution drop by drop until the meniscus reaches the 2.25 mL mark.

To measure 1.25 mL of a dye solution using a 10-mL graduate calibrated in 1-mL units and water as the diluent, you can follow these steps:

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Answer:

The mileage must the new model get so that the manufacturer meets the government requirement is 59 mpg.

Step-by-step explanation:

Consider the provided information.

Here it is given that the average mpg for the four models manufactured is 24 mpg.

The total mpg for the four models = 4 × 24 = 96

The manufacturer will add a model and the government standard is 31 mpg.

This can be written as:

[tex]\frac{96+x}{5}=31[/tex]

[tex]96+x=155[/tex]

[tex]x=155-96[/tex]

[tex]x=59[/tex]

Hence, the mileage must the new model get so that the manufacturer meets the government requirement is 59 mpg.

Final answer:

The new model car must achieve 59 mpg in order for the manufacturer to meet the government requirement of an average of 31 mpg when this model is added to the existing four models that average 24 mpg.

Explanation:

The student has asked what mileage a new model car must get so that the manufacturer meets the government requirement of an average of 31 miles per gallon (mpg). Currently, the average mpg for the manufacturer's four models is 24 mpg.

To calculate the required mpg for the fifth model, we use the formula for the average of a set of numbers, which is the sum of all numbers divided by the number of items. In this case, if x is the mpg the new model needs to achieve, the equation is:

(4 ×24 + x) / 5 = 31

Which simplifies to:

96 + x = 155

Therefore:

x = 155 - 96

x = 59 mpg

The new model will need to have an efficiency of 59 mpg for the manufacturer to meet the Corporate Average Fleet Efficiency (CAFE) standard of 31 mpg.

Answer:

True

Step-by-step explanation:

Yes, In Descriptive Statistics we are most often interested in the summary of data in words. It is done by describing the features of the data. Mostly we explain the five-number summary of the data, central tendency of data, dispersion, skewness, etc. Overall we summarize the quantitative data in words.

People who didn't travel by any mode = 45

In the question,

Total number of people included in survey = 250

People who traveled by plane but not by train = 70

i.e.

a + e = 70

People who traveled by train but not by plane = 70

i.e.

c + d = 70

People who traveled by Bus but not by Plane or Train = 20

i.e.

f = 20 ..........(1)

People who traveled by bus and plane both = 45

i.e.

e + g = 45

People who traveled by all three = 20

i.e.

g = 20

People who traveled by Plane or Train = 185

i.e.

a + b + c + d + e + g = 185 ........(2)

So,

e = 45 - g = 45 - 20 = 25

e = 25

Now, on putting in eqn. (2) we get,

a + b + c + d + 25 + 20 = 185

a + b + c + d = 140 .......(3)

Now,

We need to find out,

Number of people travelling with any of these three is,

a + b + c + d + e + f + g

So,

On putting from eqn. (3) and (1), we get,

a + b + c + d + e + f + g = 140 + 25 + 20 + 20 = 205

So,

Number of people who didn't travel by any mode =Total people - Number of people travelling by any three

People who didn't travel by any mode = 250 - 205 = 45

Final answer:

By analyzing the given data and using set theory, we determined that 90 adults did not travel by plane, train, or bus in the last year.

Explanation:

To solve this question, we need to find out how many adults did not travel by plane, train, or bus. We are given several subsets of people who use various combinations of these modes of transportation, and we can use a principle in set theory to determine the answer.

We know that 185 adults traveled by plane or train. This number includes those who traveled by both modes. There is an overlap of people who used all three modes, which is 20. So, to find the sole plane and train travelers, we subtract the people who used all three modes from those who traveled by plane but not by train and vice versa.

We calculate the number of plane-only and train-only travelers: 70 (plane only) + 70 (train only) - 20 (all three) = 120.

Since 185 traveled by either plane or train, the number that traveled by either without the bus is 185 - 20 (all three) = 165. The number that traveled by plane or train only is 165 - 45 (bus and plane) = 120.

Adding those who traveled by bus but not by plane or train (20) to those who traveled by all three (20) gives us 40. Therefore, 120 (plane or train only) + 40 = 160. To find out the number of adults who did not travel by any of these three modes, we subtract 160 from the total number of surveyed adults (250).

Finally, the number of adults who did not travel by any mode is 250 - 160 = 90.

Therefore, 90 adults did not travel by plane, train, or bus.

Answer:

Non representative

Step-by-step explanation:

A representative sample would be a subset of a population that accurately describes some characteristic from a larger group of people.

A sample of 40 patients from a local hospital is not a big enough sample to estimate the mean cholesterol level of U.S citizens (citizens from an entire country).

Therefore, this sample would be non representative.

Answer:

a. [tex]y=\frac{2}{3}x^7+cx^4[/tex]

b. [tex]y=2e^{7x}-ce^{5x}[/tex]

c. [tex]y=e^{-2x}arctan(e^{2x})+ce^{-2x}[/tex]

d. [tex]i=e^{-2t}\left(\frac{8\left(3e^{2t}\sin \left(3t\right)+2e^{2t}\cos \left(3t\right)\right)}{13}+C\right)[/tex]

Step-by-step explanation:

a) xy' -4y = 2 x^6

[tex]xy'-4y=2x^6\\y'-\frac{4}{x}y=2x^5\\p(x)=\frac{-4}{x}\\Q(x)=2x^5\\\mu(x)=\int P(x)dx=\int \frac{-4}{x}dx=Ln|x|^{-4}\\y=e^{-\mu(x)}\int {e^{\mu(x)}Q(x)dx}\\y=x^4 \int {x^{-4}2x^6}dx\\y=\frac{2}{3}x^7+cx^4[/tex]

b) y' - 5y = 4e^7x

[tex]y'-5y=4e^{7x}\\p(x)=-5\\Q(x)=4e^{7x}\\\mu(x)=\int P(x)dx=\int-5dx=-5x\\y=e^{-\mu(x)}\int {e^{\mu(x)}Q(x)dx}\\y=e^{5x}\int {e^{-5x}4e^{7x}}dx\\y=2e^{7x}-ce^{5x}[/tex]

c) dy/dx + 2y = 2/(1+e^4x)

[tex]\frac{dy}{dx}+2y=\frac{2}{1+e^{4x}}\\p(x)=2\\Q(x)=\frac{2}{1+e^{4x}}\\\mu(x)=\int P(x)dx=\int 2dx=2x\\y=e^{-\mu(x)}\int {e^{\mu(x)}Q(x)dx}\\y=e^{-2x}\int {e^{2x}\frac{2}{1+e^{4x}}}dx\\y=e^{-2x}arctan(e^{2x})+ce^{-2x}[/tex]

d) 1/2 di/dt + i = 4cos(3t)

[tex]\frac{1}{2}\frac{di}{dt}+i=4cos(3t)\\\frac{di}{dt}+2i=8cos(3t)\\p(t)=2\\Q(t)=8cos(3t)\\\mu(t)=\int P(t)dt=\int 2dt=2t\\i=e^{-\mu(t)}\int {e^{\mu(t)}Q(t)dt}\\i=e^{-2t}\int {e^{2t}8cos(3t}dt\\i=e^{-2t}\left(\frac{8\left(3e^{2t}\sin \left(3t\right)+2e^{2t}\cos \left(3t\right)\right)}{13}+C\right)[/tex]

Answer with Step-by-step explanation:

Part 1)

we know that

[tex]e^{i\theta }=cos(\theta )+isin(\theta )[/tex]

thus [tex]-i=e^{\frac{-i\times (4n-1)\pi }{2}}[/tex]

thus [tex](-i)^i=(e^{\frac{-i\times (4n-1)\pi }{2}})^i\\\\(-i)^i=e^{\frac{-i^2\times (4n-1)\pi }{2}}=e^{\frac{(4n-1)\pi }{2}}\\\\\therefore (-i)^i=e^{\frac{(4n-1)\pi }{2}}[/tex] where 'n' is any integer

Part 2)

We have [tex]-1=e^{(2n+1)\pi }\\\\\therefore (-1)^{i}=(e^{i(2n+1)\pi })^{i}\\\\(-1)^i=(e^{i^2(2n+1)\pi })\\\\(-1)^i=e^{-(2n+1)\pi }[/tex] where 'n' is any integer

Answer:

  (a) yes, (-2√2, 1) is on the graph

  (b) f(2) = 4/5, the point is (2, 4/5)

  (c) (-2√2, 1), and (2√2, 1)

  (d) all real numbers

  (e) (0, 0)

  (f) (0, 0)

Step-by-step explanation:

You want various points on the graph of the function ...

  [tex]\displaystyle f(x)=\frac{16x^2}{x^4+64}[/tex]

Yes, this point is on the graph. The value of f(x) can be found easily by realizing -2√2 = -√8, so ...

and the function value is ...

  [tex]f(-2\sqrt{2})=\dfrac{16\cdot8}{64+64}=\dfrac{128}{128}=1[/tex]

Substituting 2 for x, we have ...

  [tex]f(2) = \dfrac{16\cdot 2^2}{2^4+64}=\dfrac{64}{16+64}=\dfrac{64}{80}\\\\\boxed{f(2)=\dfrac{4}{5}}[/tex]

The point on the graph is (2, 4/5).

The answer to part (a) tells you that one of the points where f(x) = 1 is ...

  (-2√2, 1)

Since the sign of x is irrelevant, another point where x=1 is ...

  (2√2, 1)

There are no values of x that make the denominator of this rational function zero, so its domain is all real numbers.

The only x-value where f(x) = 0 is x = 0.

The x-intercept is (0, 0).

The function crosses the y-axis at the origin.

The y-intercept is (0, 0).

Answer:

x = -1

Step-by-step explanation:

5x - 6 = 3x - 8

-3x        -3x

2x - 6 = -8

+6         +6

2x = -2

----   ----

2      2

x = -1

Hey!

-------------------------------------------------

Steps To Solve:

5x - 6 = 3x - 8     ​

~Subtract 3x to both sides

5x - 6 - 3x = 3x - 8 - 3x

~Simplify

2x - 6 = -8

~Add 6 to both sides

2x - 6 + 6 = -8 + 6

~Simplify

2x = -2

~Divide 2 to both sides

2x/2 = -2/2

~Simplify

x = -1

-------------------------------------------------

Answer:

[tex]\large\boxed{x~=~-1}[/tex]

-------------------------------------------------

Hope This Helped! Good Luck!

Answer:

Square of a rational number is a rational number.

Step-by-step explanation:

Let m be a rational number. Thus, m can be written in the form of  fraction [tex]\frac{x}{y}[/tex], where x and y are integers and [tex]y \neq 0[/tex].

The square of m = [tex]m\times m = m^2[/tex]

[tex]m^2 = \frac{x}{y} \times\frac{x}{y} = \frac{x^2}{y^2}[/tex]

It is clearly seen, that [tex]m^2[/tex], can be easily written in the form of fraction and the denominator is not equal to zero.

Hence, [tex]m^2[/tex] is a rational number.

This can also be understood with the help of the fact that rational numbers are closed under multiplication that is product of a rational number is also a rational number.

Answer:

462 000mg

Step-by-step explanation:

1gram = 1000milligrams

Hence...462 grams...,

; 462 × 1000 = 462 000mg

462 grams is equal to 462,000 milligrams .

To convert grams ( g ) to milligrams ( mg ), you need to use the following conversion factor:

1 gram ( g ) = 1000 milligrams ( mg )

milligram (mg) is equal to 1/1000 grams (g).

1 mg = (1/1000) g = 0.001 g

The mass m in grams (g) is equal to the mass m in milligrams (mg) divided by 1000:

m(g) = m(mg) / 1000

This means that there are 1000 milligrams in 1 gram. Now, let's use this conversion factor to calculate 462 grams in milligrams:

462 grams * ( 1000 mg / 1 g ) = 462,000 milligrams

So, 462 grams is equal to 462,000 milligrams.

To know more about milligrams click here :

https://brainly.com/question/29827935

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Answer:

The improper integral converges and [tex]\int_0^{-\infty} e^{7x}dx = -\frac{1}{7}[/tex].

Step-by-step explanation:

First, I assume that the integral in question is

[tex]\int_0^{-\infty} e^{7x}dx[/tex].

Now, the integral is improper because, at least, one of the limits is [tex]\pm\infty[/tex]. We need to recall that an improper integral

[tex]\int_0^{-\infty} f(x)dx[/tex]

converges, by definition, if the following limit exist:

[tex]\lim_{A\rightarrow -\infty} \int_0^A f(x)dx = \int_0^{-\infty} f(x)dx[/tex].

In this particular case we need to study the limit

[tex]\lim_{A\rightarrow -\infty} \int_0^A e^{7x}dx[/tex].

In order to complete this task we calculate the integral [tex]\int_0^A e^{7x}dx[/tex]. Then,

[tex]\int_0^A e^{7x}dx = \frac{e^{7x}}{7}\Big|_0^A = \frac{e^{7A}}{7} - \frac{1}{7}[/tex].

Substituting the above expression into the limit we have

[tex]\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7} - \frac{1}{7} = - \frac{1}{7}[/tex]

because

[tex]\lim_{A\rightarrow -\infty} \frac{e^{7A}}{7}=0[/tex].

Answer:[tex]x-2y-2=0[/tex]

Step-by-step explanation:

Given :

[tex]f(x) = x^2 - 4 \\ g(x) = - 3x^2 + 2x + 8[/tex]

Point of intersection :

[tex]f(x)=g(x)\\x^2-4=-3x^2+2x+8\\4x^2-2x-12=0\\2x^2-x-6=0\\2x^2-4x+3x-6=0\\2x(x-2)+3(x-2)=0\\(x-2)(2x+3)=0\\x=2\,,\,\frac{-3}{2}[/tex]

[tex]x=2\,;f(2)=2^2-4=0\\x=\frac{-3}{2}\,; f\left ( \frac{3}{2} \right )=\left ( \frac{3}{2} \right )^2-4=\frac{-7}{4}=-1.75[/tex]

So, we have points [tex]\left ( 2,0 \right )\,,\,\left ( -1.5,-1.75\ \right )[/tex]

Equation of line passing through two points [tex]\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )[/tex] is given by [tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}\left ( x-x_1 \right )[/tex]

Let [tex]\left ( x_1,y_1 \right )=\left ( 2,0 \right )\,,\,\left ( x_2,y_2 \right )=\left ( -1.5,-1.75\ \right )[/tex]

So, equation is as follows :

[tex]y-0=\frac{-1.75-0}{-1.5-2}\left ( x-2 \right )\\y=\frac{-1.75}{-3.5}\left ( x-2 \right )\\y=\frac{1}{2}(x-2)\\2y=x-2\\x-2y-2=0[/tex]

Answer:

The general solution of the differential equation y' + 3x^2 y = 0 is:

[tex]y=e^{-x^3+C}[/tex]

Step-by-step explanation:

This equation its a Separable First Order Differential Equation, this means that you can express the equation in the following way:

[tex]\frac{dy}{dx} = f_1(x)*f_2(x)[/tex], notice that the notation for y' is changed to [tex]\frac{dy}{dx}[/tex]

Then you can separate the equation and put the x part of the equation on one side and the y part on the other, like this:

[tex]\frac{1}{f_2(x)}dy=f_1(x)dx[/tex]

The Next step is to integrate both sides of the equation separately and then simplify the equation.

For the differential equation in question y' + 3x^2 y = 0 the process is:

Step 1: Separate the x part and the y part

[tex]\frac{1}{y}dy=3x^2}dx[/tex]

Step 2: Integrate both sides

[tex]\int\frac{1}{y}dy=\int 3x^2}dx[/tex]

Step 3: Solve the integrals

[tex]Ln(y)+C=-x^3+C[/tex]

Simplify the equation:

[tex]Ln(y)=-x^3+C[/tex]

To solve the Logarithmic expression you have to use the exponential e

[tex]e^{Ln(y)}=e^{-x^3+C}[/tex]

Then the solution is:

[tex]y= e^{-x^3+C}[/tex]

Answer:

0.024

Step-by-step explanation:

Given,

Red marbles = 4,

Green marbles = 7,

Total marbles = 4 +7 = 11,

Ways of choosing 3 marbles =[tex]^{11}C_3[/tex]

Ways of choosing 3 red marble = [tex]^4C_3[/tex]

Hence, the probability of 3 red marble = [tex]\frac{^4C_3}{^{11}C_3}[/tex]

[tex]=\frac{\frac{4!}{3!1!}}{\frac{11!}{3!8!}}[/tex]

[tex]=\frac{4}{165}[/tex]

≈ 0.024

Answer:

1. ∀ y ∈ Z such that ∃ x ∈ Z, ¬R (x + y)

2. ∃ x ∈ Z, ∀ y ∈ Z such that ¬R(x + y)

Step-by-step explanation:

If we negate a quantified statement, first we negate all the quantifiers in the statement from left to right, ( keeping the same order ) then we negative the statement,

Here, the given statement,

1. ∃y ∈Z such that ∀x ∈Z, R (x + y)

By the above definition,

Negation of this statement is ∀ y ∈ Z such that ∃ x ∈ Z, ¬R (x + y),

2. Similarly,

The negation of statement ∀x ∈Z, ∃y∈Z such that R(x + y),

∃ x ∈ Z, ∀ y ∈ Z such that ¬R(x + y)

Final answer:

The negation of a mathematical statement involving existential and universal quantifiers involves switching the quantifiers and negating the inner statement. An example of a statement and its negation is 'All swans are white' and its negation 'Not all swans are white.' The law of noncontradiction implies the law of the excluded middle by rejecting the possibility of a middle ground.

Explanation:

When negating mathematical statements involving existential and universal quantifiers, the negation of an existential quantifier ∃ (there exists) becomes a universal quantifier ∀ (for all), and vice versa. Additionally, the statement inside the quantifiers gets negated. Here is the negation of each statement:

1. The original statement is ∃y ∈Z such that ∀x ∈Z, R(x + y). The negation becomes ∀y ∈Z, ∃x ∈Z such that ¬R(x + y).

2. The original statement is ∀x ∈Z, ∃y∈Z such that R(x + y). The negation becomes ∃x ∈Z, ∀y∈Z such that ¬R(x + y).

For an example of a statement and its negation: "All swans are white" is a universal affirmative statement, which can be negated to "Not all swans are white," or equivalently "There exists at least one swan that is not white."

The law of noncontradiction states that a statement and its negation cannot both be true simultaneously. The law of the excluded middle asserts that for any proposition, either the proposition is true, or its negation is true. The law of noncontradiction logically implies the law of the excluded middle, as if a statement and its negation can't both be true, then one of them must be true, rejecting the possibility of a middle ground.