From the choices provided, select all the questions that represent statistical questions,
A What was the temperature at 5:00p.m. today in Boston?
B How many pieces of fruit were used to make Shawn's smoothie today?
How many states are there?
D What types of food do students in your school bring for lunch each day?
E How many books did each student check out at the library last year?

After solving the problems the answers obtained are MN ≅ QP  and   MQ ≅ NP, MP ≅ MP, Δ MQP ≅ Δ PNM, ∠ N ≅ ∠ Q.

The phrase "parallelogram" is a translation of the Greek phrase "parallelogrammon," which means "bound by the contract by parallel lines." As a consequence, a quadrilateral that is bordered by parallel lines is called a parallelogram. It has parallel and equal opposite sides on all sides.

1. According to the parallelogram's feature that its opposed sides are congruent, the opposite sides of the given parallelogram MNPQ,

So, MN ≅ QP  and  MQ ≅ NP.

2. According to the reflexive feature of congruence, a line or other geometric figure is both a reflection of and consistent with itself. Therefore, in the example of the parallelogram MNPQ,

So,

MP ≅ MP

3. The SSS congruence postulate, which stands for Side-Side-Side congruence postulate, argues that when three adjacent sides of two triangles are congruent, the two triangles are congruent. As the sides MNQ, MQ, NP, and MP are in the given parallelogram MNPQ

Hence,

Δ MQP ≅ Δ PNM

4. Part 4 demonstrated that MQP and PNM are congruent, hence in accordance with a CPCTC property, 

∠ N ≅ ∠ Q

5. The acronym CPCTC stands for congruent portions of congruent triangles.

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Answer:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex] and [tex]m=\frac{-5-3}{12+4}[/tex]

Step-by-step explanation:

The correct formula for finding the slope is:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

This formula represents the change in y over a corresponding change in x.  This is a mathematical representation of "rise" (y-axis movement) over "run" (x-axis movement).

For the second question, let

[tex](x_1,y_1)=(-4,3)\\(x_2,y_2)=(12,-5)[/tex]

So,

[tex]m=\frac{(-5)-(3)}{(12)-(-4)}\\m=\frac{-5-3}{12+4}[/tex]

Using the points A(1,5), B(1,-4), C(-3,5), D(-3,-4) the following are matched:

Distance between points B and C = √97.

Distance between points A and B = 9.

Midpoint between points  B and C = (-1, 1/2).

Midpoint between C and D = (-3, 1/2).

Slope of the line through B and D = 0.

The distance between two points of a line is given as:

Distance = [tex]\sqrt{c - a)^2 + (d - b)^2}[/tex]

We have,

A(1,5), B(1,-4), C(-3,5) and D(-3,-4).

The distance between A and B.

= [tex]\sqrt{(1 - 1)^2 + (-4 - 5)^2}[/tex]

= √81

= 9

The distance between B and C.

= [tex]\sqrt{(-3 - 1)^2 + (5 + 4)^2}[/tex]

= √(16 + 81)

= √97

The midpoint (x, y) between B and C.

x = (1 - 3)/2 = -1

y = (-4 + 5) / 2 = 1/2

The midpoint (x, y) between C and D.

x = (-3 -3) / 2 = -6/2 = -3

y = (5 - 4) / 2 = 1/2

The slope of the line through points B and D.

= (-4 + 4) / (-3 - 1)

= 0/ -4

= 0

Thus,

The distance between A and B is 9.

The distance between B and C is √97.

The midpoint between B and C is (-1, 1/2).

The midpoint between C and D is (-3, 1/2).

The slope of the line through B and D is 0.

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Answer:

a) The function has a minimum value.

b) The minimum value of the function is -27.

c) The minimum value occurs at x=3.

Step-by-step explanation:

The solution is

a) Option A.

The equation to determine the messages sent by Bria is t/4 = 20

b) Option C.

The student is 1.25 feet above sea level , and the absolute value is 1.25 ft

c) Option C.

The values are opposites of each other , and they are both 5/2 units from zero

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

a)

Let the number of texts sent by Bria be = t

The number of texts sent by April = 20

April sent one fourth of the texts sent by Bria

So ,

The number of texts sent by Bria t = 4 x number of texts sent by April

Substituting the values in the equation , we get

The number of texts sent by Bria t = 4 x 20

Divide by 4 on both sides of the equation , we get

t/4 = 20   be equation (1)

And , t = 80 texts

Therefore , the number of texts sent by Bria is given by the equation

t/4 = 20

b)

Let the sea level be represented as = 0

The number 1.25 is above the sea level of 0

The absolute value of 1.25 is = 1.25

Therefore , the expression will be the student is 1.25 ft above the sea level and the absolute value is 1.25 ft

c)

Let the first number be = -5/2

Let the second number be = 5/2

Now , the difference of the numbers from the point zero = 5/2 units

Because the distances are always positive

The numbers -5/2 and 5/2 are opposite numbers

Hence , the expression will be the numbers are opposites of each other and they are 5/2 units from 0

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Applying distributive property of multiplication the solution of the multiplication problem is 5 11/56

The distributive property states that multiplying the total of two or more addends by a number produces the same outcome as multiplying each addend separately by the number and adding the products together.

This applied mathematically in the problem

4/7 x 4 5/8

dividing by 4 for the multiplicand and multiplier

4/7 ÷ 4 = 1/7

4 5/8 ÷ 4 = 37/32

4/7 x 4 5/8

= 4 * (1/7 + 37/32) - distributive property applied

= 4 * (291/224)

= 291 / 56

= 5 11/56 as mixed number

using distributive property the solution of the multiplication is 5 11/56

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Answer:This statement is not necessarily true

Step-by-step explanation:In general, the rate at which thermal energy is transferred depends on several factors, including the temperature difference between the material and its surroundings, the type of material, and its thermal conductivity. Some materials, such as air and glass, are poor conductors of heat and are therefore considered insulators. However, other materials, such as metal, are good conductors of heat and are not considered insulators.

Table

Top right: [tex]5+194=199[/tex]

Middle right: [tex]217+13=230[/tex]

Bottom left: [tex]5+217=222[/tex]

Bottom middle: [tex]194+13=207[/tex]

Bottom right: [tex]222+207=429[/tex]

Question 1

[tex]\frac{207}{429} \approx \boxed{0.4825}[/tex]

Question 2

[tex]\frac{13}{429} \approx \boxed{0.0303}[/tex]

Question 3

[tex]\frac{194+217}{429} \approx \boxed{0.9580}[/tex]

The slope of the line is 7/9

It is important to note that the equation of a line is represented as;

y = mx + c

Where;

The formula for calculating the slope of a line is expressed as;

Slope, m = y₂ - y₁/x₂ - x₁

Now, let's substitute the values into the formula from the points given we have;

Slope, m =1 -(-6)/ -3 - (-6)

expand the bracket

Slope, m = 1 + 6/ 3 + 6

add the values

Slope, m = 7/9

Hence, the value is 7/9

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The management assertions for the given parts is explains below.

What is assertion?

Management assertions are statements made about specific elements of a business by members of management. The idea is mainly applied to the auditing of financial statements of a company, where the auditors rely on various business assertions.

Consider, the following.

(a) Recorded sales transactions have been recorded

          Category of management.              Classes of transactions                

           Name of assertion.          .               Occurance

(b)  There are no lines are other restrictions in accounts receivable

          Category of management.             Account balances                

          Name of assertion.                          Rights and obligations.          

(c) All sales transactions have been recorded

           Category of management.             Classes of transactions                

            Name of assertion.                        Completeness

(d) Receivables are appropriately classified as to trade and other receivables in the financial  statements and are clearly described.

       Category of management.              Presentation and disclosure

       Name of assertion.                        Classification and understandabilty

(e) sales transactions have been recorded in proper period

         Category of management.                   Classes of transactions.            

          Name of assertion.                               Cut off.  

(f) Account receivables recorded at correct amounts

         Category of management.                  Account balances      

         Name of assertion.                              Valuation and allocation

(g) sales transactions recorded at appropriate accounts

         Category of management.                 Class of transactions                

          Name of assertion.                            Classification

(h) All required disclosure about sales and receivables are at recorded amounts

     Category of management.                 Presentation and disclosure                  

     Name of assertion.                             Completeness

(k) Sales transactions have been recorded at correct amounts

        Category of management.                   Account balances                  

        Name of assertion.                         Completeness

(L) Recorded accounts receivable exist

        Category of management.                Account balances.                  

        Name of assertion.                            Existence

(m) Disclosure related to sales and receivable relate to entity

         Category of management.               Presentation and disclosure                

         Name of assertion.                           Occurance and obligation

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Answer:

  see the attachment for a graph

Step-by-step explanation:

You seem to want a graph of the inequalities ...

The equations of the boundary lines are given by replacing the inequality symbol with an equal sign:

These are graphed in the usual way. It is often convenient to find the y-intercept, then use the slope to locate other points on the line.

y = -3/2x -2

  The y-intercept is (0, -2). The line has a rise/run of -3/2, so goes down 3 units for each 2 to the right. Another point on the line is (2, -5).

y = 1/2x +6

  The y-intercept is (0, 6). The line has a rise/run of 1/2, so goes up 1 unit for each 2 to the right. Another point on the line is (2, 7).

When the "or equal to" symbol (≤ or ≥) is used, the boundary line is solid. When the "or equal to" case is not part of the solution, the boundary line is dashed.

The line with negative slope through (0, -2) is solid; the line with positive slope through (0, 6) is dashed.

All you need to determine shading is a variable with a positive coefficient, and the inequality symbol.

  y ≤ . . . . . tells you shading is below the solid line

  y > . . . . . tells you shading is above the dashed line

We notice x has a positive coefficient in the second inequality, so we could determine shading from ...

  > x . . . . . tells you shading is left of the line (where x values are less than those on the line)

Of course, you can rearrange the inequality so a variable of interest has a positive coefficient. For example, we could add 3/2x to the first inequality to get ...

  3/2x + y ≤ -2

Then, looking at the x-variable, we see ...

  x ≤ . . . . . tells you shading is to the left of the line

Answer:

See attachment.

Step-by-step explanation:

When graphing inequalities:

Treat the inequalities as equations (swap the inequality sign for an equals sign) to find two points on the line to help draw the lines.

Inequality 1

[tex]y \leq -\dfrac{3}{2}x-2[/tex]

[tex]\begin{aligned}x=0 \implies y&=-\dfrac{3}{2}(0)-2\\y&=-2\end{aligned}[/tex]

[tex]\begin{aligned}x=2 \implies y&=-\dfrac{3}{2}(2)-2\\y&=-5\end{aligned}[/tex]

Plot the points (0, -2) and (2, -5).

Draw a solid straight line through the points.

Shade under the line.

Inequality 2

[tex]y > \dfrac{1}{2}x+6[/tex]

[tex]\begin{aligned}x=0 \implies y&=\dfrac{1}{2}(0)+6\\y&=6\end{aligned}[/tex]

[tex]\begin{aligned}x=4 \implies y&=\dfrac{1}{2}(4)+6\\y&=8\end{aligned}[/tex]

Plot the points (0, 6) and (4, 8).

Draw a dashed straight line through the points.

Shade above the line.

The solution to the two inequalities is the overlap of the shaded parts.

Answer:

y = -x - 4

Step-by-step explanation:

m = y2-y1 / x2-x1

(-4, 0) and (1, -5)

m = -5+0 / 1+4

m = -5 / 5

m = -1

y = -x + b

0 = -(-4) + b

0 = 4 + b

b = -4

y = -x - 4

Step-by-step explanation:

4/x + 4 = x - 2/x

multiply everything by x

4 + 4x = x² - 2

x² - 4x - 6 = 0

the general solution to such a quadratic equation

ax² + bx + c = 0

is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = (4 ± sqrt((-4)² - 4×1×-6))/(2×1) =

= (4 ± sqrt(16 + 24))/2 = (4 ± sqrt(40))/2 =

= (4 ± sqrt(4×10))/2 = (4 ± 2×sqrt(10))/2 =

= 2 ± sqrt(10)

x1 = 2 + sqrt(10) = 5.16227766...

x2 = 2 - sqrt(10) = -1.16227766...

Answer:

x=5

Step-by-step explanation:

(Not sure how I was supposed to do the model..so I ended up doing it this way..)

3x-5=10

Add 5 on both sides and then you get

3x=15

divide both sides by 3 and then you get

x=5

(Hopefully you can see my model even if it's just a little...)
For the model I drew 3 rectangles and put an x inside each one of them. for the -5, I drew 5 squares with a minus symbol/sign inside. And then I put an equal sign beside and I drew 10 squares with nothing inside but you can draw positive signs in it to show that it's positive..

So basically mines was

3 rectangles with x inside -5 squares with - inside = 10 squares

Answer:

Initial balance 500, compounds 4 times at an interest rate of 15%

Step-by-step explanation:

Initial balance 500, compounds 4 times at an interest rate of 15%

The students corresponding to those 200 numbers would beasked to participate in the survey.

What is corresponding ?

Having or participating in the same relationship (such as kind, degree, position, correspondence, or function) especially with regard to the same or like wholes (such as geometric figures or sets) corresponding parts of similar triangles. : related, accompanying.

Have given,

a) The administrators could number an alphabetical list of students from 1 to 2,500.Then use a random number generator from a calculator or computer to generate 200 unique random numbers from 1 to2,500.(if you don’t say “unique”, then say “ignoring repeats”)The students corresponding to those 200 numbers would beasked to participate in the survey.

b) One advantage is that stratified random sampling guarantees that each of the school-level strata will have some representation, because it is possible that a simple random sample would miss one or more of the strata completely.(NOTE: It is NOT sufficient to say that a stratified sample “makes it MORELIKELY that all school-levels are represented in the sample”… the student must state that stratifying by school level GUARANTEES or ENSURES that all school levels are represented in the sample.)

The students corresponding to those 200 numbers would beasked to participate in the survey.

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Answer:

1. A(t) = 65t + 700

2.B(t) = 40t + 900

3. High School A and High School B will both have 1,220 students in the 8th year

Step-by-step explanation:

1. The equation for the number of students in High School A represents a linear function.

⭐ What is a linear function?

Let's write the equation for High School A in the [tex]y = mx+b[/tex] format, known as slope-intercept form:

∴ High School A: [tex]y = 65x + 700[/tex]

2. Let's write the equation for High School B in the [tex]y = mx + b[/tex] format, known as slope-intercept form:

∴ High School B: [tex]y = 40x + 900[/tex]

3. To find at what year High School A and High School B will have the same number of students, we need to solve a system of linear equations.

⭐What is a system of linear equations?

For this problem, let's set both linear equations equal to each other to see at what point will the high school populations be the same.

[tex]A(t) = B(t)[/tex]

[tex]65t + 700 = 40t + 900[/tex]

[tex]25t = 200[/tex]

[tex]t = 8[/tex]

Now we know that in the 8th year, High School A will have the same population as High School B.

We need to find what the population will be in year 8.

Thus, substitute the value of t into one of the functions and solve.

I am choosing to substitute t into A(t), but you can also do B(t).

[tex]A(8) = 65(8) + 700[/tex]

[tex]A(8) = 520 + 700[/tex]

[tex]A(8) = 1,220[/tex]

⚠️!!! CAUTION !!! ⚠️

Some people may stop at this point and write that in the 8th year, both high schools will have a population of 1,220 students.

However, you should also substitute 8 into the other function you didn't substitute it into to make sure that 8 is correct.

[tex]B(8) = 40(8) + 900[/tex]

[tex]B(8) = 320 + 900[/tex]

[tex]B(8) = 1,220[/tex]

∴ In the 8th year, both high schools will have a population of 1,220 students.

Answer:

Standard form: [tex]y^3 -3y^2 + 21y + 4[/tex]

Degree: [tex]3[/tex]

Leading term: [tex]y^3[/tex]

Step-by-step explanation:

Standard form:

⭐  Standard form is a way you can order the terms of an equation so that the exponents of each term decrease in number when reading the equation from left to right.

Degree:

⭐ The degree of an equation is the highest exponent of the equation. You should always write equations in standard form in order to find the degree.

Leading term:

⭐ The leading term is the term that has the highest exponent. In standard form, it is the first term in the equation.

Answer:

Prove that for any x0∈X and any r>0, the open ball Br(xo) is open. My attempt: Let y∈Br(x0). By definition, d(y,x0)<r

Step-by-step explanation:

Prove that for any x0∈X and any r>0, the open ball Br(xo) is open. My attempt: Let y∈Br(x0). By definition, d(y,x0)<r

The probability that Bela and Jenn pick the same part is; 0.006944

In maps, it is pertinent to note that parallels are the lines that run from east to west and never intersect with each other while meridians run from north to south and intersect at the north and south poles.

Now, since bella's globe has 12 parallels and 12 great meridians, then it means that the total number of parts in this globe will be 12 * 12 = 144 parts.

Thus, probability that they both pick the same part is;

P(same part picked) = (12/144) * (12/144)

= 0.006944

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Answer:

Step-by-step explanation:

to find mean you add all numbers and divide by the amount of numbers, 10+4+8+15+17+6+19+16 is 95, 95 divided by 8 is 11.875

Answer:

f(g(-1)) = -8

Step-by-step explanation:

⭐ The two equations we are given are:

This problem is an example of composite functions.

⭐What are composite functions?

First, we have to compute the value for the 2nd function you see. The 2nd function is inside f(x), and it is g(-1).

Essentially, we have to find the corresponding y-value for when x = -1 in g(x)

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

[tex]g(-1) = 3[/tex]

Next, take this value and substitute it as the x-value for f(x).

[tex]f(3) = 1-x^2[/tex]

Now we have to find the y-value for f(x) for when x = 3.

[tex]f(3) = 1 -3^2[/tex]

⚠️⚠️⚠️!!! CAUTION !!! ⚠️⚠️⚠️

Some people may make the mistake of computing f(3) like this:

[tex]f(3) = 1 - 3^2[/tex]

[tex]f(3) = 1+9\\f(3) = 10[/tex]

This is wrong because only 3 is being squared, not -3. Make sure to read the equations you are given carefully to avoid mistakes like this.

[tex]f(3) = 1-3^2\\f(3) = 1-9\\f(3) = -8[/tex]

∴ f(g(-1)) = -8

Answer:

  x ≤ 40 . . . miles per hour

Step-by-step explanation:

You want to know the values of x for which d(x) = 0.05x² +x is at most 120.

  d(x) ≤ 120 . . . . . . . . the required limit

  0.05x² +x ≤ 120 . . . . . substitute the expression for d(x)

  x² +20x -2400 ≤ 0 . . . . . multiply by 20, subtract 2400

  (x -40)(x +60) ≤ 0 . . . . . . . . factor. Factors are zero for x=40, x=-60.

The product will be negative for x-values between the zeros:

  -60 ≤ x ≤ 40

We only care about positive values of x:

  0 ≤ x ≤ 40 . . . . .  miles per hour

The correct interpretation of the confidence interval is: "We are 95% confident that between 50.6% and 55.4% of California adult residents say that marijuana should be legal." (Option A)

The confidence interval represents the range of values within which we are 95% confident that the true population parameter (in this case, the percentage of California adult residents who say that marijuana should be legal) lies. It is specific to the population of California adult residents that was surveyed, and cannot be generalized to other states or to all American adults.

Therefore, Option A is the correct answer that we are 95% confident that between 50.6% and 55.4% of California adult residents say that marijuana should be legal.

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The confidence interval denotes the range of values within which we are 95% certain that the true population parameter (in this case, the proportion of adult Californians who support marijuana legalization) resides. It cannot be extrapolated to adults in other states or to all Americans as a whole because it is specific to the demographic of adult inhabitants of California who were polled.

Accordingly, Option A is the right response, and we are 95% certain that between 50.6% and 55.4% of adult inhabitants of California agree that marijuana should be legal.

Answer:

  (c)  A line that goes through (0, 0) and (3.6, 7.2)

Step-by-step explanation:

You want to find the correct descriptor of the proportional relationship that includes points (x, y) = (2.1, 4.2), (3.6, 7.2), (12.2, 24.4).

The graph of the points in the table will go through points with coordinates that have the x-value of the table listed first. Points (24.4, 12.2), (7.2, 3.6) are not on the graph. (The table y-value is listed first for those ordered pairs.)

A proportional relationship graphs as a straight line through the origin, so this graph can be described as ...

  A line that goes through (0, 0) and (3.6, 7.2)

Answer:

 (c)  A line that goes through (0, 0) and (3.6, 7.2)

Step-by-step explanation:

Answer: No

Step-by-step explanation:  If A − λI has rank 1, then the geometric multiplicity of λ is 4 − 1 = 3, equal to the algebraic multiplicity of λ.

The Matrix A is not defective.

A theorem in linear algebra known as the rank-nullity theorem states that the domain dimension of a linear map is equal to the sum of the rank (or the dimension of its image) and the nullity of the map (the dimension of its kernel).

Given:

We have  A be a 4 x 4 matrix

let λ be an eigenvalue of multiplicity 3 and rank 1, then by rank nullity theorem we have

Rank(A-λI) + Nullity((A-λI) =4

dimN(N-λI) = Nullity(A-λI)

dimN(N-λI) = 4 - Rank(A-λI)

dimN(N-λI)  = 4-1

dimN(N-λI) = 3

As a result, eigenvalue is made up of 3 linearly independent eigenvectors.

This guarantees that there is another eigenvalue other than since is an eigenvalue of multiplicity 3. This provides us 4 linearly independent eigenvectors in the matrix A, in addition to the 3 linearly independent eigenvectors in the eigenspace of. Now we If and only if A has n linearly independent eigenvectors, then A is diagonalizable. A thus has no defects.

Hence, matrix A is not defective.

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No, A is not defective.

The rank–nullity theorem is a theorem in linear algebra, which asserts that the dimension of the domain of a linear map is the sum of its rank (the dimension of its image) and its nullity (the dimension of its kernel).

Given here  A be a 4 x 4 matrix and let λ be an eigenvalue of multiplicity 3 and rank 1, then by rank nullity theorem we have

    Rank(A-λI) + Nullity((A-λI) =4

⇒dimN(N-λI) = Nullity(A-λI)

⇒dimN(N-λI) = 4 - Rank(A-λI)

⇒dimN(N-λI)  = 4-1

⇒dimN(N-λI) = 3

Therefore 3 linearly independent eigenvectors belong to eigenvalue λ. Since λ is an eigenvalue of multiplicity 3, this ensures that there is a different eigenvalue different from λ. Along with 3 linearly independent eigenvectors in the eigenspace of λ, this gives us 4 linearly independent eigenvectors in the matrix A. Now we An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. Thus A is nondefective

Hence, matrix A is not defective.

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