the distribution system for the herman company consists of three plants, two warehouses, and four customers. plant capacities and shipping costs per unit (in $) from each plant to each warehouse are as follows: customer demand and shipping costs per unit (in $) from each warehouse to each customer are as follows: choose the correct network representation of this problem. (i) (ii) (iii) (iv) formulate a linear programming model of the problem. for subtractive or negative numbers use a minus sign even if there is a sign before the blank. let xij represents relation between plants to warehouses or relation between warehouses to customers. min fill in the blank 2 x14 fill in the blank 3 x15 fill in the blank 4 x24 fill in the blank 5 x25 fill in the blank 6 x34 fill in the blank 7 x35 fill in the blank 8 x46 fill in the blank 9 x47 fill in the blank 10 x48 fill in the blank 11 x49 fill in the blank 12 x56 fill in the blank 13 x57 fill in the blank 14 x58 fill in the blank 15 x59 s.t. 1) fill in the blank 16 x14 fill in the blank 17 x15 < fill in the blank 18 2) fill in the blank 19 x24 fill in the blank 20 x25 < fill in the blank 21 3) fill in the blank 22 x34 fill in the blank 23 x35 < fill in the blank 24 4) fill in the blank 25 x46 fill in the blank 26 x47 fill in the blank 27 x48 fill in the blank 28 x49 fill in the blank 29 x14 fill in the blank 30 x24 fill in the blank 31 x34

We can conclude that -

        Q = 5(280 - P)

Given is a story of Dr. Leona Williams who is a well-known plastic surgeon.

{ 1 } -

Dr. Williams estimated demand for her work is -

Q = 280 - 0.2P

Inverse of a function can be calculated by following the steps mentioned below -

So, we can write -

P = 280 - 0.2Q

P - 280 = - 0.2Q

- (280 - P) = - 0.2Q

(280 - P) = 0.2Q

Q = (280 - P)/0.2

Q = 5(280 - P)

{ 2 } -

The average variable cost function for reconstructive nose surgery is -

AVC = 2Q² – 15Q + 400

Let AVC = A{Q}

For maximum profit -

d/dQ (A{Q}) = 0

d/dQ (2Q² – 15Q + 400) = 0

d/dQ (2Q²) - d/dQ (15Q) + d/dQ (400) = 0

4Q - 15 = 0

4Q = 15

4Q = 15

Q = 3.7

Q = 4 {approx.}

Therefore, we can conclude that -

        Q = 5(280 - P)

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Using the Central Limit Theorem, it is found that the correct option is: C  

For all samples of size 400 from this population, the mean of all resulting sample proportions of cell phone screens with a screen defect is 0.05.

By the Central Limit Theorem, for a proportion p in a sample of size n, the mean of the sampling distribution of sample proportions is u=5, while the standard error is [tex]SE=\sqrt{\frac{p*(1-p)}{n} }[/tex]

In this problem, the mean of the sampling distribution of sample proportions is 0.05, that is, the mean proportion of all samples, hence, the correct option is C.

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The value of h in the algebraic equation in fraction 11/12 = 3/4h is equal to 9/11

We shall first find the lowest common multiple LCM of the denominators to help us clear the fractions, before then simplifying further for the value of h as follows:

LCM of the denominators 12 and 4h is 12h

multiply both sides by the LCM

11/12 × 12h = 3/4h × 12h

11 × h = 3 × 3

11h = 9

divide both sides by the coefficient of h

11h/11 = 9/11

h = 9/11

Therefore, the value of h in the algebraic equation in fraction 11/12 = 3/4h is equal to 9/11.

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The size of each group will be half as big.

There will be twice as many groups.

We want to find out what is going to happen if we divide by 8/4 instead of 4/5.

The total of the tape diagram is 3¹/₅ = 16/5 and as such if we are going to divide by 4/5 we will have;

16/5 ÷ 4/5 = 4

However if we are dividing by 8/5, it means that the solution is;

16/5 ÷ 8/5 = 2

Thus, the solution will be half as big.

There will be twice as many groups.

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The fraction of a pound of granola Marty buys is 3/4.

A fraction is a part of a whole number, value, and variable.

There are proper and improper fractions, including complex fractions.

Proper fractions are those with the numerator less than the denominator. An improper fraction produces a whole number with a remainder when the denominator divides the numerator.

On the other hand, complex fractions are rational expressions that have some fractions in their numerators, denominators, or both.

1 pound = 16 ounces

The quantity of granola Marty buys = 12 ounces

The fraction of a pound of granola Marty buys = 3/4 (12/16)

Thus, we can conclude that Marty buys 3/4 or 75% of a pound of granola.

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Answer:

slope = 4

Step-by-step explanation:

So, the possible fourth corners are (4,4) and (-2,2).

Given the three vertices of a parallelogram (1,1), (4,2), and (1,3), we can use vector algebra to find the fourth corner. We know that opposite sides of a parallelogram are parallel, and therefore have the same vector direction. We can use this property to find the vector pointing from one corner to another and then use this vector to find the fourth corner.

One possible method is:

Another possible method is:

So, the possible fourth corners are (4,4) and (-2,2).

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Answer:4x

Step-by-step explanation:

The value of x in the figure is calculated using the linear pair theorem to be 6 degrees

The linear pair postulate or linear pair theorem in mathematics states that the sum of the measurements of two angles that make up a linear pair is 180°.

Because o the linear pair postulate we have that

150 + 5x = 180

collecting like terms

5x = 180 - 150

5x = 30

Isolating x

5x / 5 = 30 / 5

x = 6

hence x is solved to be 6

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The area under the normal curve with a mean of 20 and a standard deviation of 5 is 0.8413 when x = 25.

The area under the normal distribution curve is calculated using the z-score, which is the number of standard deviations a value is from the mean. To calculate the area under the normal curve for a given mean and standard deviation, one must use the formula:

Area = ½[1 + erf(z)]

Where erf is the Gaussian error function, and z is the z-score.

To calculate the area under the curve given a boundary x, the z-score must be calculated first. This is done using the formula:

z = (x - mean) / standard deviation

Substituting this into the formula above gives:

Area = ½[1 + erf((x - mean) / standard deviation)]

For example, if the mean is 20 and the standard deviation is 5, and the boundary value is x = 25, the area under the normal curve is calculated as follows:

z = (25 - 20) / 5 = 1

Area = ½[1 + erf(1)] = 0.8413

Therefore, the area under the normal curve with a mean of 20 and a standard deviation of 5 is 0.8413 when x = 25.

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The required number is 42 which gives a result of −4 when 11 is subtracted from the quotient of the number and 5.

A quotient is defined as the outcome of dividing an integer by any divisor that can be said to be a quotient. The dividend contains the divisor a specific number of times.

Let n be the number in question; then, we can write the word problem as an equation like this:

⇒ (n/6) - 11 = -4

Add 11 on both sides of the equation,

⇒ (n/6) - 11 + 11 = 11 - 4

⇒ (n/6) = 7

Multiply by 6 on both sides of the above equation, and we get

⇒ n = 42

Hence, the number is 42.

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Answer:

=

Step-by-step explanation: Please mark brainliest

Substitute 3 in for m and solve

4m - 1 = 4(3) - 1 = 11

5m - 4 = 5(3) - 4 = 11

Answer:

Step-by-step explanation:

P&R3 model and solve

problems using linear

equations of the form:

ax = b *

x

a

= b, a ≠ 0*

ax + b = c *

x

a

+ b = c ,a ≠ 0*

a x( ) + b = c *

ax + b = cx + d *

* a bx ( ) + c = d ex ( ) + f

*where a, b, c, d, e, and f [C,

CN, PS, V]

1. Introduction to Solving Linear Equations (pg. 4-7)

• Model the solution of a given linear equation using concrete or

pictorial representations, and record the process.

Algebra stones and Algebra Tiles.

Solve x + 5 = 10 , x − 7 = 10 , 2x = 10 , x

3 = 10

2. Solving equations of the form ax+b=c and a/c +b=c (pg. 8-10)

• Solve a given linear equation symbolically.

• Solve a given problem using a linear equation and record the

process.

Solve. 4m+3=31 & 2

5

m − 5 = 3

3. Solving equations of the form a(x+b)=c and ax+b=cx+d(pg. 11-14)

• Determine, by substitution, whether a given rational number is a

solution to a given linear equation.

Solve. 4(m+3)=40 & 6m+3=2m+15

Is m=5 a solution to the equation 2(m + 2) = 14 ?

4. More practice with ax+b=cx+d (Pg. 15-18)

• Identify and correct an error in a given incorrect solution of a

linear equation

Solve. 2(m+1)+4m=4(m-2)+6.

5. Solve equation with fractions. (Pg. 19-23)

• Solve a given linear equation symbolically. Solve m

3 + 2m

5 − 1

2

= 2

6. Solve Linear equations without numbers. (Pg. 24-28)

• Solve a given linear equation symbolically.

Solve for m. )( =+ BnmA

P&R4 explain and illustrate

strategies to solve single

variable linear inequalities with

rational coefficients

within a problem-solving

context.

7. Introduction to linear inequalities (Pg. 29-33)

• Translate a given problem into a single variable linear inequality

using the symbols ≥, >, < , or ≤ .

• Determine if a given rational number is a possible solution of a

given linear inequality.

Write an expression to represent the following

statement: Melanie needs at least $280 for snow

boarding.

8. Inequalities that Include Addition and Subtraction (Pg. 34-37)

• Generalize and apply a rule for subtracting a positive or

negative number to determine the solution of a given inequality.

• Generalize and apply a rule for multiplying or dividing by a

positive or negative number to determine the solution of a given

inequality.

• Solve a given linear inequality algebraically and explain the

process orally or in written form.

• Verify the solution of a given linear inequality using

substitution for multiple elements in the solution.

Solve 2x + 5 < 25 and verify your solution.

True or False.

If −2x > −10 then x > 5 .

9. Solving Problems with Linear Inequalities (Pg. 38-41)

• Compare and explain the process for solving a given linear

equation to the process for solving a given linear inequality.

• Graph the solution of a given linear inequality on a number line.

• Compare and explain the solution of a given linear equation to

the solution of a given linear inequality.

• Solve a given problem involving a single variable linear

inequality and graph the solution.

Vertical Wireless charges $50/ month plus $0.25 for

all minutes above 400 minutes per month. Frue Gal has

decided that she does not want to pay more than $70

per month. Write an inequality to represent how many

minutes she can use per month without going over her

$70 limit. Approximate your solution on the number

line

10. Chapter Review and Practice Test

• Help students develop sound study habits.

• Many students will graduate high school saying they do not

know how to study for math tests.

11. Go over Practice Test

12. Unit Evaluation

The amount that you will have in 3 years, given the interest rate and the amount won, is 1, 195, 618. 17 currency units

To find the amount earned in three years as a result of your lottery being invested, use the future value formula which is:

= Amount won x ( 1 + periodic  rate ) ^ number of periods

Periodic rate :

= 6 % / 4 quarters per year

= 1. 5 %

The number of periods is:

= 3 years x 4 quarters per year

= 12 quarters

The amount you will have in 3 years is:
= 1,000,000 x ( 1 + 1. 5 %) ¹²

= 1, 195, 618. 17 currency units

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The value of z in the given linear equation in one variable is -1.

By equating it to zero, a linear polynomial over a field, from which the coefficients are extracted, can be converted into a linear equation.

The values that, when employed as stand-ins for the unknowns in such an equation, lead to the equality being true are the solutions.

When there is just one variable, there is just one solution. The term "linear equation" is generally understood to relate to this particular circumstance, when the variable is accurately referred to as the "unknown."

It is possible to interpret each solution as the Cartesian coordinates of a specific point on the Euclidean plane when there are only two variables.

-6z -18z + 2z +6z = 16

-16z = 16

z = -1

Hence, the value of z is -1

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The infinite product measure [tex]$\mu$\\[/tex] is uniquely defined on the product measure space[tex]$(\Omega, \mathcal{A}, \mu)$[/tex] by Dynkin's theorem, since [tex]$\mu$\\[/tex] is finitely additive, countably subadditive, and [tex]$\sigma$[/tex]-finite.

Let [tex]$\mathcal{A}_1, \mathcal{A}_2, \ldots$[/tex] be a sequence of sigma algebras defined on [tex]$\Omega_1, \Omega_2, \ldots$[/tex] respectively. Let [tex]$(\Omega, \mathcal{A}, \mu)$[/tex] be the product measure space, where [tex]\Omega = \prod_{i=1}^{\infty} \Omega_i$ and $\mathcal{A} = \bigotimes_{i=1}^{\infty} \mathcal{A}_i$[/tex].

We want to prove that the infinite product measure [tex]$\mu$[/tex] is uniquely defined. To do this, we will use Dynkin's theorem.

According to Dynkin's theorem, a measure [tex]$\mu$[/tex] is uniquely determined on a product measure space [tex]$(\Omega, \mathcal{A}, \mu)$[/tex] if the following conditions are satisfied:

1. [tex]$\mu$[/tex] is finitely additive on [tex]$\mathcal{A}$[/tex]

2. [tex]$\mu$[/tex] is countably subadditive on [tex]$\mathcal{A}$[/tex]

3. [tex]$\mu$[/tex] is [tex]$\sigma$[/tex]-finite on [tex]$\mathcal{A}$[/tex]

It is easy to see that the infinite product measure [tex]$\mu$[/tex]satisfies these conditions, since it is defined as a product of finitely additive, countably subadditive, and [tex]$\sigma$[/tex]-finite measures. Thus, by Dynkin's theorem,[tex]$\mu$[/tex] is uniquely determined on the product measure space [tex](\Omega, \mathcal{A}, \mu)$.[/tex]

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Answer:

A. 881

C. 353

Step-by-step explanation:

In order to form 8 lines with each line having equal length, the total number of members has to be divisible by 8:

440/8 = 55, so 440 isn't the answer

224/8 = 28, so 224 isn't the answer

881/8 = 110.125. In addition, even numbers can't be divided evenly by odd numbers like 881, so A. 881 is the answer.

Same goes to 353 since 353 is also an odd number that can't evenly divide the even number 8. Hence, C. 353 is the answer as well.

The equation of the regression line is Y=2.83+0.52x .

Linear regression is the statistical technique that is most frequently used to simulate a relationship between two sets of variables. As a result, a data-applicable linear regression equation is produced.

Plotted your points and drew a linear regression line.

y = a + bₐₓx

bₐₓ = n∑xy-∑x-∑y

        n∑x²- ∑x²

x = 8.81

y = 7.45

∑x² = 947

∑xy=770.95

bₐₓ = 0.52

a = 2.83

y=1.3082X + 3.3436.

Its equation is

y = 2.83+0.52x .

Therefore, the regression line in a scatter plot ignores non-linear properties, so the equation of the reggresion line will be y = 2.83+0.52x .

Your question is incomplete but most probably your full question was

Use the given data to find the equation of the regression line. Examine the scatterplot and identify a characteristic of the data that is ignored by the regression line.

x y

9 7.38

9 6.75

12 13.08

9 7.42

11 7.56

13 8.53

7 5.93

4 5.33

12 8.1

6 6.46

5 5.44

Required:

Create a scatterplot of the data.

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a) Using the summation notation to write the sum is 3∑n=1 3n.

b) The total amount that Jummy will save during the first 3 weeks of his new job is $18.

The summation notation is the representation of a series in a compact form, using the capital sigma symbol (Σ) to indicate the sum of a number of similar terms.

Using the summation notation, one can write a long sum in a single expression.

In this situation, the expression can be read as the sum of 3n as n goes from week 1 to week 3, representing the first three weeks Jimmy will spend on the job.

3∑n=1 3n

= 3 x 1 + 3 x 2 + 3 x 3

= 3 + 6 + 9

= 18

Thus, we have used the summation notation to find that Jimmy will save $18 during the first three weeks at his new job.

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Answer:

Step-by-step explanation:

The given sample of the randomly selected players from the teams as described is a stratified sample. Hence, the right answer is option A.

In a stratified sample, the population is divided into groups (or strata) based on some characteristic, and a random sample is selected from each group. In this case, the population is the football players, and the groups are the varsity and junior varsity teams. The coach randomly selects one of these teams and then asks all players from that team their opinion on a new logo.

This type of sampling is used when the population is divided into subgroups that are different in some way, and the researcher wants to ensure that these subgroups are represented in the sample. In this case, the coach wants to ensure that both varsity and junior varsity players are represented in the sample, as they may have different opinions on the new logo.

It's not a simple random sample because he selected the team first and then the players. Also, it's not a voluntary response sample because the coach is selecting the players.

It's not a cluster sample because the coach is not selecting the players based on location or geographic proximity, and it's not a systematic sample because he is not selecting the players based on the order in which they appear in a list or a database.

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The notion of an equation in algebra is a statistical statement that demonstrates the equality of two mathematical expressions.

For the stated question;

365 days are included in a year.

There are 256 days a year on average when it is foggy.

Determine the approximate number of clear days by using the formula:

We round the figures.

To the nearest tens, 365 days

= 370 days

The nearest tenth of 256 days

= 260 days

370 - 260 days, then

= 110 days

The precise number of days without fog is =

= 365 − 256

= 109

Thus, the precise number of days without fog is found as 109 days.

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The complete question is -

Solve the problem by using​ addition, subtraction,​ multiplication, or division. First use front end rounding to estimate the answer. Then find the exact answer. The foggiest place in a certain country is City A. It is foggy there an average of 256days each year. How many days is it not foggy each​ year? Find the estimated and the exact number of days.

c=170/30 is the equation that represents the number of calories,c, burned per min.

A division is a process of splitting a specific amount into equal parts.

Given that Tarmu run 30 min and burned 170 calories

We need to find the equation that represents the number of calories, c, burned per min.

To find the calories burned in each minute we have to divide one hundred seventy by thirty.

c=170/30

c=5.66 cal/min

Hence, c=170/30 is the equation that represents the number of calories,c, burned per min.

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Answer: Match each equation with an equivalent equation. Some of the answer choices are

not used.

A. 3x + 6 = 4x+7

B. 3(x + 6) = 4x+7

C. 4x + 3x = 7-6

1.9x = 4x +7

2. 3x + 18 = 4x+7

3.3x = 4x + 7

4. 3x1 = 4x

5.7x = 1

A. 3x + 6 = 4x+7 = 3

B. 3(x + 6) = 4x+7 = 2

C. 4x + 3x = 7-6 = 5

1.9x = 4x +7 = 3

3x + 18 = 4x+7 = 2

3.3x = 4x + 7 = 1

3x1 = 4x = not used

5.7x = 1 = not used

So,

3x + 6 = 4x+7 matches with 3

3(x + 6) = 4x+7 matches with 2

4x + 3x = 7-6 matches with 5

1.9x = 4x +7 matches with 3

3x + 18 = 4x+7 matches with 2

3x = 4x + 7 matches with 1

3x1 = 4x is not used

7x = 1 is not used

Step-by-step explanation:

The graph with the slope of 2 is the second graph.

The slope of a line is a measure of its steepness. The slope of a line is the change in dependent variable with respect to the change in the independent variable.

Therefore, let's find the slope of the graph to know the one with the slope of 2.

Hence,

slope = y₂ - y₁ / x₂ - x₁

Using the second graph,

(-1, -1)(0, 1)

slope = 1 + 1 / 0 + 1

slope = 2 / 1

slope = 2

Therefore, the graph with slope of 2 is the second graph.

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The proportional relationship is given by the equations -

y = mx     or     y = kx     or     y = mx + b     or    y = kx + b.

Given are the statements describing proportional relationship.

        {K} = y/x = constant.

        y = mx     or     y = kx     or     y = mx + b     or    y = kx + b

Therefore, the proportional relationship is given by the equations -

y = mx     or     y = kx     or     y = mx + b     or    y = kx + b.

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Answer:11.15

Step-by-step explanation:

sry if its wrong :)

just add them

Answer:

here are the answers to the questions in order

Step-by-step explanation:

7. -7

8. 73

9. -8

10. 1