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2 years ago

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The price of a box of 15 cloud markers is $12.70. The price of a box of 42 cloud markers is $31.60. All prices are without tax,

and the price of the packaging is the same for any size. How much would 50 cloud markers in a box cost? Write an equation which will tell the price P for the number N markers in the box.

2 answers:

spin [16.1K]2 years ago

6 0

Answer: Cost of 50 markers is $37.2

Equation is

P = 0.7N+2.2

Step-by-step explanation:

Let the cost Packaging be 'c' and x be the cost of one marker and P be the total cost

So the equation for the total cost is

P = Nx +c

then cost of 15 markers is 15x +c = $12.70 ........ equation (1)

similarly cost of 42 markers is 42N+c =$ 31.60 ....... equation (2)

Subtracting equation 2 from equation 1 ,we get

42N+c = 31.60

-15x -c = 12.70

___________________

27x =18.9

x = 0.7

Plugging x = 0.7 in equation 1

15(0.7) + c = 12.7

10.5 +c = 12.7

c = 2.2

Therefore cost equation is

P = 0.7N +2.2

Cost of 50 markers will be

P = 0.7(50) +2.2

= 35.0 + 2.2

=$ 37.2

stich3 [128]2 years ago

3 0

Answer:

Here, The price of one box having N markers = Price of N markers + packaging charge

Let M be the price of one Marker and P' be the price of packaging of one box which is same for any size.

Since, The price of a box of 15 cloud markers is $12.70.

That is , 15 M + P' = 12.70 ---------(1)

Also, The price of a box of 42 cloud markers is $31.60,

That is, 42 M + P' = 31.60 ---------(2),

Equation (2) - Equation (1),

27 M = 18.90,

⇒ M = 0.7

By putting the value of P in equation (1),

We get,

10.5 + P' = 12.70 ⇒ P' = 2.2,

Thus, the price of one marker, M = 0.7

And, the price of packaging one box, P' = 2.2

Thus, the price of a box of 50 marker = 50 × M + P' = 50×0.7 + 2.2 = 35 + 2.2 = $ 37.2

And, the equation that shows the price(P) of a box of N markers,

P = 0.7 N + 2.2

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2 years ago

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Answer:

For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected $p=\frac{1}{N}$

So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.

Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.

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Step-by-step explanation:

For this case we can clasify this study as an enumerative study and inferential since they want to identify the average distance between the hometowns of students and their campuses.

For this case the sampling frame represent all the 23 campuses, and is known for the researcher,

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2 years ago

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Answer:

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length: 86 feet, width: 7 feet

Both solutions are valid.

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1. First assumption is that the shape of the fence is <u>rectangular</u>.

2. Second, assum the length parallel to the wall measure y feet, so the other two lengths, y, together with x will add up 100 feet

2x + y = 100

3. The, the area of the fence will be:

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4. Now you have two equation with two variables which you can solveL

Solve for y in the first equation: y = 100 - 2x

Substitute the value of y into the second equation: x (100 - 2x) = 600

5. Solve the last equation by completing squares:

Distributive property: 100x - 2x² = 600
Divide both sides by - 1: 2x² - 100x = - 600
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Add the sequare of the half of 50 to both sides: x² - 50x + 625 = 325
Factor the left side: (x - 25)² = 325
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2 years ago

Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

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Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

,

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

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