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2 years ago

Find the markup and the cost of the following item. Round answers to the nearest cent.

Mathematics

1 answer:

scoray [572]2 years ago

8 0

Answer:

1st question: M=22.62 while C=75.38

2nd question: M=.22 while C=1.97

Step-by-step explanation:

If a mirror costing x dollars is marked up 30%, then we have to find x such that 30%x+x is 98 dollars.

We are solving:

.3x+x=98

Combine like terms:

1.3x=98

Divide both sides by 1.3:

x=75.38

M=98-75.38=22.62

C=75.38

So M=22.62 while C=75.38.

If ream of paper cost x and is marked up 11%, then we have to find x such that 11%x+x is 2.19.

We are solving:

.11x+x=2.19

1.11x=2.19

x=1 97

M=2.19-1.97=.22

So M=.22 while C=1.97

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Drag each tile to the correct box.

Umnica [9.8K]

Answer:

Emma ⇒ IQR = 3.5

Sally ⇒ IQR = 3

Olivia ⇒ IQR = 2.5

Emma - Sally - Olivia

Step-by-step explanation:

Emma's Scores:

6 , 7 , 8 , 8 , 9 , 9 , 9 , 10 , 10 , 11 , 11 , 11 , 11 , 11 , 12 , 13 , 13 , 13 , 13 , 14

LQ = (9 + 9) ÷ 2 = 9

UQ = (12 + 13) ÷ 2 = 12.5

IQR = 12.5 - 9 = 3.5

Sally's Scores:

6 , 6 , 6 , 7 , 7 , 8 , 8 , 8 , 9 , 9 , 10 , 10 , 10 , 10 , 10 , 10 , 11 , 11 , 12 , 12 , 14

LQ = (7 + 8) ÷ 2 = 7.5

UQ = (10 + 11) ÷ 2 = 10.5

IQR = 10.5 - 7.5 = 3

Olivia's Scores:

6 , 6 , 7 , 7 , 7 , 8 , 8 , 8 , 8 , 9 , 9 , 9 , 9 , 9 , 10 , 10 , 12 , 12 , 13 , 14

LQ = (7 + 8) ÷ 2 = 7.5

UQ = (10 + 10) ÷ 2 = 10

IQR = 10 - 7.5 = 2.5

Emma - Sally - Olivia

6 0

2 years ago

Read 2 more answers

“Great job, you've already earned 1300 points, but there are 5 more days in this month. How many points a day do you need to rea

sergiy2304 [10]

You would have to get 56 points everyday.
If you subtract 1300 from 1580 you get 280. divide 280/5 which equals 56.
56*5=280
280+1300=1580.
56 is the amount of points you need

6 0

2 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

Look at picture, answer quick please

Tpy6a [65]

Step-by-step explanation:

1. C = the graphing is slowly increasing, then he walks across the top so it is flat

2. D = it started out slow and gradually got higher, which is what the story said

3. H = he had to stop walking, and there is a part of no movement in the graph H

6 0

2 years ago

this squared based pyramid is cut horizontally at a height of 15cm to leave this frustum base area=10cm

scoundrel [369]

Answer:

583.33 cm³

Step-by-step explanation:

From the image attached, the total height (H) of the pyramid is 30 cm, the height of the frustrum (h) = 15 cm, the Frustrum base = lower base = B = 10 cm . To find the length of the upper base (b), we use:

$\frac{b}{H-h}=\frac{B}{H}\\ b=(H-h)\frac{B}{H}\\b=(30-15)\frac{10}{30}=5\\b=5\ cm$

The volume of the frustrum is given by:

$V=\frac{1}{3}h(B_1+B_2+\sqrt{B_1B_2} )\\ Where\ B_1\ is \ the\ area \ of\ the\ upper\ base= 5cm*5cm=25cm^2\\B_1\ is \ the\ area \ of\ the\ lower\ base= 10cm*10cm=100cm^2\\V=\frac{1}{3}(10)(25+100+\sqrt{25*100} )=583.33\\V=583.33\ cm^3$

6 0

2 years ago