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2 years ago

15

Carlos purchased an antique chair for $15 he later so this year for $27 to an antique dealer what was the percent markup of the

chair

2 answers:

Rama09 [41]2 years ago

6 0

Okay. So, to find the percent change, we have to do change/original. In this case, it would be 12/15, because 15 and 27 are apart by 12. 12/15 is 0.8. Multiply that by 100 and you get 80. There was an 80% markup of the chair sold to the antique dealer.

jasenka [17]2 years ago

3 0

Answer: The required percent markup is 80%.

Step-by-step explanation: Given that Carlos purchased an antique chair for $15 and later he sold it this year for $27 to an antique dealer.

We are to find the percent markup of the chair.

The markup price of the chair is given by

$m_p=\$(27-15)=\$12.$

Since Carlos purchased the chair for $15, so the markup percent is given by

$\dfrac{m_p}{15}\times100\%\\\\\\=\dfrac{12}{15}\times100\%\\\\\\=\dfrac{4}{5}\times100\%\\\\=80\%.$

Thus, the required percent markup is 80%.

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A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

<u>Rate Out</u>

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

$20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000$

Therefore, the amount of salt in the tank at any time t is:

$A(t)=(50+t)-75000(50+t)^{-2}$

After 15 minutes, the amount of salt in the tank is:

$A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds$

8 0

1 year ago

After college graduation, Mario received employment offers from two companies. The first company offered a yearly salary of $31,

Alexandra [31]

The first thing we must do for this case is to define a variable:
x: number of years
y: total salary
We have then:
For first company:
y = 1500x + 31000
For second company:
y = 2000x + 28500
Equaling both equations we have:
1500x + 31000 = 2000x + 28500
Clearing x:
2000x - 1500x = 31000 - 28500
500x = 2500
x = 2500/500
x = 5
Answer:
It will take for the salaries to be the same about:
x = 5 years

3 0

1 year ago

The time for a professor to grade a student's homework in statistics is normally distributed with a mean of 12.6 minutes and a s

Rus_ich [418]

Answer:

37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 12.6, \sigma = 2.5$

What is the probability that randomly selected homework will require between 8 and 12 minutes to grade?

This is the pvalue of Z when X = 12 subtracted by the pvalue of Z when X = 8. So

X = 12

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{12 - 12.6}{2.5}$

$Z = -0.24$

$Z = -0.24$ has a pvalue of 0.4052

X = 8

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8 - 12.6}{2.5}$

$Z = -1.84$

$Z = -1.84$ has a pvalue of 0.0329

0.4052 - 0.0329 = 0.3723

37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade

7 0

2 years ago

Simplify the expression to a form in which 2 is raised to a single integer power. fraction numerator open parentheses 2 to the p

anzhelika [568]

Answer:

2^27

Step-by-step explanation:

Given the following expression:

[(2^10)^3 x (2^-10)] ÷ 2^-7

This can be easily simplified. Let us simplify the numerator first. To do that, we have

(2^10)^3 making use of the power rule of indices that says:

(A^a)^b = A^ab where a and b are powers, we have:

2^(10x3) = 2^30

Therefore the numerator becomes:

2^30 x 2^-10. Also making use of the multiplication rule that says:

A^a x A^b = A^(a + b), we have

2^30 x 2^-10 = 2^(30 – 10) = 2^20.

Now we have:

(2^20) ÷ (2^-7)

To simplify this, we need the division rule of indices which says:

A^a ÷ A^b = A^(a – b)

Therefore we have:

(2^20) ÷ (2^-7) = 2^[20 – (–7)] = 2^(20+7) = 2^27

5 0

1 year ago

On the first day a total of 40 items were sold for $356. Define the variables and write a system of equations to find the number

Alina [70]

<h3><u><em>Question:</em></u></h3>

On the first day, a total of 40 items were sold for $356. Pies cost $10 and cakes cost $8. Define the variables, write a system of equations to find the number of cakes and pies sold, and state how many pies were sold.

<h3><em><u>Answer:</u></em></h3>

The variables are defined as:

"c" represent the number of cakes sold and "p" represent the number of pies sold

The system of equations used are:

c + p = 40 and 8c + 10p = 356

18 pies and 22 cakes were sold

<h3><em><u>Solution:</u></em></h3>

Let "c" represent the number of cakes sold

Let "p" represent the number of pies sold

Cost of 1 pie = $ 10

Cost of 1 cake = $ 8

Given that total of 40 items were sold

number of cakes + number of pies = 40

c + p = 40 ------ eqn 1

<u><em>Given items were sold for $356</em></u>

number of cakes sold x Cost of 1 cake + number of pies sold x Cost of 1 cake = 356

$c \times 8 + p \times 10 = 356$

8c + 10p = 356 ----- eqn 2

<u><em>Let us solve eqn 1 and eqn 2</em></u>

From eqn 1,

p = 40 - c ---- eqn 3

Substitute eqn 3 in eqn 2

8c + 10(40 - c) = 356

8c + 400 - 10c = 356

-2c = - 44

c = 22

<em>Substitute c = 22 in eqn 3</em>

p = 40 - c

p = 40 - 22

p = 18

Thus 18 pies and 22 cakes were sold

3 0

2 years ago