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1 year ago

Jen bought 30 shares of eBay at the close price of $20.95. She bought 20

Mathematics

2 answers:

Tom [10]1 year ago

7 0

Answer: The answer is $1407.50 APEX

pashok25 [27]1 year ago

4 0

Answer:

The answer is: C. $1407.50

Step-by-step explanation:

Buying:

30 * $20.95 + 20 * $25.50 =

$628.50 + $510 = $1138.

Selling:

50 * $28.75 = $1437.50

Remaining Cash:

$1437.50 - $30 = $1407.50

(This answer assumes Jen paid the other $30 per transaction when she bought the original shares, but had to pay $30 when she sold the shares.)

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The total operating cost to run the organic strawberry farm varies directly with its number of acres. What is the constant of va

kow [346]

Answer:

$16\ acres$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Let

x ----> the number of acres

y ---> total operating costs

we have the ordered pair (20,551,520)

For x=20 acres, y=$551,520

<em>Find the value of the constant of proportionality k</em>

$k=\frac{y}{x}$

substitute the values of x and y

$k=\frac{551,520}{20}=\$27,576\ per\ acre$

The linear equation is equal to

$y=27,576x$

Find out how many acres are on a farm with a total operating cost of $441,216

For y=$441,216

substitute in the linear equation the value of y and solve for x

$441,216=27,576x$

$x=441,216/27,576$

$x=16\ acres$

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2 years ago

Read 2 more answers

Jim Ryan, an owner of a Burger King restaurant, assumes that his restaurant will need a new roof in 7 years. He estimates the ro

Lelu [443]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$9000\\
P=\textit{original amount deposited}\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &7
\end{cases}
\\\\\\
9000=P\left(1+\frac{0.06}{4}\right)^{4\cdot 7}\implies 9000=P(1.015)^{28}\implies \cfrac{9000}{1.015^{28}}=P$

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2 years ago

PLEASE HELP !!! The cost of a pottery painting party is $9 per person, plus a one-time supply fee of $89. Which of the following

stiv31 [10]

c(x)=9x+89

I know that you don't have the answer listed, but it is the correct answer.

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2 years ago

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The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a

maxonik [38]

Given that:

Total number of fish = 140

Fish are green swordtails female = 44

Fish are green swordtails male = 36

Fish are orange swordtails female = 36

Fish are orange swordtails male = 24

Solution:

A. We have to find the probability that the selected fish is a green swordtail.

$\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}$

$\text{P(green swordtail)}=\dfrac{80}{140}$

$\text{P(green swordtail)}=\dfrac{4}{7}$

Therefore, the probability that the selected fish is a green swordtail is $\dfrac{4}{7}.$

B. We have to find the probability that the selected fish is male.

$\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}$

$\text{P(Male fish)}=\dfrac{36+24}{140}$

$\text{P(Male fish)}=\dfrac{60}{140}$

$\text{P(Male fish)}=\dfrac{3}{7}$

Therefore, the probability that the selected fish is a male, is $\dfrac{3}{7}.$

C. We have to find the probability that the selected fish is a male green swordtail.

$\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}$

$\text{P(Male green swordtail)}=\dfrac{36}{140}$

$\text{P(Male green swordtail)}=\dfrac{9}{35}$

Therefore, probability that the selected fish is a male green swordtail is $\dfrac{9}{35}.$

We have to find the probability that the selected fish is either a male or a green swordtail.

$\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}$

$\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}$

$\text{P(Male or green swordtail)}=\dfrac{96}{140}$

$\text{P(Male or green swordtail)}=\dfrac{24}{35}$

Therefore, the probability the selected fish is either a male or a green swordtail is $\dfrac{24}{35}.$

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1 year ago

Jing spent 1/3 of her money on a pack of pens, 1/2 of her money on a pack of markers, and 1/8 of her money on a pack of pencils.

Reika [66]

As per the problem

Jing spent $\frac{1}{3}$ of her money on a pack of pens.

$\frac{1}{2}$ of her money on a pack of markers.

and $\frac{1}{8}$ of her money on a pack of pencils.

Total fraction of money spent cab be given as below

Fraction of Money Spent = $\frac{1}{3} +\frac{1}{2}+\frac{1}{8}$

Take the LCD of denominator, we get LCD of (3,2,8)=24

Fraction of Money Spent = $\frac{8+12+3}{24} =\frac{23}{24} \\\\$

$\\ \text{Hence fraction of Money Spent }=\frac{23}{24} \\ \\ \text{Fraction of Money left}=1-\frac{23}{24} \\ \\ \text{Simplify, we get}\\ \\ \text{Fraction of Money left}=\frac{24-23}{24} \\ \\ \text{Fraction of Money left}=\frac{1}{24}$

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1 year ago

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