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labwork [276]
3 years ago
8

Kyle buys $30,000 of company XYZ's shares, at a share price of $50. A year later, XYZ's share price is $55, and Kyle sells all h

is shares. How many dollars did Kyle's investment gain
Mathematics
2 answers:
Ivan3 years ago
5 0

Answer:

10%

Step-by-step explanation:

x - amount of shares bought

x * 50[$] = 30000[$]

x = 30000/50 = 600

if he bought 600 shares then he sold earning in total:

600 * 55[$] = 33000[$]

that means investmant gain can be calculate as:

return on investment =  (gain from investment – cost of investment) / cost of investment

return on investment = (33000 - 30000) / 30000 = 3000/30000 =0.1 = 10%

creativ13 [48]3 years ago
4 0

Answer:

$3,000

Step-by-step explanation:

Given that the shares are $50 per share and that Kyle buys $30,000 worth of shares,

Number of shares bought = total price of shares ÷ price per share

= $30,000 ÷ $50

= 600 shares

We are also give that a year later, the shares are worth $55

Total value of shares 1 year later = Price per share one year later x number of shares

= $55 x 600

= $33,000

Hence the investment gained $33,000 - $30,000 = $3,000

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aniked [119]

I'm not sure if this is right, but I multiplied 115 by 4 and got 460- Try that? ;^;

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3 years ago
Which choice is equivalent to the expression below?
alekssr [168]

Answer: OPTION A

Step-by-step explanation:

We need to remember that Product of powers property, which states that:

(a^m)(a^n)=a^{(m+n)}

Let's check the options:

A. For 5^9*5^{\frac{9}{10}}*5^{\frac{6}{100}}*5^{\frac{9}{1000}} you can  apply the property mentioned before. Then:

5^{(9+\frac{9}{10}+\frac{6}{100}+\frac{9}{1000})=5^{9.969}

(It is the equivalent expression)

 B. Add the exponents:

 5^9*5^{(\frac{9}{10}+\frac{9}{10}+\frac{6}{1000})=5^{10.806}

(It is not the equivalent expression)

C. For 5^9*5^{\frac{96}{10}}*5^{\frac{9}{100}}} you can  apply the property mentioned before. Then:

 5^{(9+\frac{96}{10}+\frac{9}{100})=5^{18.69}

(It is not the equivalent expression)

D. We know that 5^{9.969}=9,290,347.808 and we maje the addition indicated in this option, we get:

 5^9+5^{\frac{9}{10}}+6^{\frac{6}{100}}=1,953,130.37

(It is not the equivalent expression)

6 0
3 years ago
Read 2 more answers
Form a polynomial f(x) with real coefficients having the given degree and zeros.
dimaraw [331]
There are many polynomials that fit the bill,
f(x)=a(x-r1)(x-r2)(x-r3)(x-r4)  where a is any real number not equal to zero.
A simple one is when a=1.
where r1,r2,r3,r4 are the roots of the 4th degree polynomial.
Also note that for a polynomial with *real* coefficients, complex roots *always* come in conjugages, i.e. in the form a±bi  [±=+/-]

So a polynomial would be:
f(x)=(x-(-4-5i))(x-(-4+5i))(x--2)(x--2)
or, simplifying
f(x)=(x+4+5i)(x+4-5i)(x+2)^2
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6 0
3 years ago
B Christian installed an electric pump to pump water from a borehole into a 30 000 litre cement dam. If the water is pumped at a
skelet666 [1.2K]

Given the amount of water needed to fill the cement dam and the water pump rate, the time needed to fill the dam is 400 minutes or 6hours 40minutes.

Option B)6h 40min is the correct answer.

<h3>How long does it take to fill the dam?</h3>

Given that;

  • Amount of water needed to fill the dam A = 30000 litres
  • Pump rate r = 75 litres per minute
  • Time needed to fill the dam T = ?

To determine how long it take to fill the dam, we say;

Time need = Amount of water needed ÷ Pump rate

T = A ÷ r

T = 30000 litres ÷ 75 litres/minute

T = 400 minutes

Note that; 60min = 1hrs

Hence,

T = 6hours 40minutes

Given the amount of water needed to fill the cement dam and the water pump rate, the time needed to fill the dam is 400 minutes or 6hours 40minutes.

Option B)6h 40min is the correct answer.

Learn more word problems here: brainly.com/question/2610134

#SPJ1

8 0
2 years ago
Ignoring leap days, the days of the year can be numbered 1 to 365. Assume that birthdays are equally likely to fall on any day o
faust18 [17]

Answer:

Follows are the solution to the given question:

Step-by-step explanation:

They can count the days of the year 1 to 365. The random project consists of drawing a sample of n objects from D where elements are n people's birth in a group but instead, D = {1,....365}. And then there's the issue.

S=365^n

This because the list of future birthdays of n people was its test point; therefore m points will be in the sequence so each point contains 365 distinct outcomes. The probability function P for \Omega is that any event is likely to happen in 365 days.

P(x)=\frac{1}{365^{n}}

if x is between 1 and 365 as well as the occurrence is just all similarly possible

In point i:

That somebody mentions their birthday throughout the party

Guess I was born on day b. Therefore the consequence of "x is in A" is "b is now in the series of x," which is to say, b = bk for some amount k approximately 1 and n.

In point ii:

Any 2 persons share the same birthday at this party". A result x is in B" means which "two of entries in x are same." This means that perhaps the outcome x is in B if or only if bj = bk is in B of two numbers j, and k of 1, of two. , no, n.

In point iii:

Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.

6 0
2 years ago
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