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puteri [66]
2 years ago
15

Assume that you own 1,500 shares of $10 par value common stock and the company has a 5-for-1 stock split when the market price p

er share is $30. How many shares of common stock will you own after the stock split?
Mathematics
1 answer:
pochemuha2 years ago
8 0

After the stock split, I would have 7500 shares

A stock split is a method by which a company increases the number of its shares outstanding. A stock split leads to a fall in the price per share of the company. A stock split does not have any effect on the market capitalisation of a firm.

A five for one split means that for every one stock a shareholder owns, it is multiplied by 5

Number of stocks a shareholder has after a 5-for-1 stock split = 5 x number of stocks the shareholder had before the split

5 x 1500 = 7500 shares

To learn more about stock splits, please check: brainly.com/question/15685682?referrer=searchResults

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a store has apples on sale for $3.00 on two pounds . how many pounds of apples can you buy for $9 if an apple is approximately 5
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a circle is inscribed in a square. the circumference of the circle is increading at a constant rate of 6 inches per second. As t
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Answer:

The rate at which Perimeter of the square is increasing is \frac{24}{\pi} \ in/secs.

Step-by-step explanation:

Given:

Circumference of the circle = 2\pi r

Rate of change of in circumference = 6 in/secs

We need to find the rate at which the perimeter of the square is increasing

Solution:

Now we know that;

\frac{d(2\pi r)}{dt} =6\\\\2\pi\frac{dr}{dt}=6\\\\\frac{dr}{dt}=\frac{6}{2\pi}\\\\\frac{dr}{dt}=\frac{3}{\pi}

Now we know that;

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side of the square = 2r

Now Perimeter of the square is given by 4 times length of the side.

P=4\times 2r =8r

Now we need to find the rate at which Perimeter is increasing so we will find the derivative of perimeter.

\frac{dP}{dt}= \frac{d(8r)}{dt}\\\\\frac{dP}{dt}= 8\times\frac{dr}{dt}

But \frac{dr}{dt} =\frac{3}{\pi}

So we get;

\frac{dP}{dt}= 8\times\frac{3}{\pi}\\\\\frac{dP}{dt}= \frac{24}{\pi}\  in/sec

Hence The rate at which Perimeter of the square is increasing is \frac{24}{\pi} \ in/secs.

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