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Bas_tet [7]
3 years ago
10

the top x shareholders in a corporation each own y shares of a certain stock. the corporations ownership is represented by a tot

al of w shares of stock. express the percent of the corporation owned by the top x shareholders
Mathematics
2 answers:
lord [1]3 years ago
8 0

Answer:

\frac{x*y}{w} *100

Step-by-step explanation:

To know how much shares have each shareholder you have to multiply the shareholders by the shares they own, so:

x*y

if we want to know the amount comparing to the total of shares of stock, you get:

\frac{x*y}{w}

And to know the percentage you have to multiply by 100:

\frac{x*y}{w} *100

BabaBlast [244]3 years ago
6 0
The total number of shares owned by x shareholders
= x*y
=xy shares

The percentage of shares=\frac{xy}{w} *100
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Can someone please help ?
N76 [4]
This is right solution:

12^2+9^2=The third side^2
144+81=The third side^2
225=The third side^2
√225=The third side
15=The third side

Hope it helps....
5 0
3 years ago
-5x+7y=-34<br> -5x+3y=-6
lisov135 [29]

Answer:

<h2>         x = -3,  y = -7</h2>

Step-by-step explanation:

-5x + 7y = -34     ⇒  -5x = -34  - 7y

    -5x + 3y = -6

-34  - 7y + 3y = -6

-34 - 4y = -6          {add 34 to both sides}

  -4y = 28             {divide both sides by (-4)}

     y = - 7

-5x + 3y = -6

-5x + 3(-7) = -6

-5x - 21 = -6        {add 21 to both sides}

-5x = 15            {divide both sides by (-5)}

  x = -3

7 0
2 years ago
The table shows the minutes students spent studying for their history exam and the scores they received.
Novosadov [1.4K]
Is there a graph maybe post a picture??
6 0
3 years ago
Jacob needs less than 5 C's on his transcript to qualify for UGA. He already has 1 C. At most, how many more C's can he get and
victus00 [196]
<span>C' x + 1 < 5 ; x < 4 ...............................................................................................</span>
6 0
3 years ago
Read 2 more answers
A recent survey based on a random sample of n=470 voters, predicted that independent candidate for the mayoral election will get
julsineya [31]

I have an expression

\sigma = \sqrt{p(1-p)/n}

floating around in my head; let's see if it makes sense.

The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

E( (b-p)^2 ) =  (-p)^2(1-p) + (1-p)^2p = p(1-p)

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

\sigma = \sqrt{p(1-p)/n}

Plugging in the numbers,

\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.

So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.

7 0
3 years ago
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