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murzikaleks [220]

3 years ago

11

a company has 95,000 shares of stock to sell. A stock broker decides to buy 150 shares of the stock for each of her 250 clients.

How many shares will the company have left to sell after the stockbroker makes her purchase?

2 answers:

Ivahew [28]3 years ago

8 0

There would be 57,500 shares left to sell

Nat2105 [25]3 years ago

3 0

Answer:

57,500 shares will the company have left to sell after the stockbroker makes her purchase.

Step-by-step explanation:

A company has 95,000 shares of stock to sell.

A stock broker decides to buy 150 shares for each of her 250 clients.

For each 250 clients 150 shares = 250 × 150 = 37,500

Stock broker needs 37,500 shares of stock

After broker's purchase = 95,000 - 37,500 = 57,500

57,500 shares

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mihalych1998 [28]

Answer:

$=4.+2.$

Step-by-step explanation:

<u>Linear Combination Of Vectors </u>

One vector $\vec b$ is a linear combination of $\vec a_1$ and $\vec a_2$ if there are two scalars $x_1, x_2$ such as

$\vec b=x_1\vec a_1+x_2\vec a_2$

In our case, all the vectors are given in $R^3$ but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.

We have

$\vec a_1=,\ \vec a2=,\ \vec b=$

We set the equation

$=x_1.+x_2.$

Multiplying both scalars by the vectors

$=+$

Equating each coordinate, we get

$4x_1-4x_2=8$

$5x_1+3x_2=26$

$-4x_1+3x_2=-10$

Adding the first and the third equations:

$-x_2=-2$

$x_2=2$

Replacing in the first equation

$4x_1-4(2)=8$

$4x_1=8+8$

$x_1=4$

We must test if those values make the second equation become an identity

$5(4)+3(2)=20+6=26$

The second equation complies with the values of $x_1$ and $x_2$ , so the solution is

$=4.+2.$

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