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djyliett [7]
2 years ago
10

10. Jan purchased 140 shares of stock in the ABC Company at a price of $18.75 per share. During the

Mathematics
1 answer:
andre [41]2 years ago
5 0

\bold{\huge{\orange{\underline{ Solution }}}}

\bold{\underline{ Given :- }}

  • <u>Jan </u><u>purchased </u><u>1</u><u>4</u><u>0</u><u> </u><u>shares </u><u>of </u><u>stock </u><u>in </u><u>ABC </u><u>company </u><u>at </u><u>a </u><u>price </u><u>of </u><u>$</u><u>1</u><u>8</u><u>.</u><u>7</u><u>5</u><u> </u><u>per </u><u>share </u>
  • <u>During </u><u>the </u><u>next </u><u>3</u><u> </u><u>days</u><u>, </u><u> </u><u>the </u><u>value </u><u>of </u><u>share </u><u>declined </u><u>by </u><u>$</u><u>1</u><u>.</u><u>0</u><u>0</u><u> </u><u>,</u><u> </u><u>$</u><u>1</u><u>.</u><u>7</u><u>5</u><u> </u><u>and </u><u>$</u><u>1</u><u>.</u><u>5</u><u>0</u>

\bold{\underline{ To \: Find :- }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>a </u><u>share </u><u>of </u><u>ABC </u><u>stock </u><u>at </u><u>the </u><u>end </u><u>of </u><u>3</u><u> </u><u>days </u>

\bold{\underline{ Let's \: Begin :- }}

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

Cost of 1 share of ABC company = $18.75

<u>Therefore</u><u>, </u>

Cost of 140 shares purchased by Jan in the ABC company

\sf{ = 140 × 18.75 }

\sf{ = 2625\: dollars }

<u>Now</u><u>, </u>

For next 3 days, the value of share declined

\sf{ = (1 + 1.75 + 1.50)dollars }

\sf{ = 4.25 \: dollars }

<u>Therefore</u><u>, </u>

The value of shares after 3 days will be

\bold{ = }{\bold{\frac{2625 - 4.25}{ 140}}}

\bold{ = }{\bold{\frac{ 2620.75}{ 140}}}

\bold{\red{ = 18.75\: dollars}}

Hence, The value of share after 3 days will be $18.75 .

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  • To answer 90% of calls instantly, the organization needs four extension lines.
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A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$$

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The likelihood that 2 servers will be busy may be calculated using the formula below.

P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$

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The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$

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The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$

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The probability that a call will receive a busy signal if four extensions lines are used is,

P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$

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