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attashe74 [19]
3 years ago
14

Luc Do purchased stocks for $6,000. He paid $4,000 in cash and borrowed $2,000 from the brokerage firm. He bought 100 shares at

$60.00 per share ($6,000 total). The loan has an annual interest rate of 8 percent. Six months later, Luc Do sold the stock for $65 per share. He paid a commission of $120 and repaid the loan. What was the net profit?
Mathematics
1 answer:
Elina [12.6K]3 years ago
5 0

Answer:His net profit is $300

Step-by-step explanation:

Total amount borrowed is $2000. The loan has an annual interest rate of 8 percent. It was borrowed for only six months. This amount of interest paid will be determined using the simple interest formula,

I = PRT/100

Where

P = principal = $2000

R = rate = 8%

T = time = 6 months = 0.5 years

I = interest

I = (2000× 8 × 0.5)/100 = 80

The interest that will be paid for $2000 is $80

Total amount paid back is 2000 + 80 = $2080

He bought 100 shares at $60.00 per share and the total is $6,000

He sold the stock for $65 per share. This means that the total amount for which he sold the stock is 65×100 = $6500.

To determine his profit, we would subtract the total amount paid back due to the loan, the commission and his initial cash from the total amount recovered from selling his stock.

It becomes

6500 - ( 2080 + 120 + 4000)

= 6500 - 6200 = 300

His net profit is $300

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The exam scores for 200 students are normally distributed with a mean of 72 and a standard deviation of 10. Which answer choice
Paha777 [63]

Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

\mu = 72, \sigma = 10

The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.

X = 82:

Z = \frac{X - \mu}{\sigma}

Z = \frac{82 - 72}{10}

Z = 1

Z = 1 has a p-value of 0.84.

X = 72:

Z = \frac{X - \mu}{\sigma}

Z = \frac{72 - 72}{10}

Z = 0

Z = 0 has a p-value of 0.5.

0.84 - 0.5 = 0.34.

Out of 200 students, the number is given by:

0.34 x 200 = 68 students with scores between 72 and 82.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

6 0
2 years ago
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