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2 years ago

Calculate the loss on selling 50 shares of stock originally bought at 13 and 3/4 and sold at 12

1 answer:

4 0

Note that 13 3/4 = 13.75.

The cost of purchasing the stock is
(50 shares)*(13.75 $/share) = $687.50

The revenue generated by selling the 50 shares at $12 per share is
(50 shares)*(12 $/share) = $600.00

Gain = 600.00 - 687.50 = - $87.50
The loss is $87.50

Answer: $87.50

You might be interested in

What is 1345/99 rounded to the nearest integer

ludmilkaskok [199]

1345/99 = 15.5858
15.5858 round up because 5+ is round up while 4- is round down.
Your answer is 16.

7 0

2 years ago

Read 2 more answers

Suppose that 3\%3%3, percent of over 200{,}000200,000200, comma, 000 books borrowed from a library in a year are downloaded. The

sleet_krkn [62]

Answer:

The mean of the sampling distribution of the proportion of downloaded books is 0.03 and the standard deviation is 0.0197.

Step-by-step explanation:

Central Limit Theorem

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

3% of books borrowed from a library in a year are downloaded.

This means that $p = 0.03$

SRS of 75 books.

This means that $n = 75$

What are the mean and standard deviation of the sampling distribution of the proportion of downloaded books

By the Central Limit Theorem

Mean: $\mu = p = 0.03$

Standard deviation: $s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.03*0.97}{75}} = 0.0197$

The mean of the sampling distribution of the proportion of downloaded books is 0.03 and the standard deviation is 0.0197.

4 0

2 years ago

A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter

faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let c (t ) denote the amount of chlorine (in grams) in the tank at time t .

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(c (t )/(1000 + (10 - 25)t ) g/L) * (25 L/s) = 5c (t ) /(200 - 3t ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after t s there will be (1000 + 10t ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time t is (1000 - 15t ) L. Then the concentration of chlorine per unit volume is c (t ) divided by this volume.

So the amount of chlorine in the tank changes according to

$\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}$

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5t )^(5/3), then integrate both sides to solve for c (t ):

$\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0$

$\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0$

$\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0$

$\dfrac{c(t)}{(200-3t)^{5/3}}=C$

$c(t)=C(200-3t)^{5/3}$

There are 20 g of chlorine at the start, so c (0) = 20. Use this to solve for C :

$20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}$

$\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}$

7 0

2 years ago

A carload of steel rods has arrived at Cybermatic Construction Company. The car contains 50,000 rods. Claude Ong, manager of Qua

amid [387]

Answer:

There is a 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch. This means that $\mu = 120, \sigma = 0.05$ .

Claude Ong, manager of Quality Assurance, directs his crew measure the lengths of 100 randomly selected rods. This means that $n = 100, s = \frac{\sigma}{\sqrt{n}} = \frac{0.05}{\sqrt{100}} = 0.005$ .