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Vaselesa [24]
4 years ago
11

How many different groups of students can show up for a seminar with an enrollment of 12 ​?

Mathematics
1 answer:
nexus9112 [7]4 years ago
8 0

The solution would be like this for this specific problem:

 

12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

= 479,001,600

 

<span>So there are 479,001,600 <span>different group of students that can show up for a seminar with an enrollment of 12.</span></span>
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Consider that your profit-maximizing quantity of 500 can be produced at an average cost per unit of $9 and sold for a market pri
timama [110]

Based on the quantity sold and the price, the profit per unit is <u>$5</u> and the total profit is<u> $2,500</u>

The profit made from each unit sold can be found as:

<em>= Market price - Average cost </em>

= 14 - 9

= $5

The total profit would therefore be:

<em>= Profit per unit x Number of units sold </em>

= 5 x 500

= $2,500

In conclusion, the profit per unit is $5 and the total profit is $2,500

Find out more about profit per unit at brainly.com/question/25531124.

4 0
2 years ago
While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that vary between
aleksandr82 [10.1K]

Answer:

The probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

Step-by-step explanation:

Let the random variable <em>X</em> denote the water depths.

As the variable water depths is continuous variable, the random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 2.00 m and <em>b</em> = 7.00 m.

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{b-a};\ a

Compute the probability that a randomly selected depth is between 2.25 m and 5.00 m as follows:

P(2.25

                               =\frac{1}{5.00}\int\limits^{5.00}_{2.25} {1} \, dx\\\\=0.20\times [x]^{5.00}_{2.25} \\\\=0.20\times (5.00-2.25)\\\\=0.55

Thus, the probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

6 0
3 years ago
A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now
algol [13]

The answers to the bearing problems are listed below:

  1. How far North is the ship now ___________ 38.82 km
  2. How far East is the ship now ____________ 144.89 km
  3. How far is the ship from its starting point_____241.48 km
  4. On what bearing is it now from its original position____035°

<h3>Meaning of bearing.</h3>

Bearing can defined as branch of mathematics that describes the accurate location of an object at any point in time.

<h3>Analysis</h3>

The answers to the problem from 1-3 can be gotten by using sin and cos to find the missing sides.

The final value can be gotten using the cosine rule

4). bearing from original position = cos^{-1} (a^{2} + b^{2}﹣c^{2}) / 2ab = 35°

In conclusion, the answers for each is given in the list above.

Learn more about Bearing: brainly.com/question/24142612

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8 0
2 years ago
The first term of an arithmetic sequence is -5, and the tenth term is 13. find the common difference
Vilka [71]

Answer:

2

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Statistics
user100 [1]

Answer:

Step-by-step explanation:

The lower bound is 10.5 and the upper bound is 15.9

Hope it helps!

8 0
2 years ago
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