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marishachu [46]
3 years ago
12

There are 15 research doctors participating in the study and the research board needs to be established with the offices of dire

ctor, assistant director, quality control analyst, and correspondent. (Doctors can only hold one office on the research board.) Determine how many ways this research board can be chosen and explain your process.
Mathematics
2 answers:
VikaD [51]3 years ago
7 0

Answer: There are 32,760 ways that research board can be chosen.

Step-by-step explanation:

Since we have given that

Number of research doctors participating in the study = 15

the research board requires to be established with offices of director, Assistant director, Quality control analyst, and Correspondent.

Since there are 15 choices for the Offices of director,

There are 14 choices for the Assistant director,

There are 13 choices for Quality control analyst,

There are 12 choices for Correspondent.

So, by applying "Fundamental theorem of counting", we get that

Number of ways that research board can be chosen is given by

15\times 14\times 13\times 12\\\\=32,760

Hence, there are 32,760 ways that research board can be chosen.

Nat2105 [25]3 years ago
3 0

Answer:

32760

Step-by-step explanation:

There are 4 positions on the board.  There are 15 possible candidates for the first position.  That leaves 14 candidates for the second position.  Which leaves 13 for the third position, and 12 for the fourth position.

15 × 14 × 13 × 12 = 32760

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Question 1 (3 points)
Basile [38]

Answer:

The mean is 12, The median is 13, The mode is 14, and the range is 8.

The second choice is the correct answer.

Explanation:

  1. I found the mean by adding up all the numbers and dividing by the number of numbers, or sun divided by count. The sum of the numbers was 132, and the count was 11. I divided 132 by 11 to get 12.
  2. I listed the numbers in order from least to greatest. The data set was: 7, 8, 10, 12, 12, 13, 13, 14, 14, 14, 15
  3. I found the median by crossing off the number on each end of the data set, and kept going. For example, I crossed off 8, and then 15, and then 10, and then 14 and so on until I got to the middle number. (Keep in mind, if there are two middle numbers, find the mean of them [add them and divide by two])
  4. The mode is the number that shows up the most in the data set, so I looked at the data set and saw that 14 showed up 3 times, more times than any other number.
  5. The range is the difference between the greatest and least number (maximum and minimum) so I just subtracted 15 - 7 to get a difference of 8.
6 0
3 years ago
Kaya rented a limousine for prom. There was a one-time charge of $100 plus an hourly rate of the total
kherson [118]

Answer:

3 hours

Step-by-step explanation:

it is $100 for 1 hour right. Ok so forget about the hourly rates. Then you are left with $300. $300 total÷100 per hour = 3 hours.

3 0
2 years ago
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I need help like need it pls help
lara [203]
What’s the question tho?
5 0
3 years ago
Please help! :( i don't know what to do!
tigry1 [53]

Answer:

it is to small to see sorry

8 0
2 years ago
Read 2 more answers
A store sells Brazilian coffee for $10 per lb. and Columbian coffee for $14 per lb. If the store decides to make a 150-lb. blend
Artemon [7]

Answer:

The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.

Step-by-step explanation:

Let "b" be the amount of Brazilian coffee, in pounds, required for the blend and "c" the amount of Colombian coffee required, in pounds.

Since there are two unknown variables a two-equation system is needed to solve the problem, we can set up one equation for weight and another for price as follows:

b+c=150\\10*b+14*c=11*150

Solve for "c" by multiplying the first equation by -10 and adding it to the second one:

b+c=150\\10b+14c -10b-10c=11*150 -(10*150)\\4c=150\\c = 37.5

Now, solve for b by replacing the value obtained into the first equation

b+c=150\\b= 150 - 37.5\\b=112.5

The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.

6 0
3 years ago
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