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murzikaleks [220]
3 years ago
12

There are six professors teaching the introductory discrete mathematics class at someuniversity. The same final exam is given by

all five professors. The exam score is aninteger between 0 and 100 (inclusive, that is, both 0 and 100 are possible). How manystudents must there be to guarantee that there are two students with the same professorwho earned the same final examination score?
Mathematics
2 answers:
musickatia [10]3 years ago
5 0

Answer:

Step-by-step explanation:

zysi [14]3 years ago
5 0

Answer:

607 students to guarantee that there are two students with the same professorwho earned the same final examination score.

Step-by-step explanation:

This problem is an example of the Pigenhole principle.

The first step is finding the number of boxes and objects:

For each score, we have a box which contains the student who got that score.

If there were only one professor grading, there would need to be 101+1 = 102 students to ensure that that there are two students with the same professorwho earned the same final examination score.

However, for each student, there are a combination of six to five = 6 possible combinations for a score.

So there should be at least 6*101+1 = 607 students to guarantee that there are two students with the same professorwho earned the same final examination score.

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This figure represents a platform on a stage. The stage crew needs to paint it black. One gallon of paint will cover 40 square y
ICE Princess25 [194]

Answer:

14

Step-by-step explanation:

I took the test.

5 0
2 years ago
Having a small brain moment, can you please help and show work
ankoles [38]

Answer:

The value if x is 0.9875.

Step-by-step explanation:

First, we have to get rid of brackets by expanding :

0.5(2x +  \frac{3}{4} ) -  \frac{1}{3} (0.1 + x) = 1

0.5(2x) + 0.5( \frac{3}{4} ) -  \frac{1}{3} (0.1) -  \frac{1}{3} (x) = 1

x +  \frac{3}{8}  -  \frac{1}{30}   -   \frac{1}{3} x = 1

Next you have to collect like terms :

x  -  \frac{1}{3} x +  \frac{3}{8}  -  \frac{1}{30}  = 1

\frac{2}{3} x +  \frac{41}{120}  = 1

Lastly, you can solve x :

\frac{2}{3} x = 1 -  \frac{41}{120}

\frac{2}{3} x =  \frac{79}{120}

x =  \frac{79}{120}  \div  \frac{2}{3}

x =  \frac{79}{80}

x = 0.9875

8 0
3 years ago
Use the distributive property to simplify the expression: 5(-3x-7)
mixer [17]
5(-3x-7) (first multiply 5 to -3x then to -7)
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The answer is D) -15x-35
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3 years ago
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kirill115 [55]
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Rama09 [41]
$80

add what you took out
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