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Ganezh [65]
3 years ago
9

Which expression is equivalent to 54n-20m+6n

Mathematics
2 answers:
azamat3 years ago
6 0

Answer:

C, or

the 3rd answer, or

594 > -11n

Step-by-step explanation:

Eva8 [605]3 years ago
4 0

Answer:

C. 594 > -11n

Step-by-step explanation:

i took the test on edgen

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In the field of quality​ control, the science of statistics is often used to determine if a process is​ "out of​ control." Suppo
Usimov [2.4K]

Answer:

a) P=0.0225

b) P=0.057375

Step-by-step explanation:

From exercise we have that  15​% of items produced are defective, we conclude that  probabiity:

P=15/100

P=0.15

a)  We calculate the probabiity  that two items are​ defective:

P= 0.15 · 0.15

P=0.0225

b)  We calculate the probabiity  that  two  of  three items are​ defective:

P={3}_C_{2} · 0.85 · 0.15 · 0.15

P=\frac{3!}{2!(3-2)!} · 0.019125

P=3 · 0.019125

P=0.057375

6 0
3 years ago
I need help with this question I'm not sure what the answer is
Tpy6a [65]

Answer:15/17

Step-by-step explanation: cos = adjacent/hypotenuse

3 0
3 years ago
‼️ WILL LIST BRAINLIEST ‼️ -5/6 x -3/4​
ella [17]

Answer:

5/8

that's the answer:)

6 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE IM STRUGGLING
zepelin [54]
I think the answer is either B or C I am not completely sure but I got 36
4 0
2 years ago
If the Math Olympiad Club consists of 11 students, how many different teams of 3 students can be formed for competitions?
vivado [14]

Answer:

165 different teams of 3 students can be formed for competitions

Step-by-step explanation:

Combinations of m elements taken from n in n (m≥n) are called all possible groupings that can be made with the m elements so that:

  • Not all items fit
  • No matter the order
  • Elements are not repeated

That is, a combination is an arrangement of elements where the place or position they occupy within the arrangement does not matter. In a combination it is interesting to form groups and their content.

To calculate the number of combinations, the following expression is applied:

C=\frac{m!}{n!*(m-n)!}

It indicates the combinations of m objects taken from among n objects, where the term "n!" is called "factorial of n" and is the multiplication of all the numbers that go from "n" to 1.

In this case:

  • n: 3
  • m: 11

Replacing:

C=\frac{11!}{3!*(11-3)!}

Solving:

C=\frac{11!}{3!*8!}

being:

  • 3!=3*2*1=6
  • 8!=8*7*6*5*4*3*2*1=40,320
  • 11!=39,916,800

So:

C=\frac{39,916,800}{6*40,320}

C= 165

<u><em>165 different teams of 3 students can be formed for competitions</em></u>

3 0
3 years ago
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