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Agata [3.3K]
3 years ago
15

-4

Mathematics
1 answer:
zavuch27 [327]3 years ago
8 0
Is this even a question
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LM is the diameter of OP. N L *not drawn to scale IfmXLMN = 48° what is mZNLM? A. 24° B. 42° C. 48° D. 66° E. 84​
Sindrei [870]

Answer:

c

Step-by-step explanation:

3 0
3 years ago
Mr. Harrow has eight boys and six girls in his Honors Pre-calculus class. If he randomly chooses two students, one at a time, wh
Sergeu [11.5K]

Answer: 15/91 which is choice B

=================================================================

There are two methods to find this answer.

Method 1) We have 6 girls and 8+6 = 14 students. The probability of picking a girl is 6/14 = 3/7. After the first girl is chosen, we have 5 girls left out of 14-1 = 13 students overall. The probability of picking another girl (assuming the first selection was a girl) is 5/13. Multiply these probabilities: (3/7)*(5/13) = (3*5)/(7*13) = 15/91

--------------------

Method 2) We can use the nCr combination formula. Order does not matter.

We have nCr = 6 C 2 = 15 ways to pick 2 girls. See the attached image below for the steps (figure 1)

Out of nCr = 14 C 2 = 91 ways to pick 2 students. See the attached image below for the steps (figure 2)

So that's another way to get the answer 15/91.

5 0
3 years ago
Read 2 more answers
The formula for finding the perimeter of equilateral triangle is P=3s. how can you transform the formula so that it could be use
Mandarinka [93]

Your answer is D. Hope this helps -John



5 0
3 years ago
Make r the formula t=r/r-3
Dafna11 [192]
R = dfrac3 t-1 + t. Brainliest please?
7 0
3 years ago
Read 2 more answers
What is the solution of |x – 2| > –3
Amanda [17]
Answer: The solution is the set of all real numbers (there are infinitely many solutions).

The reason why this is the case is because |x| is never negative. The smallest it can get is 0, which is larger than -3. That applies to |x-2| as well. So |x-2| is ALWAYS larger than -3 no matter what you pick for x. The smallest |x-2| can get is 0 and that happens when x = 2. Otherwise, the result is some positive value which is larger than -3.

So that's why |x-2| > -3 has infinitely many solutions. We can replace x with any real number we want, and the inequality would be true. 
6 0
3 years ago
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