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Novosadov [1.4K]
3 years ago
13

NEED HELP ON HOMEWORK ASAP

Mathematics
1 answer:
Alexandra [31]3 years ago
4 0

Answer:c

Step-by-step explanation:

3*1= 3 3*2=6 and so on

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Please help with numbers 1 and 2 !!!
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Well, this is a stem and leaf problem. 

The numbers are 20, 28, 29, 29, 31, 32, 36, 41, and 42. 

In order to solve for the mean, we find the average and we add the numbers together and divide by 9(there are 9 numbers). 

The mean is 32.
The median is the middle number, or 31. 
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3 years ago
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Suppose Colby rolls the number cube 1000 times, about how many times can she expect to roll an odd number?​
DiKsa [7]

Since there are 3 odd numbers ( 1, 3 and 5) and 3 even numbers ( 2, 4 and 6) in the cube ,

The chance of getting an odd number is 50/50 or 50%

So out of 1000 she will probably get 500 odd since 1000 × 0.5 (which is 50%) is 500

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4 years ago
Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv
Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number x such that taking it mod 4, 5, and 7 leaves the desired remainders:

x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6

  • Taken mod 4, the last two terms vanish and we have

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4

so we multiply the first term by 3.

  • Taken mod 5, the first and last terms vanish and we have

x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5

so we multiply the second term by 2.

  • Taken mod 7, the first two terms vanish and we have

x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7

so we multiply the last term by 7.

Now,

x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147

By the CRT, the system of congruences has a general solution

n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}

or all integers 27+140k, k\in\mathbb Z, the least (and positive) of which is 27.

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3 years ago
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Answer:

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mart [117]
Answer: 11, 7

11 + 7 = 18
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3 years ago
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