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Otrada [13]
4 years ago
12

Anthony got a new iPhone. He needs to create a 4-digit passcode. How many passcodes can be created if REPETITION OF DIGITS IS AL

LOWED? Please show your work.
Mathematics
1 answer:
Snezhnost [94]4 years ago
7 0
An infinite amount...since there ARE and infinite amount of number in the universe.

1111

2222

3333

4444

5555

6666

7777

8888

9999


1010

1234

4321

5432

2345

5678

8765

0987

7890

4509

9054

Etc
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Plz answer will get brainliest
rjkz [21]

Answer:

4

Step-by-step explanation:

1/3 of 30 is 10

2/5 of 10 is 4

8 0
3 years ago
Dan rolls 2 fair dice and adds the results from each.
cricket20 [7]

Answer:

2/36

Step-by-step explanation:

He will have a probability of getting a total of 3 at 5.56% out of 100%.

3 0
3 years ago
Write the five rational number which can be expressed has terminating decimal​
Tatiana [17]

Answer:

1/3

1/7

1/11

1/13

1/17

Step-by-step explanation:

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8 0
3 years ago
What percentage is 15 out of 17?
Ket [755]
88.23%

Simply plug 15/17 into you calculator 

and it will give you a decimal, in this case it should be .88235294...

simply move the decimal back to placements and you should have

88.235294... thats still a lot of numbers so just reduce it to 2 places after the decimal and add a percent sign

88.23%

and your done :) this works for any similar problems

I hope this helps

8 0
3 years ago
Read 2 more answers
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.

3 0
3 years ago
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