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koban [17]
2 years ago
12

how many different 3 digit numbers can be formed using the digits 9,3,4,7 and 6? assume no number can be used more than once

Mathematics
1 answer:
RSB [31]2 years ago
3 0

Answer:

57

Step-by-step explanation:

976/974/973/967/964/963/947/946/943/937/936/934/796/794/793/769/764/763/749/746/743/739/736/734/697/694/693/679/674/673/649/647/643/639/637/634/497/496/493/479/476/473/439/437/436/397/396/394/379/376/374/369/367/364/349/347/346

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Read 2 more answers
2: A bag contains 5 black counters, 4 blue counters, and
drek231 [11]

Answer:

The probability that the counter  was blue is \mathbf{\frac{2}{5}}

Step-by-step explanation:

Number of black Counters = 5

Number of blue Counters = 4

Number of white Counters = 1

We need to write down the probability that the counter  was blue.​

First find Total Counters

Total Counters = Number of black Counters + Number of blue Counters + Number of white Counters

Total Counters = 5+4+1

Total Counters = 10

Now, we need to find probability that the counter taken was blue

The formula used is:

Probability= \frac{Number\:of\:favourable\:outcomes}{Total\:outcomes}

There are 4 blue counters in the back, so Favourable outcomes = 4

Probability= \frac{Number\:of\:favourable\:outcomes}{Total\:outcomes}\\Probability= \frac{4}{10}\\Probability= \frac{2}{5}

The probability that the counter  was blue is \mathbf{\frac{2}{5}}

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3 years ago
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X+y+z=12<br> 6x-2y+z=16<br> 3x+4y+2z=28<br> What does x, y, and z equal?
lianna [129]

Answer:

x = 20/13 , y = 16/13 , z = 120/13

Step-by-step explanation:

Solve the following system:

{x + y + z = 12 | (equation 1)

6 x - 2 y + z = 16 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Swap equation 1 with equation 2:

{6 x - 2 y + z = 16 | (equation 1)

x + y + z = 12 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/6 × (equation 1) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+(4 y)/3 + (5 z)/6 = 28/3 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Multiply equation 2 by 6:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/2 × (equation 1) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+5 y + (3 z)/2 = 20 | (equation 3)

Multiply equation 3 by 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+10 y + 3 z = 40 | (equation 3)

Swap equation 2 with equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+8 y + 5 z = 56 | (equation 3)

Subtract 4/5 × (equation 2) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+(13 z)/5 = 24 | (equation 3)

Multiply equation 3 by 5:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+13 z = 120 | (equation 3)

Divide equation 3 by 13:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract 3 × (equation 3) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y+0 z = 160/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 2 by 10:

{6 x - 2 y + z = 16 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Add 2 × (equation 2) to equation 1:

{6 x + 0 y+z = 240/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract equation 3 from equation 1:

{6 x+0 y+0 z = 120/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 1 by 6:

{x+0 y+0 z = 20/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Collect results:

Answer:  {x = 20/13 , y = 16/13 , z = 120/13

8 0
3 years ago
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