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Otrada [13]
3 years ago
15

Correct answers only please!

Mathematics
2 answers:
Svetlanka [38]3 years ago
7 0

Answer:

  D.  7

Step-by-step explanation:

Divide the number into odd-numbered and even-numbered digits (starting counting digit number 1 from the right).

  odd digits: d, 0, 5, 6, 3, 1, 8, 3 . . . . sum of 26+d

  even digits: 8, 1, 3, 2, 0, 3, 2, 4 . . . sum of 23

<em>Multiply the sum of even-numbered digits by 2, and add 1 for each even-numbered digit* that is 5 or greater: 23×2 +1 = 47</em>

<em>Add this to the sum of odd-numbered digits, and choose "d" so the result is a multiple of 10.</em>

  47 + (26+d) = 73+d

The next higher multiple of 10 is 80. For the total to be 80, d = 7.

_____

* The actual algorithm is described as "for each product of a digit and 2 that is greater than 9, subtract 9 from the product." Doing that, the number added to the sum will be ...

  • for 5: 2×5 - 9 = 1
  • for 6: 2×6 - 9 = 3
  • for 7: 2×7 - 9 = 5
  • for 8: 2×8 - 9 = 7
  • for 9: 2×9 - 9 = 9

In each case, the number contributed to the sum is 1 more than twice the digit, mod 10.

Novay_Z [31]3 years ago
3 0

Answer:

D.) 7

Step-by-step explanation:

To calculate the check digit, multiply every even-position digit (when counted from the right) in the number by two. If the result is a two digit number, then add these digits together to make a single digit (this is called the digital root):

Odd numbers:  (1.)= 8 (3.)= 4 (5.)= 6 (7.)= 0 (9.)= 4 (11.)= 6 (13.)= 2 (15.)=(1 + 6)= 7

Even numbers: (2.)= 3 (4.)= 8 (6.)= 1 (8.)= 3 (10.)= 6 (12.)=5 (14.)=0

To this total, we then add every odd-position digit:

Odd = 37

Even = 26

This will result in a total:

37 + 26 = 63

The check-digit is what number needs to be added to this total to make the next multiple of 10:

Next multiple of 10 is 70.

70 - 63 = 7

Answer:

7

Hope this helps : D

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anyanavicka [17]

The ratio is that of one term to the one before:

... 12/6 = 2

It is "common" because it applies to every pair of adjacent terms:

... 24/12 = 48/24 = 96/48 = 2

The appropriate choice is ...

... B 2

3 0
3 years ago
A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e
My name is Ann [436]

Answer:

\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)}  \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

Step-by-step explanation:

If the system contains n atoms, we can arrange r quanta of energy in

                         \binom{n}{r} = \dfrac{n!}{r!(n-r)!}

ways.

\mathbf{a)}

In this case,

                                n  = 2, r=1.

Therefore,

                    \binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2

which means that we can arrange 1 quanta of energy in 2 ways.

\mathbf{b)}

In this case,

                                n  = 20, r=10.

Therefore,

                    \binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756

which means that we can arrange 10 quanta of energy in 184 756 ways.

\mathbf{c)}

In this case,

                                n = 2 \times 10^{23}, r = 10^{23}.

Therefore, we obtain that the number of ways is

                    \binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

3 0
3 years ago
D square=3h+s<br>change of subject formula<br>make h the leader<br>​
Ksju [112]

Answer: h=(D²-s)/3

Step-by-step explanation: D² =3h+s

Subtract a from both sides of the equation

D²-s =3h+s-s

Simplifying,

D²-s =3h

Divide both sides by 3( to isolate 'h')

(D²-s)/3 =3h/3

Simplifying,

(D²-s)/3 =h

Hence, h= (D²-s)/3

Hope it helps!

7 0
2 years ago
Question 1: Use the image and your knowledge of the isosceles triangle to find the value of x
pochemuha

Answer:

35 degrees

Step-by-step explanation:

since both sides are equal, so are the angles involved.

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3 years ago
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