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morpeh [17]
3 years ago
7

9. To order books from an online site, the buyer must open an account. The buyer needs a username and a password.

Mathematics
1 answer:
slega [8]3 years ago
7 0

Answer:

a)

i)208827064576

ii)62990928000

b)

i)28179280429056

ii)15214711438080

c)

i)899194740203776

ii)607681858331520

Step-by-step explanation:

a)

i)The alphabet consists of 26 letters.

For the first letter of the username there are 26 options.

For the second letter of the username there are 26 options.

For the third letter of the username there are 26 options.

For the fourth letter of the username there are 26 options.

For the fifth letter of the username there are 26 options.

For the sixth letter of the username there are 26 options.

For the seventh letter of the username there are 26 options.

For the eighth letter of the username there are 26 options.

To get the number of the possible usernames we multiply the number of options for each letter: 26 x 26 x 26 x 26 x 26 x 26 x 26 x 26 =26^8=208827064576

ii) If letter cannot be repeated.

For the first letter of the username there are 26 options.

For the second letter of the username there are only 25 options.

For the third letter of the username there are only 24 options.

For the fourth letter of the username there are only 23 options.

For the fifth letter of the username there are only 22 options.

For the sixth letter of the username there are only 21 options.

For the seventh letter of the username there are only 20 options.

For the eighth letter of the username there are only 19 options.

To get the number of the possible usernames we multiply the number of options for each letter: 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 =62990928000

b) 26 letters+10digits+12special characters=48 options

i) For the first letter of the password there are 48 options.

For the second letter of the password there are 48 options.

For the third letter of the password there are 48 options.

For the fourth letter of the password there are 48 options.

For the fifth letter of the password there are 48 options.

For the sixth letter of the password there are 48 options.

For the seventh letter of the password there are 48 options.

For the eighth letter of the password there are 48 options.

To get the number of the possible usernames we multiply the number of options for each letter: 48 x 48 x 48 x 48 x 48 x 48 x 48 x 48= 48^8=28179280429056

ii)  If letter cannot be repeated.

For the first letter of the password there are 48 options.

For the second letter of the password there are only 47 options.

For the third letter of the password there are only 46 options.

For the fourth letter of the password there are only 45 options.

For the fifth letter of the password there are only 44 options.

For the sixth letter of the password there are only 43 options.

For the seventh letter of the password there are only 42 options.

For the eighth letter of the password there are only 41 options.

To get the number of the possible usernames we multiply the number of options for each letter: 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 =15214711438080

c) 26 uppercase letters+26 lowercase letters +10digits+12special characters=74 options

i) For the first letter of the password there are 74 options.

For the second letter of the password there are 74 options.

For the third letter of the password there are 74 options.

For the fourth letter of the password there are 74 options.

For the fifth letter of the password there are 74 options.

For the sixth letter of the password there are 74 options.

For the seventh letter of the password there are 74 options.

For the eighth letter of the password there are 74 options.

To get the number of the possible usernames we multiply the number of options for each letter: 74 x 74 x 74 x 74 x 74 x 74 x 74 x 74 =74^8=899194740203776

ii) If letter cannot be repeated.

For the first letter of the password there are 74 options.

For the second letter of the password there are only 73 options.

For the third letter of the password there are only 72 options.

For the fourth letter of the password there are only 71 options.

For the fifth letter of the password there are only 70 options.

For the sixth letter of the password there are only 69 options.

For the seventh letter of the password there are only 68 options.

For the eighth letter of the password there are only 67 options.

To get the number of the possible usernames we multiply the number of options for each letter: 74 x 73 x 72 x 71 x 70 x 69 x 68 x 67 =607681858331520

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saul85 [17]

Answer:

Option B is the correct answer

Step-by-step explanation:

<u>A line segment is a line drawn with two end points.</u>

1.) Option A is a point

2.) <u>Option B is a line segment</u>

3.) Option C is a ray

4.) Option D is a line

Hope this helps!

8 0
3 years ago
The set S contains some real numbers, according to the following three rules. (i) 1 1 is in S. (ii) If a b is in S, where a b is
pogonyaev

Answer:

The solution is given in the pictures below

Step-by-step explanation:

4 0
3 years ago
How do I find the area of a circle and the radius is 4
Virty [35]
Area= pi(radius)squared
6 0
3 years ago
Read 2 more answers
Hi I have no idea how to do this question :)
erik [133]

Answer:

A. 36

Step-by-step explanation:

Because the two angles at the bottom of both the triangles are congruent, the two triangles are similar. Therefore, the corresponding side lengths of the triangles must be proportional to each other

if the area of a triangle is hb/2,

(with h=length of height, b=length of base)

the height of the first triangle is:

25 = 10h/2

50 = 10h

h=5

you can write ratios representing the proportions of the two triangles:

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then find the area of the second triangle with the height (6) and base (12)

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7 0
2 years ago
Let $\overline{AB}$ and $\overline{CD}$ be chords of a circle, that meet at point $Q$ inside the circle. If $AQ = 6,$ $BQ = 12$,
Cloud [144]

9514 1404 393

Answer:

  2

Step-by-step explanation:

The products of chord lengths are the same for the intersecting chords:

  AQ×BQ = CQ×DQ

  6×12 = CQ×(38 -CQ)

This gives a quadratic in CQ:

  CQ² -38CQ +72 = 0 . . . . . write in standard form

  (CQ -2)(CQ -36) = 0 . . . . . factor the quadratic

  CQ = 2 or 36 . . . . . . . values of CQ that make the factors zero

The minimum length of CQ is 2 units. (DQ will be 36.)

3 0
3 years ago
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