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3 years ago

A key code must contain 2 letters and 3 numbers

Mathematics

2 answers:

Alchen [17]3 years ago

7 0

26 choices for each letter, 10 for each number, do 26×26×10×10×10

d1i1m1o1n [39]3 years ago

5 0

26 letters {A; B; C; D;...;Z}

10 numbers {0; 1; 2; 3;...;9}

L-letter; N-number

code
LLNN
LNLN
LNNL
NNLL
NLNL
NLLN

26 · 26 · 10 · 10 · 6 = 135,200

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2(x-8) + 4x = 6(x - 2) - 4

mrs_skeptik [129]

0=0, all numbers are solutions

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3 years ago

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Select all ratios equivalent to 5:3. 13:9 9:15 20:12

torisob [31]

Answer:

20:12

brainlest if it helped

Step-by-step explanation:

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2 years ago

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Hfuuf help me plsssss

denpristay [2]

Answer:

a) 26

b) 26/3

Step-by-step explanation:

1) First, you have to turn 4 1/3 into an improper fraction, so you get 13/3

2) Then you do 13/3 *6/1 =78/3 (so you multiply both numerators and denominators)

3) Lastly, 78/3 can be simplified as 26

1) First you turn both fractions into improper fractions, so you get 13/5 and 10/3

2) Then you do 13/5*10/3 (so you multiply both numerators and denominators)

3) You get 130/15, which can be simplified as 26/3

4 0

3 years ago

HELP QUICK! You must find all intervals where f(x) is negative. f(x)=2(x+3)(x-7)

beks73 [17]

Answer:

any # where -2<=x<7

Step-by-step explanation:

(x-7) would be negative and (x+3) positive, resulting in f(x) being negative.

4 0

2 years ago

A deck of 52 cards contains 12 picture cards. If the 52 cards are distributed in a random manner among four players in such a wa

Mkey [24]

Answer:

The probability that each player will receive three picture cards = 0.0324

Step-by-step explanation:

As given,

A deck of 52 cards contains 12 picture cards

Remaining card = 52 - 12 = 40

So,

Total number of ways in which 12 picture card is distributed = $\frac{12!}{3! 3! 3! 3!}$

Now,

The Total number of ways in which Remaining cards are distributed = $\frac{40!}{10! 10! 10! 10!}$

So,

Total number of ways of getting 3 picture card and remaining card = $\frac{12!}{3! 3! 3! 3!}$ × $\frac{40!}{10! 10! 10! 10!}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$

Now,

Total number of ways to distribute 52 cards so that each people get 13 card = $\frac{52!}{13! 13! 13! 13!} = \frac{52!}{ (13!)^{4} }$

∴ The probability = $\frac{\frac{12! 40!}{(3!)^{4} (10!)^{4} }}{\frac{52!}{(13!)^{4} }}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13!)^{4} }{ 52! }$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13.12.11.10!)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41.40! }$

= $\frac{12!}{(3!)^{4} }$ × $\frac{(13.12.11)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= $\frac{479,001,600}{(6)^{4} }$ × $\frac{(1716)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= 0.0324

∴ we get

The probability that each player will receive three picture cards = 0.0324

6 0

3 years ago