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Masteriza [31]
3 years ago
13

A password has to use the following format: LDDDDLLL, where L can be any of the upper case letters in the alphabet, and D can be

any digit including 0. None of the digits or letters can repeat. What is the probability that a random password uses only vowels (A,E,I,O,U) and odd numbers (1,3,5,7,9)? Show work but do not evaluate. What is the probability that the password spells the word “MATH?”
Mathematics
1 answer:
maw [93]3 years ago
6 0

Answer:

Only vowels and odd numbers:

P = \frac{5}{26} *\frac{4}{25}* \frac{3}{24}* \frac{2}{23} *\frac{5}{10}* \frac{4}{9} *\frac{3}{8} *\frac{2}{7}

Spells math:

P = \frac{1}{26} *\frac{1}{25}* \frac{1}{24}* \frac{1}{23}

Step-by-step explanation:

We have four letters, so the probability that one letter is a vowel is 5/26 (we have 5 vowels in a total of 26 letters), then the second letter has a probability of 4/25 of being a vowel (1 vowel used), and so on (third letter being vowel = 3/24 and fourth letter being vowel = 2/23)

Then, for the digits, we do the same, one digits has 5/10 probability of being odd, then the second digit has 4/9, the third has 3/8 and the fourth has 2/7.

So the final probability would be:

P = \frac{5}{26} *\frac{4}{25}* \frac{3}{24}* \frac{2}{23} *\frac{5}{10}* \frac{4}{9} *\frac{3}{8} *\frac{2}{7}

To find the probability that the password spells the word “MATH", each letter has to be a specific letter, so the first letter has 1/26 probability, the second has 1/25, and so on:

P = \frac{1}{26} *\frac{1}{25}* \frac{1}{24}* \frac{1}{23}

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50 Points!
Advocard [28]
Well, 50 points total given, 25 points per user  and 13 bonus for brainliest

anyway


1.
add them equations, the y's will cancel
x+2y=4
<span>3x-2y=4 +</span>
4x+0y=8

4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution


2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
<span>-4x+14y=-6 +</span>
0x+0y=0
0=0
infinite solutions

that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions



3.
multiply first equation by 2 and add to first
4x+2y=-6
<span>1x-2y=-4 +</span>
5x+0y=-10

5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1

x=-2
y=1
(x,y)
(-2,1)


4.
coincident means they are the same line
so

we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16


5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'


EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a


40 tickets, c+a
40=c+a

the equations are
150=3c+5a and
40=c+a

eliminate
multiply 2nd equaton by -3 and ad to first one

150=3c+5a
<span>-120=-3c-3a +</span>
30=0c+2a

30=2a
divide by 2
15=a

sub back
40=c+a
40=c+15
minus 15
25=c

25 children tickets and 15 adult tickets were sold


ANSWERS:

1.
(2,1) is solution

2.
infinite solutions

3.
(-2,1)

4.
B=10 and C=16

5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
 'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
the equations are
150=3c+5a and
40=c+a
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