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2 years ago

10

a building security code has 2 numeric digits 0 through 9 followed by 2 letters. what is the probability that the first digit is

nine an last letter is A?

2 answers:

sleet_krkn [62]2 years ago

7 0

If we remove human choices in creating the code the answer is 10 (the possible choices) times 26 (the possible choices) so 260 is the probability

svp [43]2 years ago

5 0

Answer: $\dfrac{1}{260}$

Step-by-step explanation:

Given : A building security code has 2 numeric digits 0 through 9 followed by 2 letters.

Total digits (0,1,2,3,4,5,6,7,8,9)= 10

Total letters in English Alphabet = 26

Then, the total number of ways to make a code if repetition is allowed :-

$10\times10\times26\times26=67600$

The number of ways to make a code with first digit is nine an last letter is A (we fix first position for 9 as 1 and the last position for a as 1) , if repetition is allowed :-

$1\times10\times26\times1=260$

Then, the probability that the first digit is nine an last letter is A :-

$=\dfrac{260}{67600}=\dfrac{1}{260}$

Hence, the probability that the first digit is nine an last letter is A= $\dfrac{1}{260}$

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In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
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P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
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2 years ago

Lydia runs an experiment to determine if a coin is fair by counting the number of times a coin lands heads up. The table shows h

kakasveta [241]

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Step-by-step explanation:

5 0

1 year ago

Read 2 more answers

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dimaraw [331]

Answer:

part A 1.68

part B 29.67

Step-by-step explanation:

First you multiply $39.99 dollars and 0.30 = $11.9970 you round this to $12.00

After you apply the discount $39.99 - $12.00 = 27.99

Next you must multiply 27.99*0.06 =1.6794 and round to 1.68 to get tax

Last you add $27.99 and 1.68 to get a total of $29.67 for the headphones

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4 0

2 years ago

an artist is arranging tiles in rows to decorate a wall. each new row has 2 fewer tiles than the row below it. if the first row

valentinak56 [21]

we know that

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and

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$an=a1+d*(n-1)$

where

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d is the common difference

so

$a1=23\\\\ d=-2$

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<u>the answer is</u>

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1 year ago

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Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll

JulijaS [17]

Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

$Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So, the possibility of getting a total of 7 = $\frac{6}{36}=\frac{1}{6}$

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = $\frac{2}{36}$ = $\frac{1}{18}$

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = $\frac{28}{36}$ = $\frac{7}{9}$

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

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$\frac{11}{3}+\frac{11}{3}-\frac{77}{9}$

$\frac{22}{3}-\frac{77}{9}$

$\frac{66-77}{9}$

$-\frac{11}{9}$

$-1.22$

Therefore the expected payoff for this game is -$1.22.

4 0

1 year ago