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Setler [38]
3 years ago
9

A card is Chosen at random from a deck of 52 cards it is replaced and a second card is chosen what is the probability both cards

will be an ace?
A) 1/169
B) 2/13
C) 1/26
D) 1/13
Mathematics
2 answers:
ankoles [38]3 years ago
6 0
There are 4 aces in a deck of 52 cards. the odds of drawing one is 1/13. the probability of doing it again is still 1/13. the odds of doing it twice, as described, is then 1/13 * 1/13, or (1/13)², or 1/169 option A
Gemiola [76]3 years ago
3 0

Answer:

option A

Step-by-step explanation:

A card is Chosen at random from a deck of 52 cards

there are 4 aces in the deck of 52 cards

Probability of picking a ace = total ace cards divide by total number of cards

P(first ace)=\frac{4}{52} =\frac{1}{13}

it is replaced and a second card is chosen

So total cards = 52 and ace =4

P(second ace)=\frac{4}{52} =\frac{1}{13}

P(both are ace)=P(first ace) \cdot P(second ace)

=\frac{1}{13}  \cdot \frac{1}{13} =\frac{1}{169}

option A

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Find the value of x such that 365 based seven + 43 based x = 217 based 10.
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We need to find the base x in the following equation:

365_7+43_x=217_{10}

First, lets convert 365 from base 7 to base 10. This is given by

365_7=3\times7^2+6\times7^1+5\times7^0

where the upperindex denotes the position of eah number. This gives

\begin{gathered} 365_7=3\times49+6\times7+5\times1 \\ 365_7=147+42+5 \\ 365_7=194_{10} \end{gathered}

that is, 365 based 7 is equal to 194 bases 10.

Now, lets do the same for 43 based x. Lets convert 43 based x to base 10:

43_x=4\times x^1+3\times x^0

where again, the superindex 0 and 1 denote the position 0 and 1 in the number 43. This gives

43_x=(4x+3)_{10}

Now, we have all number in base 10. Then, our first equation can be written in base 10 as

194_{10}+(4x+3)_{10}=217_{10}

For simplicity, we can omit the 10 and get

194+4x+3=217

so, we can solve this equation for x. By combining similar terms. we have

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and by moving 197 to the right hand side, we obtain

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Finally, we get

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Therefore, the solution is x=5

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