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choli [55]
4 years ago
7

If You choose a card at random, what is the possibility of choosing an ace or a heart?

Mathematics
2 answers:
natima [27]4 years ago
8 0
RE-POSTING MY PREVIOUS ANSWER
***************************************************************
there are 13 hearts in every deck as well as 4 aces making 17 cards. However, you are counting the ace of hearts twice so you are trying to choose 16 cards out of 52.
So, the probability equals
16 / 52 = 4 / 13.

ASHA 777 [7]4 years ago
5 0

Answer:

4/13

Step-by-step explanation:

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The sum of two numbers is twenty-four. The difference of the two numbers is four.
RSB [31]

Answer:

x + y = 24

x - y = 4

Now, add these two equations.

You get,

2x = 28

x =  \frac{28}{2}

x = 14

Given,

x + y = 24

14 + y = 24

y = 24 - 14

y = 10

You can test this to the other equation as well.

x - y = 4

14 - 10 = 4

Hence, the two numbers are 14 and 10.

3 0
3 years ago
How many solutions can the equation 6x = 48 have?
ziro4ka [17]
Only one. 8

Divide by 6 on both sides. x = 8. That's it.
7 0
3 years ago
Please help I’ll give you points and mark you brainliest
elixir [45]
ANSWER:

i think it’s c
5 0
3 years ago
51-54 I dont understand help please
allsm [11]

Answer:

54

Step-by-step explanation:

54

8 0
3 years ago
What is the factorization of 729^15+1000​
Nesterboy [21]

\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 729=27^2\\ \qquad (3^3)^2\\ 1000=10^3 \end{cases}\implies 729^{15}+1000\implies ((3^3)^2)^{15}+10^3 \\\\\\ ((3^2)^{15})^3+10^3\implies (3^{30})^3+10^3\implies (3^{30}+10)~~[(3^{30})^2-(3^{30})(10)+10^2] \\\\\\ (3^{30})^3+10^3\implies (3^{30}+10)~~~~[(3^{60})-(3^{30})(10)+10^2]

now, we could expand them, but there's no need, since it's just factoring.

3 0
3 years ago
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