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il63 [147K]
3 years ago
14

If two cards are drawn without replacement from a deck, find the probability that the second card is a spade, given that the fir

st card was a spade.
Mathematics
2 answers:
Dominik [7]3 years ago
6 0
We have 13 spades in a standard pack of cards.
Let's take cases.
Case 1: First card -- spade
Our sampling space is 52 cards, it isn't changed because we still haven't drawn anything.

Taken that it's a spade, our sampling space is reduced because it is dependent of our first event.

Case 2: Second card -- spade
Given that we don't replace the card, the second card is dependent of the first card. If we pick up a spade on the first draw, we have 51 cards to choose 12 spades.

(ie 4/17)
madam [21]3 years ago
5 0
There are 13 spades in a standard deck of cards, so if the first one is a spade, there are 12 spades left.

So the chances of getting another one would be 12/52 because there are 52 cards in a standard deck of cards.

Reduced, your answer is 3/13
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Find the mean of the integers<br> -16,-27,21,-19,14,-3
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2 years ago
Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.
Mice21 [21]

Answer:

<em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

Step-by-step explanation:

Making factors of 8y^{6}+ 144y^{5}+ 640y^{4}

Taking 8y^{4} common:

\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)

Using <em>factorization</em> method:

\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times  4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)

Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)

Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is 2y^{2} (y+ 10).

We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

Hence, <em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

5 0
2 years ago
determinati cel mai mic multiplu comun al urmatoarelor numere naturale a-36,48,54;b-30,45,75;c-48,60,72;d-90,24,72;e-75,100,150;
fenix001 [56]
<span>f-72,108,144 is the answer i hop it helps you</span>
8 0
2 years ago
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