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Tanya [424]
3 years ago
6

If you can buy one can of pineapple chunks for $2 then how many can you buy with $10?(WITH PROPORTIONS PLEASE)

Mathematics
2 answers:
Alex73 [517]3 years ago
7 0

Answer:

5

Step-by-step explanation:

10 divided by 2 equalls 5

Vadim26 [7]3 years ago
3 0

Answer:

5 cans

Step-by-step explanation:

$2/1 can = $10/x cans

x = 5

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484 oranges and 350 grapefruit

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because 350 plus 134 equals 484

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Simplify the expression
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\bf \left( \cfrac{2}{5m^5} \right)^{-4}\implies \left( \cfrac{5m^5}{2} \right)^{4}\implies \cfrac{5^4m^{5\cdot 4}}{2^4}\implies \cfrac{625m^{20}}{16}
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Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

5 0
3 years ago
Solve for d<br> s = n /2 [ 2 a + ( n − 1 ) d ]
ludmilkaskok [199]

The answer is d= -2an+2s/n^2-n

That answer should be in a fraction form.

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