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Naily [24]
3 years ago
6

Adam goes to the grocery store to buy cereal. the shelves contain 9 boxes of Brand A and 6 boxes of Brand B. He selects one bran

d at random and then put it back. Another person does the same thing. What is the probability they both selected Brand A?
Mathematics
1 answer:
Paladinen [302]3 years ago
7 0

We can first find the probability of the first person picking Brand A, which would be:

9/6+9

=9/15

=3/5

We can then find the probability of the second person choosing brand A:

(9-1)/(6+9-1)

=8/14

To find the probability they both select Brand A, we could multiply them together:

3/5 x 8/14

=12/35

Therefore the answer would be 12/35.

Hope it helps!


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143.8% is the answer

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three friends earned more than $200 washing cars. they paid there parents $28 for supplies and divided the rest of money equally
makvit [3.9K]
So
x=earned
x>200
they paid their parrents 28 for supplies (this assumes that they paied 28 altogether and not 28 each)
so subtract 28 from pay
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Can you help me with the question please?
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Answer:

\frac{1}{4} y \geqslant 8 =

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a father is 10 times as old as his son. in 5 years he will be 5 times as old as his son. how old is the father and the son now
inn [45]

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The father is 40 and the son is 4.

Step-by-step explanation:

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5 0
3 years ago
How do the values in Pascal’s triangle connect to the coefficients?
damaskus [11]

Explanation:

Each row in Pascal's triangle is a listing of the values of nCk = n!/(k!(n-k)!) for some fixed n and k in the range 0 to n. nCk is <em>the number of combinations of n things taken k at a time</em>.

If you consider what happens when you multiply out the product (a +b)^n, you can see where the coefficients nCk come from. For example, consider the cube ...

  (a +b)^3 = (a +b)(a +b)(a +b)

The highest-degree "a" term will be a^3, the result of multiplying together the first terms of each of the binomials.

The term a^b will have a coefficient that reflects the sum of all the ways you can get a^b by multiplying different combinations of the terms. Here they are ...

  • (a +_)(a +_)(_ +b) = a·a·b = a^2b
  • (a +_)(_ +b)(a +_) = a·b·a = a^2b
  • (_ +b)(a +_)(a +_) = b·a·a = a^2b

Adding these three products together gives 3a^2b, the second term of the expansion.

For this cubic, the third term of the expansion is the sum of the ways you can get ab^2. It is essentially what is shown above, but with "a" and "b" swapped. Hence, there are 3 combinations, and the total is 3ab^2.

Of course, there is only one way to get b^3.

So the expansion of the cube (a+b)^3 is ...

  (a +b)^3 = a^3 + 3a^2b +3ab^2 +b^3 . . . . . with coefficients 1, 3, 3, 1 matching the 4th row of Pascal's triangle.

__

In short, the values in Pascal's triangle are the values of the number of combinations of n things taken k at a time. The coefficients of a binomial expansion are also the number of combinations of n things taken k at a time. Each term of the expansion of (a+b)^n is of the form (nCk)·a^(n-k)·b^k for k =0 to n.

6 0
3 years ago
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