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Simora [160]
3 years ago
12

Please Help, Thank you!

Mathematics
1 answer:
Paha777 [63]3 years ago
8 0

let x be the number of quarters, and y be the number of dimes.

we know that x + y = 29

then we are also given that

y = 5 + 3x

so we can solve for x and y:

plug y into the first equation:

x + 5 + 3x = 29

4x = 24

x = 6

and

y = 5 + 3*6 = 23

I have 6 quarters and 23 dimes

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Suppose that P(A)=0.5, P(B)= 0.4 and P(B/A) =0.6. Find each of the following.
Lady_Fox [76]

Part (a)

P(A) = 0.5

P(B) = 0.4

P(B/A) = 0.6

P(A and B) = P(A)*P(B/A)

P(A and B) = 0.5*0.6

P(A and B) = 0.3

<h3>Answer: 0.3</h3>

==========================================

Part (b)

We'll use the result from part (a)

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 0.5 + 0.4 - 0.3

P(A or B) = 0.6

<h3>Answer: 0.6</h3>

===========================================

Part (c)

A and B are not independent since P(B) does not equal P(B/A). The fact that event A happens changes the probability P(B). Recall that P(B/A) means "probability P(B) based on event A already happened". A and B are independent if P(B) = P(B/A).

Events A and B are not mutually exclusive since P(A or B) is not zero.

<h3>Answer: Neither</h3>
7 0
3 years ago
242 X 38 1 9 36 9,1 96 Which digits belong in the blue boxes shown on the multiplication problem?
babymother [125]

Answer:

C. 726

Step-by-step explanation:

9196-1936=7260 and take away the 0 so your answer will be 726

4 0
2 years ago
Read 2 more answers
Which undefined term can contain parallel lines?
dusya [7]
Point I think is the anwser
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Q= QUADRADO
Lubov Fominskaja [6]
Answer
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6 0
3 years ago
The sample space listing the eight simple events that are possible when a couple has three children is {bbb, bbg, bgb, bgg, gbb,
NNADVOKAT [17]

Answer:

bbbb,

bbbg, bbgb, bgbb, gbbb,

bbgg, bgbg, bggb, gbgb, gbbg, ggbb,

bggg, gbgg, ggbg, gggb,

gggg

Step-by-step explanation:

3 children: bbb, bbg, bgb, gbb, bgg, gbg, ggb, ggg

4 children: bbbb, bbbg, bbgb, bgbb, gbbb, bbgg, bgbg, bggb, gbgb, ggbb, bggg, gbgg, ggbg, gggb, gggg

You need to take a methodological approach;

The 2 easiest are the possibility of all boys and all girls;

Then consider 3 boys and 1 girl:

bbbg, bbgb, bgbb, gbbb

Then 2 boys and 2 girls:

bbgg, bgbg, bggb, gbgb, gbbg, ggbb

Lastly, 1 boy and 3 girls:

bggg, gbgg, ggbg, gggb

In total, there are 16 possibilities

P.S. interesting to note is that the number of possibilities here follows the pattern of Pascal's triangle:

       1

     1   1

   1  2  1

 1  3  3  1

1  4  6  4  1

The last row is relevant here;

There is 1 possibility where there are 4 boys;

There are 4 possibilities (in terms of the order of birth) where there are 3 boys and 1 girl;

There are 6 possibilities where there are 2 boys and 2 girls

There are 4 possibilities where there is 1 boy and 3 girls;

There is 1 possibility where there are 4 girls;

The pattern ∴ is 1 4 6 4 1, as the 5th row of Pascal's triangle reads;

If your talking about 3 children, it would match the 4th row of Pascal's triangle;

So, 1 possibility of 3 boys;

3 possibilities of 2 boys and 1 girl;

3 possibilities of 1 boy and 2 girls;

And 1 possibility of 3 girls;

If your talking about 10 children, it would match the 11th row of Pascal's triangle.

(Maths can be so cool XD)

7 0
3 years ago
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