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Leno4ka [110]
4 years ago
9

The ratio of the number of apples to the number of pears in a basket was 3:5. After 7more apples were added to the basket and 9

pears were removed ,there were an equal number of apples and pears .how many apples and how many pears were originally in the basket
Mathematics
1 answer:
atroni [7]4 years ago
5 0

Answer:

Apples = 24

Pears = 40

Step-by-step explanation:

Let u be the unit of fruit in the basket,

number of apples: 3u +7

number of pears: 5u -9

now to find u, we equate the two equations.

3u+7=5u-9

Now simplify.

5u-3u=7+9

2u=16

u=8

Now we can find the original number of apples and pears by substituting u.

so original number of apples = 3u = 3(8) = 24

and original number of pears = 5u = 5(8) = 40

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After a late night of studying, Ebony decides to grab a latte before class so she can stay awake through her morning lecture. Sh
Rzqust [24]

Answer:

P(Same\ Bill) = \frac{1}{3}

P(Second

P(Both\ Even) = \frac{1}{9}

Pr(One\ Odd) = \frac{4}{9}

P(Sum < 10) = \frac{1}{3}

Step-by-step explanation:

Given

Bills: \$1, \$5, \$10

Selection = 2\ bills

The sample space is as follows:

This implies that we construct possible outcome that Ebony selects a bill, returns the bill and then select another.

This means that there are possibilities that the same bill is selected twice.

So, the sample space is as follows:

S = \{(1,1), (1,5), (1,10), (5,1), (5,5), (5,10), (10,1), (10,5), (10,10)\}

n(S) = 9

Solving (a): P(Same\ Bill)

This means that the first and second bill selected are the same.

The outcome of this are:

Same = \{(1,1),(5,5),(10,10)\}

n(Same\ Bill) = 3

The probability is:

P(Same\ Bill) = \frac{n(Same\ Bill)}{n(S)}

P(Same\ Bill) = \frac{3}{9}

P(Same\ Bill) = \frac{1}{3}

Solving (a): P(Second  < First\ Bill)

This means that the second bill selected is less than the first.

The outcome of this are:

Second < First = \{(1,5), (1,10), (5,10)\}

n(Second < First) = 3

The probability is:

P(Second

P(Second

P(Second

Solving (c): P(Both\ Even)

This means that the first and the second bill are even

The outcome of this are:

Both\ Even = \{(10,10)\}

n(Both\ Even) = 1

The probability is:

P(Both\ Even) = \frac{n(Both\ Even)}{n(S)}

P(Both\ Even) = \frac{1}{9}

Solving (e): P(Sum < 10)

This question has missing details.

The correct question is to determine the probability that, the sum of both bills is less than 10

The outcome of this are:

One\ Odd = \{(1,10), (5,10), (10,1), (10,5)\}

n(One\ Odd) = 4

The probability is:

Pr(One\ Odd) = \frac{n(One\ Odd)}{n(S)}

Pr(One\ Odd) = \frac{4}{9}

 

Solving (d): P(One\ Odd)

This question has missing details.

The correct question is to determine the probability that, exactly one of the bills is 0dd

The outcome of this are:

Sum < 10 = \{(1,1), (1,5), (5,1)\}

n(Sum < 10) = 3

The probability is:

P(Sum < 10) = \frac{n(Sum < 10)}{n(S)}

P(Sum < 10) = \frac{3}{9}

P(Sum < 10) = \frac{1}{3}

 

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3 years ago
Mr. Wilkinson is building a stone path through his backyard. Each stone is 1 3 of a foot long. He wants the path to be 8 feet lo
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Answer:

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4 0
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Hi!! Please explain how you got the answer, thanks!! :)) ​
Slav-nsk [51]

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Since all we have to look at is 1/4 of this square, multiply the area by 1/4.

(16)1/4

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The area of the triangle is 4.

Now we need the area of the circle.

a = πr²

Sub: a = π(2)²

a = π(4)

Since the portion of the circle that covers the triangle is 1/4 of the circle, subtract 1/4 of the area of the circle from the area of the circle.

a = 4 - 1/4(π(4))

a = 4 - 1/4π x 1

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Now this is where it's supposed to go, considering the fact that π isn't given a value.

Normally, I'd use 3.14 for π, as the entire population would, and then it would be easy, but obviously that isn't an option.

Now this answer isn't up there, so my best option is to tell you it's most likely c, and I hope I'm right.

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Answer:

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