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sergey [27]
3 years ago
14

Two thirds of the fruit in the basket are apples. One fourth of the apples are green apples. What fraction of the fruit in the b

asket are green apples?
Mathematics
2 answers:
Schach [20]3 years ago
6 0

<u>Answer:</u>

The green apples are one sixth of the total apples in the basket.

<u>Solution: </u>

Let us assume that total fruits in the basket are x

In the basket, two third of the fruit are apples

Lets say total apple in the basket = y  

So, in the basket total amount of apples y=\left(\frac{2}{3} \text { of } x\right)=\frac{2 x}{3}

Now, in that apple one fourth apples are green.

So green apples are \left(\frac{1}{4} \text { of } y\right)=\left(\frac{1}{4} \times \frac{2 x}{3}\right) (//substituting value of y)

\Rightarrow \frac{x}{6}    

\Rightarrow \left(\frac{1}{6} \ of \ x\right)

BabaBlast [244]3 years ago
3 0

Answer:

1/6

Step-by-step explanation:

if 2/3 of the fruit are apples, of these 2/3, 1/4 of them are green apples so:

2/3 x 1/4 = 2/12 = 1/6

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marysya [2.9K]

Melanie: 15 years old

Kevin: 5 years old

Assume Kevin's present age is x.

That means that 3 years ago, his age would be x - 3.

Melanie's age would be 6(x - 3).

Same thing for 5 years into the future:

Kevin's age would be x + 5.

Melanie's age would be 2(x + 5).

The two equations we have here are:

\left \{ {{x + 5 = 2(x + 5)} \atop {x - 3 = 6(x - 3)}} \right.

By subtracting both equations, we are simplified to:

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By solving the equation, we get x = 5.

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Melanie's age could be calculated by substituting Kevins'.

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3 0
2 years ago
How to I solve problem 1?
Verdich [7]
#1

The uniforms are numbered 0, 1, 2, ..., 99. That's 100 numbers. Half of them are odd and half of them are even. So the probability that any one of the uniforms is odd is 1/2 just like the probability that any one uniform is even is 1/2.

(a) The numbers on the uniforms are independent of one another. That is, the number of her cross-country uniform does not in any way determine the number on her basketball uniform and vice versa. This means that we can find the probability that each is odd and multiply these together using what is called the counting principle. The probability that all are odd is:
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(b) This is done the same way we did part (a). Since the probability of any one uniform being odd is the same as it being even (1/2), the answer here is the same: (1/2)(1/2)(1/2)=1/8

(c) This problem differs from that in (a) and (b). There is only one way for all three uniforms to be odd numbers: (odd, odd, odd) or all even (even, even, even). However, there are multiple ways for the uniforms to be two odd and one even. If the uniforms are listed in order: cross-country, basketball, softball we can get exactly one even in any of three ways:
even, odd, odd
odd, even, odd
odd, odd, even
The probability for any one of these possibilities is (1/2)(1/2)(1/2)=1/8 but since there are three way the probability that we get even exactly once is equal to (3)(1/8) = 3/8
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