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sergey [27]
2 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]2 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]2 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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Step-by-step explanation:

This is just an addition question. If you have a calculator just add the numbers.

3.25 + 1.625 = 4.875 cups in the bowl.

Sometimes you can use another method. Sometimes it is shorter.

3 1/4 + 1 5/8 = 3 2/8 + 1 5/8 = 4 7/8.

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Slope: 2

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A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January
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Answer:

a) 27.89% probability that two have ever boycotted goods for ethical reasons

b) 41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) 41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) The expected number is 2.3 and the standard deviation is 1.33.

Step-by-step explanation:

We use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

23% of the respondents have boycotted goods for ethical reasons.

This means that p = 0.23

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

This is P(X = 2) when n = 6. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{6,2}.(0.23)^{2}.(0.77)^{4} = 0.2789

27.89% probability that two have ever boycotted goods for ethical reasons

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Either less than two have, or at least two. The sum of the probabilities of these events is decimal 1. So

P(X < 2) + P(X \geq 2) = 1

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084

P(X = 1) = C_{6,1}.(0.23)^{1}.(0.77)^{5} = 0.3735

P(X < 2) = P(X = 0) + P(X = 1) = 0.2084 + 0.3735 = 0.5819

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5819 = 0.4181

41.81% probability that at least two respondents have boycotted goods for ethical reasons

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

Now n = 10.

P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343

P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225

P(X = 5) = C_{10,5}.(0.23)^{5}.(0.77)^{5} = 0.0439

P(X = 6) = C_{10,6}.(0.23)^{6}.(0.77)^{4} = 0.0109

P(3 \leq X \leq 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.2343 + 0.1225 + 0.0439 + 0.0109 = 0.4116

41.16% probability that between 3 and 6 have boycotted goods for ethical reasons

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

E(X) = np = 10*0.23 = 2.3

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{10*0.23*0.77} = 1.33

The expected number is 2.3 and the standard deviation is 1.33.

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