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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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Answer:

t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349      

p_v =P(t_9>10.349)=1.34x10^{-6}  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation      

\bar X=51.6 represent the sample mean

s=1.1 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =48 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean score is higher than 48, the system of hypothesis would be:      

Null hypothesis:\mu \geq 48      

Alternative hypothesis:\mu > 48      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)      

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic      

We can replace in formula (1) the info given like this:      

t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349      

Calculate the P-value      

First we find the degrees of freedom:

df=n-1=10-1=9

Since is a one-side upper test the p value would be:      

p_v =P(t_9>10.349)=1.34x10^{-6}  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.        

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