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slava [35]
3 years ago
7

Sarah plants 760 vines in rows containing either 20 vines or 25 vines. There are 3 times as many rows containing 25 vines as the

re are rows containing 20 vines. How many rows contain 25 vines
Mathematics
1 answer:
bogdanovich [222]3 years ago
6 0

Answer: 27 rows contain 25 vines.

Step-by-step explanation:

Let x represent the number of rows that contain 20 vines.

Let y represent the number of rows that contain 25 vines.

Sarah plants 760 vines in rows containing either 20 vines or 25 vines. This means that

20x + 25y = 760- - - - - - - - - - -1

There are 3 times as many rows containing 25 vines as there are rows containing 20 vines. This means that

y = 3x

Substituting y = 3x into equation 1, it becomes

20x + 25 × 3x = 760

20x + 75x = 760

95x = 760

x = 760/95

x = 9

y = 3x = 3 × 9

y = 27

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The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machi
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Answer:

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

p_v =P(t_{(72)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=88.8 represent the sample mean

s=26.6 represent the sample standard deviation for the sample  

n=73 sample size  

\mu_o =93 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 check if the true mean i lower than 93 min, the system of hypothesis would be:  

Null hypothesis:\mu \geq 93  

Alternative hypothesis:\mu < 93  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{s}{\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{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=73-1=72  

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

p_v =P(t_{(72)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

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3 years ago
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