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liq [111]
3 years ago
12

Find the unit rate. 6,460 liters in 323 minutes liters per minute

Mathematics
2 answers:
Sloan [31]3 years ago
8 0

Answer:

20

Step-by-step explanation:

6,460 divided by 323 = 20

Maurinko [17]3 years ago
6 0
Answer is 20.
Just divide the number of liters (6,460) by the amount of minutes (323).
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The average yield of the standard variety off soybeans per acre on a farm in a region is 48.8 bushels per acre, with a standard
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Answer:

z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3  

p_v =P(Z>2.3)=1-P(Z  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=53.4 represent the sample mean

\sigma=12 represent the population standard deviation assumed

n=36 sample size  

\mu_o =48.8 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 48.8, the system of hypothesis would be:  

Null hypothesis:\mu \leq 48  

Alternative hypothesis:\mu > 48  

Since we assume that know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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

z-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".  

3) Calculate the statistic  

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

z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3  

4)P-value  

Since is a right tailed test the p value would be:  

p_v =P(Z>2.3)=1-P(Z  

5) Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.  

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