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3 years ago

Can someone explain to me how to do question number 6 please?

Mathematics

1 answer:

kirill115 [55]3 years ago

6 0

Do the crossing out method. so if it is 2.50 per pound for peaches, try A, 11 pounds. 2.50 times 11 is 27.5, and now we have the rest, 154 pounds as apples so we multiply it by 1.75 which is 269.5, we add them up and we dont get 337.50, so we try out b, 18 times 2.5 is 45 so we still have 147 pounds left for apples, multiply it by 1.75 which is 257.25 and if we add up 45 and 257.25 we dont get 337.5 so now we try c which is 65, multiply it by 2.5 its 162.5 and we have 100 pounds of apples left. we do the same and we get 175, 175 plus 162.5 is 337.5 so C/3 is the correct answer

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2 years ago

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AnnZ [28]

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Step-by-step explanation:

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Suppose a local manufacturing company claims their production line has a variance of less than 9.0. A quality control engineer d

Alexxx [7]

Answer:

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

$p_v =P(\chi^2$

If we compare the p value and the significance level provided we see that $p_v$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=35$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s =2.12$ represent the sample deviation obtained

$\sigma_0 =3$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is lower than 9 and the deviation lower than 3, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 9$

Alternative hypothesis: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df= n-1= 35-1=34. And since is a left tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(16.979,34,TRUE)"

Conclusion

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3 years ago