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valentinak56 [21]
3 years ago
5

I NEED HHELP ASAP STOP SENDING ME LINK

Mathematics
2 answers:
spayn [35]3 years ago
8 0
I’m pretty sure the answer is B
Ymorist [56]3 years ago
5 0

Answer:

The answer is B

Step-by-step explanation:

hope that helps

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Find the slope of the line given the table of values?
wariber [46]

Answer:

2/-3

Step-by-step explanation:

(y1 - y2)/(x1 - x2) = (-6 - (-8))/(1 - 4) = 2/-3

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4 years ago
The graph shows a line joining two complex numbers. What are the two complex numbers?
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3i and 3i minus 3 is the answer?
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Tommy purchased a new vacuum for $1,150 at the Best Buy. He has a 15% coupon. How much will he save?
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Step-by-step explanation:

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3 years ago
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An automotive part must be machined to close tolerances to be acceptable to customers. Production specifications call for a maxi
Vilka [71]

Answer:

H0: \sigma \leq 0.0004

H1: \sigma >0.0004

t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25

For this case since we have a right tailed test the p value is given by:

p_v = P(\Chi^2_{29}>36.25)=0.166

If we compare the p value and the significance level given we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal  than 0.0004 at 5% of significance.  

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 30 sample size

s^2 =0.0005 represent the sample variance

s= 0.0224 represent the sample deviation

\sigma_o =0.095 the value that we want to test

p_v represent the p value for the test

t represent the statistic

\alpha=0.05 significance level

State the null and alternative hypothesis

On this case we want to check if the population variance is higher than 0.0004 (the specification), so the system of hypothesis are:

H0: \sigma \leq 0.0004

H1: \sigma >0.0004

In order to check the hypothesis we need to calculate th statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with df=30-1=19 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,29)". And our critical value would be \Chi^2 =42.56

Since our calculated value is lower than the critical value we to reject the null hypothesis.

What is the approximate p-value of the test?

For this case since we have a right tailed test the p value is given by:

p_v = P(\Chi^2_{29}>36.25)=0.166

If we compare the p value and the significance level given we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal  than 0.0004 at 5% of significance.  

6 0
3 years ago
Ryan bought a total of 40 juice boxes. He bought 8 more boxes of apple juice than of grape juice. How many of each kind did he b
postnew [5]

we take away the extra 8 boxes of apple juice first:

40 - 8 = 32


Now that we have equal number of apple juice and graph juice, we divide them into 2:

32 ÷ 2 = 16


There are 16 boxes of graph juice.


We add back the 8 boxes of apple juice we took away just now

16 + 8 =24


There are 24 boxes of apple juice.


Answer: 24 boxes of apple juice and 16 boxes of graph juice.



3 0
3 years ago
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