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Hatshy [7]
3 years ago
10

the light bulbs you use are on sale. but you only get the lower price if you buy 3 packsages of bulbs. you want to know how much

you'll save. write a formula that will help you determine how much you'll save by buying 3 packages on sale. s=sale price per package r=regular price per package c=cost saving
Mathematics
1 answer:
Brilliant_brown [7]3 years ago
8 0
(R*3)-(S*3)=C
The regular price of 3 light bulbs minus the sale price of 3 lightbulbs equal the amount you will save on 3 lights
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Peter has decided to save $30 each week to buy a new stereo system.Write a direct variation equation for the amount of money, m,
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M=w-30 is the answer if you want to find how much you want to find each week.
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A human resource manager wants to see if there is a difference in the proportion of minority applicants who get the job and the
mash [69]

Answer:

Error Bound = 0.04

Step-by-step explanation:

Whenever we want to estimate parameter from a subset (or sample) of the population, we need to considerate that your estimation won't be a 100% precise, in other words, the process will have a random component that prevents us from always making the exact decision.

With that in mind, the objective of a confidence interval is to give us a better insight of where we expect to find the "true" value of the parameter with a certain degree of certainty.

The estivamative of the true difference between proportions was -0.19 and the confidence interval was [-0.23 ; -0.15].

The question also defines the error bound, as the right endpoint of the confidence interval minus the sample mean difference, so it's pretty straight foward:

Error Bound = -0.15 -(-0.19) = -0.15 + 0.19 = 0.04

The interpretation of this would be that we expect that the estimative for the difference of proportions would deviate from the "true" difference about \pm 0.04 or 4%.

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3 years ago
Help me plot please.
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Find the height using the pythagorean theorem <br>​
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3

Step-by-step explanation:

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Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne
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Answer:

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

p_v =P(t_{(15)}>1.507)=0.076

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, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

\bar X_{CTS}=2.35 represent the mean for the sample CTS

\bar X_{N}=1.83 represent the mean for the sample Normal

s_{CTS}=0.88 represent the sample standard deviation for the sample of CTS

s_{N}=0.54 represent the sample standard deviation for the sample of Normal

n_{CTS}=10 sample size selected for the CTS

n_{N}=7 sample size selected for the Normal

\alpha=0.01 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 mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis:\mu_{CTS} \leq \mu_{N}

Alternative hypothesis:\mu_{CTS} > \mu_{N}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

P-value

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

df=n_{CTS}+n_{N}-2=10+7-2=15

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

p_v =P(t_{(15)}>1.507)=0.076

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

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, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0
3 years ago
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