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mr Goodwill [35]
3 years ago
13

Shawn drew a rectangle that was 2 units wide and 6 units long. draw a different rectangle that has the same perimeter but a diff

erent area?
Mathematics
1 answer:
sleet_krkn [62]3 years ago
6 0
4 units wide as well as 4 units long
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Step-by-step explanation:

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An office supply company produces yellow document envelopes. The envelopes come with a variety of sizes, but the length is alway
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P= 2(2w+4)+2w
P=4w+8+2w
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How much material is needed to construct a triangular tent that is 10 feet wide, 12 feet tall, and 18 feet long with side measur
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The answer is " 768 square feet. ".

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A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of
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Answer:

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

Step-by-step explanation:

Data given and notation

\bar X=250000 represent the sample mean  

s=37500 represent the standard deviation for the sample

n=5 sample size  

\mu_o =210250 represent the value that we want to test  

\alpha=0.1 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 to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:  

Null hypothesis:\mu \leq 210250  

Alternative hypothesis:\mu > 210250  

Compute the test statistic  

We don't know the population deviation, so for this case 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".  

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

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

Conclusion

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

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

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Triple a number, minus the sum of the number and six
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\large \mathfrak{Solution : }

let the number be x

According to above statement :

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