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Aloiza [94]
3 years ago
5

Find the arc length

Mathematics
1 answer:
umka21 [38]3 years ago
7 0

Answer:

39/2 π ft^2

Step-by-step explanation:

2(13)(270)/360

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Which type of graph is useful for demonstrating how something changes or fluctuates in value over time?
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Line graphs (or line charts) are best when you want to show how the value of something changes over time, or compare how several things change over time relative to each other. Whenever you hear that key phrase “over time,” that's your clue to consider using a line graph for your data.

-BBBM

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Illusion [34]

Answer:

The answer is b

Step-by-step explanation:

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4 years ago
Can someone please explain this to me???
Softa [21]

If it costs $60 for 200 minutes you can do $60 divided by 200 to find how much money it costs per minute which equals .3 or 3 cents a minute. so if you have 1272 minutes you can multiply that by .3 and it equals $381.60

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3 years ago
X = −3 y = 4 and z = −5 <br><br> Work out the value of x + 2y − 3z
solniwko [45]

Answer:

20

Step-by-step explanation:

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4 0
3 years ago
A professor believes the class he is teaching in the current semester is above the average of the classes he has taught. In orde
MakcuM [25]

Answer:

1) Since we know the info from all the students that he teaches and we know the population deviation from past data we can use a z test to check the hypothesis

2) t=\frac{82-78}{\frac{15}{\sqrt{120}}}=2.921    

3) p_v =P(Z>2.921)=0.0017  

Since the p value is lower than the significance level of 0.01 we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis

4) For this case since we reject the null hypothesis we have enough evidence ot conclude that the scores for this semester are above the historical value of 78 so then the claim stated by the teacher makes sense

Step-by-step explanation:

Part 1

Since we know the info from all the students that he teaches and we know the population deviation from past data we can use a z test to check the hypothesis

Part 2

\bar X=82 represent the sample mean  for the scores

\sigma=15 represent the population standard deviation

n=120 represent the sample selected

\alpha=0.01 significance level  

System of hypothesis

He wants to test if the group for this current semester is above the average of the classes he has taught (mean 78), the system of hypothesis are:

Null hypothesis:\mu \leq 78  

Alternative hypothesis:\mu > 78  

Since we know the population deviation we can use the following statistic

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

Replacing the info given we got:

t=\frac{82-78}{\frac{15}{\sqrt{120}}}=2.921    

Part 3

We can calculate the p value for this test with this probability taking in count the alternative hypothesis:

p_v =P(Z>2.921)=0.0017  

Since the p value is lower than the significance level of 0.01 we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis

Part 4

For this case since we reject the null hypothesis we have enough evidence ot conclude that the scores for this semester are above the historical value of 78 so then the claim stated by the teacher makes sense

5 0
4 years ago
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