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lisov135 [29]
3 years ago
11

A ball is dropped from a height of 10 meters. Each time it bounces, it reaches 50 percent of its previous height. The total vert

ical distance the ball has traveled when it hits the ground the fifth time is ______ meters.
How many meters is it?
Mathematics
2 answers:
Gwar [14]3 years ago
6 0
A ball falls from the height of 10 m.
1st bounce: 0.5 * 10 = 5 m
2nd bounce: 5 * 0.5 = 2.5 m
3rd bounce: 2.5 * 0.5 = 1.25 m
4th bounce: 1.25 * 0.5 = 0.625 m
After that the ball hits the ground the 5th time.
Total distance:
10 + ( 5 * 2 ) + ( 2.5 * 2 ) + ( 1.25 * 2 ) + ( 0.625 * 2 ) =
= 10 + 10 + 5 + 2.5 + 1.25 = 28.75 m
Answer: The total vertical distance the ball has traveled when it hits the ground the 5th time is  28.75 m.
tester [92]3 years ago
4 0

Answer: 28.75

Step-by-step explanation: You honestly just need to divide each distance by 2 then add them together. First time it drops it goes down 10 meters. Then up five and back down. Then 2.5 meters then back down. Then 1.25, back down then .625 the back down.

At the end if you add together the total vertical distance you should get 28.75. If that is wrong well then the questions needs to be worded differently.

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Answer:

The estimated weight is in the range:

75.6 pounds \leq Weight \leq 92.4 pounds

Step-by-step explanation:

Since, the maximum error is 10%.

Therefore, the maximum and minimum vales will be 10% more and 10% less than 84 pounds, respectively.

<u>For Maximum Limit</u>:

Weight_{max} = (1.1)(84 pounds)

Weight_{max} = <u>92.4 pounds</u>

<u>For Minimum Limit</u>:

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3 years ago
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Answer:

k=1

Step-by-step explanation:

3k + 8 = 11\\

Subtract 8 from both sides

3k = 3

Divide

\frac{3k}{3} = \frac{3}{3}

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3 years ago
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Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n
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Answer:

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If we compare the p value obtained and the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

\hat p=\frac{209}{377}=0.554 estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

p_o=0.5 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

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Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

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