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aniked [119]
3 years ago
10

Jethro used dimensional analysis to convert one rate of speed to another. Two of the conversion factors he used are 5280 ft/1 mi

and 1 hr/60 min. What could be the units of the initial rate of speed and the final rate of speed? Justify your answer.
Mathematics
1 answer:
NNADVOKAT [17]3 years ago
3 0

The units of the initial and final rates of speed are;

<u><em>Initial rate of Speed = ft/hr</em></u>

<u><em>Final rate of speed = mi/min</em></u>

  • We are told that the conversion factors he used are;

5280 ft/1 mi and 1 hr/60 min

  • Now, the meaning of 5280 ft/1 mi denotes that;

5280 ft mile is equivalent to 1 mile

This means that the initial distance <em>unit </em>used was ft while the final one was mile

Also, 1 hr/60 min denotes that;

1 hour is equivalent to 60 minutes.

This means that the initial time units used was hour while the final one was minute.

  • In conclusion, since we know the initial time and distance units used, we can say that the initial rate of speed ft/hr while final speed rate was miles/minutes.

Read more at; brainly.com/question/17734333

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Answer: |3-(-8)|


Step-by-step explanation:

Given: Paige lives 3 miles east of her school. Diego lives 8 miles west of the school.

From the picture , the number line represents the position of the houses of the two friends such that position of Paige is 3 but position of Diego is (-8) [since east and west are opposite directions]

We know that, Distance between any two points from A to B= B-A

Since distance is a positive quantity, thus

The distance between them = |3-(-8)| miles=11 miles


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Answer:

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A national park covers 212375 acres
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20. Joaquin tak 4-hours to clean a pool; Sherrie takes 1--hours. Calculate the time difference in hours. Express your answer in
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In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were re
11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μemployed(μ1) be the sample mean of students who are employed and μnot employed(μ2) be the sample mean of students who are not employed

The random variable is μ1 - μ2 = difference in the mean of the employed and unemployed students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 3.22

μ2 = 3.33

s1 = 0.475

s2 = 0.524

n1 = 172

n2 = 116

t = (3.22 - 3.33)/√(0.475²/172 + 0.524²/116)

t = - 1.81

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.475²/172 + 0.524²/116]²/[(1/172 - 1)(0.475²/172)² + (1/116 - 1)(0.524²/116)²] = 0.00001353363/0.00000005878

df = 230

We would determine the probability value from the t test calculator. It becomes

p value = 0.036

Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis.

Therefore, at a 5% significant level, this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed

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3 years ago
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