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2 years ago

Francis ran 2 1/2 miles in 7/12 of an hour. How many miles can francis run in one hour?

Mathematics

2 answers:

nevsk [136]2 years ago

8 0

7/12 of 1 hour:

1 hour = 60 min.

7/12 * 60

7 * 5 = 35

Francis ran 2 1/2 miles in 35 min.

2 1/2 = 5/2

5/2 miles = 35 min

x miles = 60 min

(Cross Multiply)

5/2 * 60 = 35x

5 * 30 = 35x

150 / 35 = x

x = 30 / 7 = 4 2/7 miles

Francis ran 4 2/7 miles in one hour.

Artyom0805 [142]2 years ago

4 0

Since a fraction multiplied by the the numbers flipped (in this case, it's 12/7) gets you one, you have to multiply both 2 and 1/2 and 7/12 by 12/7 you get 4 and 4/14 miles ran in 2 hour

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What is the formula for the confidence interval? Then, define what each variable represents.

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The confidence interval for a (1−α)% confidence level is given by

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A confidence interval (CI) is desired for the true average stray-load loss u (watts) for a certain type of induction motor when

Genrish500 [490]

Answer:

A) CI = (57.12 , 59.48)

B) CI = (57.71 , 58.89)

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D) n = 239.63

Step-by-step explanation:

given data:

mean, $\bar X = 58.3$

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mean, $\bar X = 58.3$

standard deviation, σ = 3

sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

$ME = zc * σ \sqrt{n}$

$ME = 1.96 * 3\sqrt{100}$

ME = 0.59

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 1.96 * 3\sqrt{100} , 58.3 + 1.96 * 3\sqrt{100})$

CI = (57.71 , 58.89)

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sample standard deviation, σ = 3

sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

$ME = Zc * σ \sqrt{n}$

$ME = 2.58 * 3\sqrt{100}$

ME = 0.77

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Significance Level, α = 0.01,

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σ = 3

The critical value for α = 0.01 is 2.58.

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