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3 years ago

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A model rocket is launched with an initial upward velocity of 60 m/s. The rocket's height h (in meters) after t seconds is given

by the following.
h=60t-512
Find all values of t for which the rocket's height is 27 meters.
Round your answer(s) to the nearest hundredth.

1 answer:

Vedmedyk [2.9K]3 years ago

3 0

Answer:

0.47 and 11.53

Step-by-step explanation:

h = 60t − 5t²

27 = 60t − 5t²

5t² − 60t + 27 = 0

Quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a

t = [ -(-60) ± √((-60)² − 4(5)(27)) ] / 2(5)

t = (60 ± √3060) / 10

t = 0.47 or 11.53

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Answer:

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b) We need a sample size of at least 4145.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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In which

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n is found when $M = 0.02, \pi = 0.25$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{2.575\sqrt{0.25*0.75}}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.575\sqrt{0.25*0.75}}{0.02})^{2}$

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(b) he does not use any prior estimates?

When we do not use any prior estimate, we use $\pi = 0.5$

So

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$0.02 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.5*0.5}$

$\sqrt{n} = \frac{2.575\sqrt{0.5*0.5}}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.575\sqrt{0.5*0.5}}{0.02})^{2}$

$n = 4144.1$

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