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mote1985 [20]
3 years ago
14

A falling stone is at a certain instant 250 feet above the ground and 3 seconds later it is only 10 feet above the ground. From

what height was it dropped?
Mathematics
1 answer:
hjlf3 years ago
6 0
Let h =  the height (ft) from which the stone was dropped.

Assume that g = 32 ft/s² and ignore air resistance.

At time t seconds, the stone is 250 ft above ground. Therefore
(1/2)gt² = h - 250                 (1)

After another 3 seconds, the stone is 10 ft above ground. Therefore
(1/2)g(t+3)² = h - 10              (2)

Subtract (1) from (2).
(1/2)g[(t+3)² - t²] = 240
32(t+3)² = 480
(t+3)² = 15
t = 0.873 s

From (1), obtain
16*(0.873)² = h - 250
12.194 = h - 250
h = 262.194 ft

Answer: 262.2 ft (nearest tenth)
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Which of the diagrams below represents the contrapositive of the statement
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Answer:

The correct figure is B.

Step-by-step explanation:

The statement provided is:

"If it is an equilateral triangle, then it is an isosceles triangle"

The contra-positive of a statement is determined by switching the hypothesis and conclusion of the provided statement and negating both.

Then the contra-positive of the statement provided will be:

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        If it is an isosceles triangle - It is an equilateral triangle.

  • Negating both the statements:

        If it is not an isosceles triangle - it is not an equilateral triangle.

Thus, the contra-positive of the statement is:

"If it is not an isosceles triangle, then it is not an equilateral triangle."

Thus, the correct figure is B.

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4 years ago
Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to f
BabaBlast [244]

Answer:

153 times

Step-by-step explanation:

We have to flip the coin in order to obtain a 95.8% confidence interval of width of at most .14

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ME = \frac{width}{2}

ME = \frac{0.14}{2}

ME = 0.07

ME\geq z \times \sqrt{\frac{\widecap{p}(1-\widecap{p})}{n}}

use p = 0.5

z at 95.8% is 1.727(using calculator)

0.07 \geq 1.727 \times \sqrt{\frac{0.5(1-0.5)}{n}}

\frac{0.07}{1.727}\geq sqrt{\frac{0.5(1-0.5)}{n}}

(\frac{0.07}{1.727})^2 \geq \frac{0.5(1-0.5)}{n}

n \geq \frac{0.5(1-0.5)}{(\frac{0.07}{1.727})^2}

n \geq 152.169

So, Option B is true

Hence  we have to flip 153 times the coin in order to obtain a 95.8% confidence interval of width of at most .14 for the probability of flipping a head

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