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CaHeK987 [17]
2 years ago
14

A package delivery service divides their packages into weight classes. suppose that packages in the 14 to 20 pound class are uni

formly distributed, meaning that all weights within that class are equally likely to occur. find the 40th percentile of packages weights in this weight class.
Mathematics
1 answer:
nexus9112 [7]2 years ago
6 0

Answer: yy - \alpha \alpha = Percentile (\beta \beta -\alpha \alpha )

yy = 0.40 (20 - 14) + 14[tex] yy = 2.4 + 14

yy = 16.4

Thus, 16.4 pound class belongs to the 40th percentile of the weight class.

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Thus, Option D is the correct option.

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You are planning to take a group of friends to the movies for your birthday. The local movie theater charges $12.99 per person.
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Find the difference: square root of 20 - square root of 80
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Answer:

-2\sqrt{5}

Step-by-step explanation:

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Which description best compares the graphs of the two functions below?
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The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat
Greeley [361]

Answer:

1) L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

2) T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

3) T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

Part 2

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

Part 3

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

Replacing we got:

T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

8 0
2 years ago
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