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zheka24
2 months ago
10

A man standing at a window 5 meters tall watches a falling ball pass by the window in 0.3 seconds. From how high above the top o

f the window was the ball released?
Mathematics
1 answer:
babunello [11.8K]2 months ago
4 0

Answer:

The height is 11.76 meters.

Step-by-step explanation:

To solve this problem, the following equations are necessary:

The first one relates initial and final velocities, taking into account that gravity is 9.81 m/s² and the time is 0.3 seconds:

Vf = Vo + g * t = Vo + 9.81 * 0.3

Vf = Vo + 2.94 (1)

The second one also connects initial and final velocities, but this time with distance S, which we know to be 5 meters:

Vf² = Vo² + 2 * g * S

Vf² = Vo² + 2 * 9.81 * 5

Vf² = Vo² + 98.1 (2)

We now have two equations with two unknowns. By substituting (1) into (2):

(Vo + 2.943)² = Vo² + 98.1

Vo² + 5.886 * Vo + 8.66 = Vo² + 98.1

Canceling Vo² and rearranging gives us:

Vo = 89.44 / 5.886

Vo = 15.195 m/s

Now using the formula:

Vo² = 2 * g * h

h = Vo² / (2 * g) = (15.195²) / (2 * 9.81) = 11.76 meters

So, the height corresponds to 11.76 meters.

You might be interested in
The bar graph shows the numbers of reserved campsites at a campground for one week. What percent of the reservations were for Fr
Svet_ta [12734]

Answer:

66\frac{2}{3} \%

Step-by-step explanation:

Refer to the attached bar graph.

The bar graph illustrates the count of reserved campsites at a campground across a week.

Consequently, the total number of reserved campsites on Friday and Saturday will amount to (26 + 30) = 56.

Now, calculating the total reservations from Monday to Sunday gives us (5 + 3 + 4 + 7 + 26 + 30 + 9) = 84.

Therefore, the percentage of bookings for Friday and Saturday will be

\frac{56}{84} \times 100\% = \frac{200}{3}\% = 66\frac{2}{3}\%. (Answer)

8 0
2 months ago
Based on daily measurements, Bob's weight has a mean of 200 pounds with a standard deviation of 16 pounds, while Mary's weight h
babunello [11817]

Answer:

(C) They have the same coefficient of variation

Step-by-step explanation:

The coefficient of variation (CV) is calculated using the formula:

CV = \frac{\sigma}{\mu}

Where \sigma represents standard deviation and \mu represents the mean.

Bob's average weight is 200 pounds with a standard deviation of 16 pounds

This indicates that \sigma = 16, \mu = 200.

Thus, his coefficient of variation is

CV = \frac{16}{200} = 0.08

Mary's average weight is 125 pounds, with a standard deviation of 10 pounds.

This implies \sigma = 10, \mu = 125

Therefore, her coefficient of variation is

CV = \frac{10}{125} = 0.08

Since both have the same coefficient of variation, the accurate response is.

(C) They have the same coefficient of variation

6 0
2 months ago
Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab
Svet_ta [12734]

Response:  a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Detailed explanation:

In Problem 8-4, the computer time-sharing system experiences teleport inquiries at an average rate of 0.1 per millisecond. We are tasked with determining the probabilities of the inquiries over a specific period of 50 milliseconds:

Given that

\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5

Applying the Poisson process, we find that

(a) at most 12

probability=  P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980

(b) exactly 13

probability= P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013

(c) more than 12

probability= P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020

(d) exactly 20

probability= P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026

(e) within the range of 10 to 15, inclusive

probability=P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318

Thus, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0
2 months ago
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