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ICE Princess25 [194]
3 years ago
7

In the video, I drop a ball off a three-story balcony. We know that objects in freefall follow the equation for height h, in

Mathematics
1 answer:
Reil [10]3 years ago
4 0

Answer:

The initial height from which the ball was dropped = 7.90 meters

Step-by-step explanation:

The correct question is:

In the video, I drop a ball off a three-story balcony. We know that objects in free fall follow the equation for height h, in meters, h = -4.9t^2 + c, where t is time in seconds, and c is the initial height. A student timed the drop at 1.27 seconds.

Use this to determine the height from which the ball was dropped, to at least 2 decimal places.

Solution:

The height h of the free falling ball is represented by the equation as:

h=-4.9t^2+c where t is time in seconds, and c is the initial height.

To determine the initial height from which the ball was dropped.

The drop was timed at 1.27 seconds, we will plugin t=1.27 in the given height function to find h(1.27)

h(1.27)=-4.9(1.27)^2+c

h(1.27)=-7.90+c

Since, the drop is timed at 1.27 seconds, so in the given time the ball will reach the ground, making h(1.27)=0

So, we have.

0=-7.90+c

Adding both sides by 7.90 to solve for c.

0+7.90=-7.90+7.90+c

7.90=c

∴ c=7.90\ m

Thus, the initial height from which the ball was dropped = 7.90 meters

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Answer:

The radius of a sphere is 2 millimeters

The surface area of a sphere is 50.24 square millimeters.

The circumference of the great circle of a sphere is 12.56  millimeters.

Step-by-step explanation:

<u><em>Verify each statement</em></u>

case A) The radius of a sphere is 8 millimeters

The statement is false

we know that

The volume of the sphere is equal to

V=\frac{4}{3}\pi r^{3}

we have

V=100.48/3\ mm^{3}

\pi =3.14

substitute and solve for r

100.48/3=\frac{4}{3}(3.14)r^{3}

r^{3}=(100.48)/(4*3.14)

r=2\ mm    

case B) The radius of a sphere is 2 millimeters

The statement is True

(see the case A)

case C) The circumference of the great circle of a sphere is 9.42 square millimeters  

The statement is false

The units of the circumference is millimeters not square millimeters

The circumference is equal to

C=2\pi r

we have

r=2\ mm  

\pi =3.14

substitute

C=2(3.14)(2)

C=12.56\ mm

case D) The surface area of a sphere is 50.24 square millimeters.

The statement is True

Because

The surface area of the sphere is equal to

SA=4\pi r^{2}

we have

r=2\ mm  

\pi =3.14

substitute

SA=4(3.14)(2)^{2}

SA=50.24\ mm^{2}

case E)  The circumference of the great circle of a sphere is 12.56  millimeters.  

The statement is true

see the case C

case F) The surface area of a sphere is 25.12 square millimeters

The statement is false

because the surface area of the sphere is SA=50.24\ mm^{2}

see the case D

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3. Assume that the likelihood of a child under the age of ten watching PBS is 0.76. Three children are
natta225 [31]

Using the binomial distribution, the probabilities are given as follows:

a) 0.4159 = 41.59%.

b) 0.5610 = 56.10%.

c) 0.8549 = 85.49%.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

For this problem, the values of the parameters are:

n = 3, p = 0.76.

Item a:

The probability is P(X = 2), hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{3,2}.(0.76)^{2}.(0.24)^{1} = 0.4159

Item b:

The probability is P(X < 3), hence:

P(X < 3) = 1 - P(X = 3)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.76)^{3}.(0.24)^{0} = 0.4390

Then:

P(X < 3) = 1 - P(X = 3) = 1 - 0.4390 = 0.5610 = 56.10%.

Item c:

The probability is:

P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4159 + 0.4390 = 0.8549

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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