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Tems11 [23]
2 years ago
5

A ball is kicked 4 feet above the ground with an initial vertical velocity of 51 feet per second. The function h(t)=−16t2+51t+4

represents the height h (in feet) of the ball after t seconds. Using a graph, after how many seconds is the ball 28 feet above the ground? Round your answers to the nearest tenth. I'd like to ask for the answer and the steps to get there. Thank you!

Mathematics
1 answer:
Katyanochek1 [597]2 years ago
7 0

Check the first picture below.

now, the exercise states to get it from the graph, so we have to graph it first.

Check the second picture below.

recall, the height in feet is on the y-axis, and the time in seconds is on the x-axis.

and rounded up to the nearest tenth will be 0.6.

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This are the right steps

Step 1: first you divide the both size by 7 because there is 7 a's

7a/7 = 28/7

Step 2: You solve the equation

7a/7= a. 28/7 = 4

So, the answer is a = 4

not 7 = 4

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The proof TriangleABC ≅ TriangleDCB that is shown. Given: AngleA ≅ AngleD; CD||AB Prove: TriangleABC ≅ TriangleDCB Triangles C D
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2 years ago
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Before the distribution of certain statistical software, every fourth compact disk (CD) is tested for accuracy. The testing proc
lilavasa [31]

Answer:

a) For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

P(R) =\frac{1}{4} =0.25

The failure probability for each program are given by:

P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932

So then the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

And if we use the fact that we have 4 possible test the true probability of interest would be:

P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017

b) p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))

And replacing we got:

p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484

c) From part a we now that the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

So then if we have 100 CDs the expected number of rejected Cd's are:

100*0.068= 6.8 \approx 7

Step-by-step explanation:

Part a

For this case we have 4 programs so then if we define the event R that a CD is tested we have the following probability for each test:

P(R) =\frac{1}{4} =0.25

The failure probability for each program are given by:

P(F_1) = 0.01 , P(F_2) = 0.03 , P(F_3) = 0.02 , P(F_4) = 0.01

For this case we assume that each test is independet form the others.

We can calculate the probability that all 4 programs works properly like this:

P(4 work) = (1-0.01)*(1-0.03)*(1-0.02)*(1-0.01)= 0.932

So then the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

And if we use the fact that we have 4 possible test the true probability of interest would be:

P(R \cap F) = P(R)*P(F) = 0.25*0.068=0.017

Part b

For this case we want the probability that it failed program 2 or 3

So then we can find this probability like this:

p= P(F'_1) P(F'_4) *(1- P(F'_2)*P(F'_3))

And replacing we got:

p =(1-0.01)*(1-0.01) *[1- (1-0.03)(1-0.02)]= 0.99*0.99*[1- 0.97*0.98]= 0.0484

Part c

From part a we now that the probability that any program fails would be given by:

P(F) = 1- 0.932= 0.068

So then if we have 100 CDs the expected number of rejected Cd's are:

100*0.068= 6.8 \approx 7

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<em>   N = n/5</em>

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Answer:

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Step-by-step explanation:


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