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lina2011 [118]
3 years ago
13

A track-and-field athlete releases a javelin. The height of the javelin as a function of time is shown on the graph below. Use t

he graph to complete the statements that follow. The height of the javelin above the ground is symmetric about the line t = seconds. The javelin is 20 feet above the ground for the first time at t = seconds and again at t = seconds

Mathematics
1 answer:
Georgia [21]3 years ago
5 0
Part 1:
 
 For this case we must see in the graph the axis of symmetry of the given parabola.
 We have then that the axis of symmetry is the vertical line t = 2.
 Answer:
 
The height of the javelin above the ground is symmetric about the line t = 2 seconds:

 
Part 2:

 
For this case, we must see the time t for which the javelin reaches a height of 20 feet for the first time.
 We then have that when evaluating t = 1, the function is h (1) = 20. To do this, just look at the graph.
 Then, we must observe the moment when it returns to be 20 feet above the ground.
 For this, observing the graph we see that:
 h (3) = 20 feet
 Therefore, a height of 20 feet is again reached in 3 seconds.
 Answer:
 
The javelin is 20 feet above the ground for the first time at t = 1 second and again at t = 3 seconds
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Alenkasestr [34]
Answer:
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Step-by-step explanation:
The radius of the circle = 4 + 26i - (-6 + 2i)
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The radius will be the absolute values of this |10 + 24i|.

If a point is on the circle then it's distance from the centre must be 10+ 24i.
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3 0
2 years ago
The probability of winning a certain lottery is 1/77076 for people who play 908 times find the mean number of wins
Alex73 [517]

The mean is 0.0118 approximately. So option C is correct

<h3><u>Solution:</u></h3>

Given that , The probability of winning a certain lottery is \frac{1}{77076} for people who play 908 times

We have to find the mean number of wins

\text { The probability of winning a lottery }=\frac{1}{77076}

Assume that a procedure yields a binomial distribution with a trial repeated n times.

Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial.

n=908, \text { probability } \mathrm{p}=\frac{1}{77076}

\text { Then, binomial mean }=n \times p

\begin{array}{l}{\mu=908 \times \frac{1}{77076}} \\\\ {\mu=\frac{908}{77076}} \\\\ {\mu=0.01178}\end{array}

Hence, the mean is 0.0118 approximately. So option C is correct.

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Hi,

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