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10 days ago

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A pebble is tossed into the air from the top of a cliff. The height, in feet, of the pebble over time is modeled by the equation

y = -16x^2 + 32x + 80. What is the maximum height, in feet, reached by the pebble?
The maxium height is _____ feet.

1 answer:

PIT_PIT [9.1K]10 days ago

5 0

Answer:

96

Step-by-step explanation:

The highest height achieved by the pebble represented by the quadratic function

$h(t)=-16t^2+32t+80$ can be determined by finding the vertex.

First, we identify the t-coordinate of this vertex. This maximum height corresponds to this value of t, thus, we need to evaluate the h(t)-coordinate.

By comparing $h$ and $at^2+bt+c$ , we observe that:

$a=-16$

$b=32$

$c=80$

Next, we compute to uncover the t-coordinate of the vertex:

$t=\frac{-b}{2a}$

$t=\frac{-32}{2(-16)}$

$t=\frac{-32}{-32}$

$t=1$

Now, substituting t into $-16t^2+32t+80$ with 1 will enable us to find the corresponding h(t)-coordinate:

$-16(1)^2+32(1)+80$

$-16(1)+32+80$

$-16+32+80$

$16+80$

$96$

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Applying the multiplication principle and n-factorial number:

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Answer:

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Step-by-step explanation:

Based on the provided details:

A Poisson process is utilized to model the occurrence of this event:

Minor radio blackouts occur, on average, two times annually.

Therefore:

$\lambda = 2 \ years$

For question (a)

time t = 1/2 (i.e., six months)

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Then:

$P(x = 3) = \dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!}$

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P(x = 3) = 0.061313

b).

The expected number of radio blackouts over two years is calculated as follows:

We start with:

t = 2 years

E(x) = λ × t

E(x) = 2 × 2

E(x) = 4

Thus, the predicted number of radio blackouts = 4

c).

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In this case, time (t) =???

Thus:

P(x =1) = 0.5

$P(x = 1) = \dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!} = 0.5$

$\dfrac{e^{-\lambda t} \times (\lambda t)^x}{x!} = 0.5$

Consequently, we are unable to ascertain the probability (50%) of the next radio blackout.

d)

We can compute the probability that the time until the fourth blackout is at most 2 years as follows:

Here;

x =4, t = 2

Thus:

$P(x = 4) = \dfrac{e^{-4 } \times (4)^4}{4!}$

P(x = 4) = 0.19537

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Response:

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2

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3

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