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padilas [110]
3 years ago
9

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

e given equation. Using the equation, what height was the rocket launched from? y=-16x^2+57x+32
Mathematics
1 answer:
iren [92.7K]3 years ago
6 0

9514 1404 393

Answer:

  32 feet

Step-by-step explanation:

0 seconds after launch, the rocket is still sitting on the tower. Its height is...

  y = -16(0^2) +57(0) +32

  y = 0 + 0 + 32 = 32

The rocket was launched from a tower 32 feet high.

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Read 2 more answers
(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX
pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and  variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

P(4

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

P(X

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

P(X

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0
3 years ago
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