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Anna007 [38]
3 years ago
5

A diver jumps up and over from a cliff in order to avoid rocks and land in the deeper water. The path of the dive can be modeled

using the function f(x)=−16x2+2x+48, where f(x) is the height, in feet, of the diver at x time in seconds. What is the maximum height, in feet, the diver will reach? Round your answer to the nearest hundredth if necessary. Do not include units in your answer.
Mathematics
1 answer:
lions [1.4K]3 years ago
4 0

Answer:

48.06  to the nearest hundredth.

Step-by-step explanation:

f(x) = -16x^2 + 2x + 48

To find the maximum height we convert to vertex form:

= -16(x^2 + 1/8x) + 48

= -16[x + 1/16)^2 - 1/256] + 48

= -16(x + 1.16)^2  + 16/256 + 48

= 48.0625.

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Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t
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Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6)  =\dfrac{20}{36}

The probability distribution of this event is given as follows.

\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|

First, we determine the expected Value of this event.

Expected Value

=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning  =\dfrac{20}{36}

If the game is played 100 times

Number of times expected to win

=\dfrac{20}{36} \times 100\\=56$ times

Therefore, number of times expected to loose

= 100-56

=44 times

8 0
3 years ago
Ik it’s pretty easy but u knowh am not really sure if am right or not
gladu [14]

2,6,18

r = 6/2 = 3

18 *3 = 54

54*3 = 162

162*3=486

Answer: 54,162,486

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3 years ago
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