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3 years ago

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A rock is thrown upward from a bridge into a river below. The function f(t) = −16???? 2 + 41t + 130 determines the height of the

rock above the surface of the water (in feet) in terms of the number of seconds t since the rock was thrown.

1 answer:

Butoxors [25]3 years ago

7 0

Answer:

The bridge's height above the water is 130 feets.

Step-by-step explanation:

A rock is thrown upward from a bridge into a river below :

$h(t)=-16t^2+41t+130$

Here t is time in seconds

It is required to find the bridge's height above the water. When it reaches the height of the rock above the surface of the water, then :

h(t) = 0

$f(0)=-16t^2+41t+130\\\\f(0)=-16(0)^2+41(0))+130\\\\f(0)=130\ ft$

So, the bridge's height above the water is 130 feets.

You might be interested in

Suppose that IQs of East State University’s students can be described by a Normal model with mean 130 and standard deviation 8 p

lianna [129]

Answer:

a) $P(X \geq 125) = 1-P(X$

b) $P(R \geq 5) = 1-P(R$

c) $P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z$

d) $P(H\geq 5)= P(Z> \frac{5-10}{7.394}) = 1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the scores for the East population, and for this case we know the distribution for X is given by:

$X \sim N(130,8)$

Where $\mu_x=130$ and $\sigma_x=8$

Let Y the random variable that represent the scores for the West population, and for this case we know the distribution for Y is given by:

$X \sim N(120,10)$

Where $\mu_y=120$ and $\sigma_y=10$

Part a

For this case we want this probability:

$P(X\geq 125)$

And we can use the z score given by:

$z = \frac{X -\mu}{\sigma}$

And if we replace we got:

$P(X \geq 125) = 1-P(X$

Part b

For this case we need to define the following random variable R = X-Y and we know that the distribution of R is given by:

$R \sim N (130-120= 10, \sigma_R = \sqrt{8^2 +10^2}=12.806)$

And we want this probability:

$P(R\geq 5)$

We can use the z score given by:

$z= \frac{R -\mu_R}{\sigma_R}$

If we use this formula we got:

$P(R \geq 5) = 1-P(R$

Part c

For this case we select a sample size of n =3 for the Y distribution, the sample mean have the following distribution:

$\bar Y \sim N(120, \frac{10}{\sqrt{3}}=5.774)$

And we want this probability:

$P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z$

Part d

For this case we define the following random variable $H = \bar X -\bar Y$ and the distribution for H is given by:

$H \sim N (130-120=10, \sigma_H = \sqrt{\frac{8^2 +10^2}{3}}= 7.394)$

And the z score would be given by:

$z = \frac{H -\mu_H}{\sigma_H}$

And if we find the probability required we got:

$P(H\geq 5)= P(Z> \frac{5-10}{7.394}) = 1-P(Z$

8 0

3 years ago

What is the area of the parallelogram?<br><br>Any help will be appreciated. Thank you!☺️

andreev551 [17]

Answer:

40

Step-by-step explanation:

A = bxh

=(8)(5)

=40

4 0

3 years ago

Read 2 more answers

You toss three 6-sided dice and record the sum of the three faces facing up. a) Describe precisely a sample space S for this exp

timama [110]

Answer:

a.)Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

b.) 9/216

c.) 9/216

d.) 212/216

Step-by-step explanation:

Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]

[1,2,1] [1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],

[1,3,1] [1,3,2] [1,3,3] [1,3,4] [1,3,5] [1,3,6]

[1,4,1] [1,4,2] [1,4,3] [1,4,4] [1,4,5] [1,4,6]

[1,5,1] [1,5,2] [1,5,3] [1,5,4] [1,5,5] [1,5,6]

[1,6,1] [1,6,2] [1,6,3] [1,6,4] [1,6,5] [1,6,6]

[2,1,1] [2,1,2] [2,1,3] [2,1,4] [2,1,5] [2,1,6]

[2,2,1] [2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],

[2,3,1] [2,3,2] [2,3,3] [2,3,4] [2,3,5] [2,3,6]

[2,4,1] [2,4,2] [2,4,3] [2,4,4] [2,4,5] [2,4,6]

[2,5,1] [2,5,2] [2,5,3] [2,5,4] [2,5,5] [2,5,6]

[2,6,1] [2,6,2] [2,6,3] [2,6,4] [2,6,5] [2,6,6]

[3,1,1] [3,1,2] [3,1,3] [3,1,4] [3,1,5] [3,1,6]

[3,2,1] [3,2,2] [3,2,3] [3,2,4] [3,2,5] [3,2,6],

[3,3,1] [3,3,2] [3,3,3] [3,3,4] [3,3,5] [3,3,6]

[3,4,1] [3,4,2] [3,4,3] [3,4,4] [3,4,5] [3,4,6]

[3,5,1] [3,5,2] [3,5,3] [3,5,4] [3,5,5] [3,5,6]

[3,6,1] [3,6,2] [3,6,3] [3,6,4] [3,6,5] [3,6,6]

[4,1,1] [4,1,2] [4,1,3] [4,1,4] [4,1,5] [4,1,6]

[4,2,1] [4,2,2] [4,2,3] [4,2,4] [4,2,5] [4,2,6],

[4,3,1] [4,3,2] [4,3,3] [4,3,4] [4,3,5] [4,3,6]

[4,4,1] [4,4,2] [4,4,3] [4,4,4] [4,4,5] [4,4,6]

[4,5,1] [4,5,2] [4,5,3] [4,5,4] [4,5,5] [4,5,6]

[4,6,1] [4,6,2] [4,6,3] [4,6,4] [4,6,5] [4,6,6]

[5,1,1] [5,1,2] [5,1,3] [5,1,4] [5,1,5] [5,1,6]

[5,2,1] [5,2,2] [5,2,3] [5,2,4] [5,2,5] [5,2,6],

[5,3,1] [5,3,2] [5,3,3] [5,3,4] [5,3,5] [5,3,6]

[5,4,1] [5,4,2] [5,4,3] [5,4,4] [5,4,5] [5,4,6]

[5,5,1] [5,5,2] [5,5,3] [5,5,4] [5,5,5] [5,5,6]

[5,6,1] [5,6,2] [5,6,3] [5,6,4] [5,6,5] [5,6,6]

[6,1,1] [6,1,2] [6,1,3] [6,1,4] [6,1,5] [6,1,6]

[6,2,1] [6,2,2] [6,2,3] [6,2,4] [6,2,5] [6,2,6],

[6,3,1] [6,3,2] [6,3,3] [6,3,4] [6,3,5] [6,3,6]

[6,4,1] [6,4,2] [6,4,3] [6,4,4] [6,4,5] [6,4,6]

[6,5,1] [6,5,2] [6,5,3] [6,5,4] [6,5,5] [6,5,6]

[6,6,1] [6,6,2] [6,6,3] [6,6,4] [6,6,5] [6,6,6]

b.) Probability that the sum is 16 or more is

Pr[4,6,6] + pr[5,5,6] + pr [ 5,6,5] + pr [5,6,6] + pr [6,5,5] + pr [6,5,6] + pr [6,6,4] + pr[6,6,5] + pr [6,6,6]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is 4 or 5 is

Pr [ 1,1,2] or pr[1,2,2] or pr [1,1,3] or pr [1,2,1] or pr[2,1,2] or pr[1,3,1] or pr[3,1,1] or pr[2,1,1] or pr[2,2,1]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is less than 17

We take it as:

1- probability that the sum is 17 and above.

Now probability that the sum is 17 and above becomes

pr[5,6,6] or pr[6,5,6] or pr[6,6,5] or pr[6,6,6]

= [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 4/216

Hence, probability that the sum is less than 17 becomes:

1-4/216 = 212/216.

3 0

3 years ago

-24-8x=12y 1+5/9y=-7/18x

kipiarov [429]

Answer:

(6,-6)

Step-by-step explanation:

6 0

3 years ago

Bob played a trivia game in which he lost 3 points for each incorrect answer and gained 8 points for each correct answer. If Bob

hammer [34]

Answer:

76

Because 11*8=88 and 4*3=12

88-12=76

3 0

3 years ago