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gtnhenbr [62]
3 years ago
14

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 this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot.
y=−16x^2+146x+137
Mathematics
1 answer:
Mars2501 [29]3 years ago
3 0

Answer:

10.17

Step-by-step explanation:

Given the following Quadratic equation:

The steps to solve it are:

1. Substitute  into the equation:

2. Use the Quadratic formula .

In this case:

Substituting values into the Quadratic equation, you get (Round the values to the nearest hundreth):

3. The positive value is the time that the rocket will hit the ground.

Therefore, the rocket will hit the ground after .

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Sergio [31]

The answer is 10 because all the angles are equal making it a equilateral triangle.

Therefore, the sides are all equal to one another, 3x - 8 = 5x - 20.

You just solve for x and get 6. You then plug in 6 for x and evauluate the equation, getting 10.

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When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwis
Mademuasel [1]

Answer:

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so \mu = 22, \sigma = 4.

What is the probability that a fisherman catches a spotted seatrout within the legal limits?

They must be between 10 and 30 inches.

So, this is the pvalue of the Z score of X = 30 subtracted by the pvalue of the Z score of X = 10

X = 30

Z = \frac{X - \mu}{\sigma}

Z = \frac{30 - 22}{4}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 10

Z = \frac{X - \mu}{\sigma}

Z = \frac{10 - 22}{4}

Z = -3

Z = -3 has a pvalue of 0.00135.

This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

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Answer:

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Answer:

f(x) and r(x)

Step-by-step explanation:

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