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Oksana_A [137]
2 years ago
5

A diver jumps off a 20 foot high diving board with an initial upward velocity of 4 feet per second. Use the vertical motion mode

l
h=-16t² + vt + s.
PART A:
Using the information above, the vertical motion model of the diver is h-

PART B:
The solutions of the model are t =
and t=

PART C:
Choose the letter that best fits the scenario.
A. There are no real solutions, because the solutions found do not make sense.
B. There is only one solution, because the solutions indicate time and time cannot be negative.
C. There are two solutions, because when solving, there were two solutions that occurred.
D. None of the above because there is not enough information.
Answer: ______
Mathematics
1 answer:
Pachacha [2.7K]2 years ago
7 0

The height is represented by h = -16t² + 4t + 20 and There is only one solution, because the solutions indicate time and time cannot be negative.

<h3>How to solve vertical motion models?</h3>

We are given the vertical model of the diver's jump as;

h = -16t² + vt + s

where;

h is height

t is time

v is velocity

s is distance

A) We are given;

Distance; s = 20 ft

velocity; v = 4 ft/s

Thus;

h = -16t² + 4t + 20

B) The solution of the model is gotten from;

-16t² + 4t + 20 = 0

Solving this quadratic equation gives;

t = 1.25 seconds or -1 second

C) B. There is only one solution, because the solutions indicate time and time cannot be negative.

Read more about Vertical Motion Models at; brainly.com/question/27526920

#SPJ1

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The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches
ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

H_0: \mu = 28.76

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

H_1: \mu > 28.76

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that \mu = 28.76

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that \sigma = 7.5, X = 33, n = 32.

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}

z = 3.2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level \alpha = 0.05.

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

1 - 0.9993 = 0.0007

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

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Answer:

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Step-by-step explanation:

10 2/3 * 3.5=37 1/3

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Simplify the equation -4d - 37 = 7d + 18
exis [7]

Answer:

d=-5

Step-by-step explanation:

-4d-37=7d +18

move the terms to its correct side

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