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SVEN [57.7K]
2 years ago
3

A diver jumped from a diving platform. The image shows her height above the water at several different times after leaving the p

latform.
Part A - Find the equation of the quadratic function that describes the relationship between the diver's time and height. Round to the nearest tenth.

(Equation has to be in standard form, y = ax^2+bx+c)

Mathematics
2 answers:
Contact [7]2 years ago
6 0

It is given to us that we have to use the equation in the standard form as given as:

y=ax^2+bx+c

which is the quadratic equation of a generalized parabola.

Now, to find the answer to the Part A all that we need to do is take selected snapshots of the divers coordinates and plug them in the given equation and then solve for the parameters a,b,c which will help us complete the quadratic function. We know that the time (in seconds) is the x coordinate and the height (in ft) is the y coordinate.

We will need (any) three of the five positions of the diver given in the question because we have three unknowns a,b and c that we need to find.

Let us take the first three positions. Thus, our equation will become:

27=a(0.25)^2+b(0.25)+c

29=a(0.5)^2+b(0.5)+c and

29=a(1.5)^2+b(1.5)+c

Solving the above three equations simultaneously by using a calculator, we get the values of the three parameters (rounded) to the nearest tenths to be:

a=-6.4, y=12.8, z=24.2

Thus, the equation of the quadratic function that describes the relationship between the diver's time and height is:

y=-6.4x^2 +12.8x+24.2

MArishka [77]2 years ago
3 0

Answer:

y = -5*x^2 + 10.2*x +24.9

Step-by-step explanation:

time (s)   height (ft)

0.25       27

0.5         29

1.5          29

2            25

2.75        15

To find the quadratic equation that correlates the data you can use either a calculator or a spreadsheet software, like MS Excel. Using Excel, first you have to input data; then select it and add a scatter plot; next, select one of the dots, left click, add regression line; finally, in the new menu, choose polynomial, degree = 2. In the picture attached, the result is shown.

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riadik2000 [5.3K]

Answer:

The conclusion that both groups of overweight and non - overweight got cardiovascular benefit from playing DDR games requires Inferential statistics.

Step-by-step explanation:

Inferential statistics is simply a type of research statistic whereby a generalized conclusion is made about a larger group based on representative observations

Now,in the given question, we see that both group hearts were above the minimum recommended for cardiovascular exercise. Now we can infer that the DDR games played by both groups gave them cardiovascular benefits. This conclusion is an example of Inferential statistics where we generalize about a large population based on observations from a small sample.

Thus the conclusion that both groups of overweight and non-overweight got cardiovascular benefit from playing DDR games requires Inferential statistics.

3 0
2 years ago
Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45
vovikov84 [41]

Answer:

a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544

b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390

ii. There are no statistical significant evidence that the prep course helped

d. i. The 95% confidence interval for the change in average test scores is  3.47467 < μ₁ - μ₂ < 14.52533

ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

Step-by-step explanation:

The mean of the standardized test = 1,000

The number of students test to which the test is administered = 453 students

The mean score of the sample of students, \bar{x} = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}

At 95% confidence level, z = 1.96, therefore, we have;

CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}

Therefore, we have;

1,022.94559 < μ < 1,003.0544

b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. The parameters of the students taking the test are;

The number of students, n = 503

The number of hours preparation the students are given, t = 3 hours

The average test score of the student, \bar{x} = 1019

The number of test scores of the student, s = 95

At 95% confidence level, z = 1.96, therefore, we have;

The confidence interval, C.I., for the difference in mean is given as follows;

C.I. = \left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore, we have;

C.I. = \left (1013- 1019  \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}

Which gives;

-18.955390 < μ₁ - μ₂ < 6.955390

ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped

d. The given parameters are;

The number of students taking the test = The original 453 students

The average change in the test scores, \bar{x}_{1}- \bar{x}_{2} = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = \bar{x}_{1}- \bar{x}_{2} + 1.96 × Δs/√n

∴ C.I. = 9 ± 1.96 × 60/√(453)

i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533

ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

5 0
2 years ago
Complete the steps for solving 7 = –2x2 + 10x. Factor out of the variable terms. inside the parentheses and on the left side of
Mamont248 [21]

we have

7=-2x^{2} +10x

Factor the leading coefficient

7=-2(x^{2} -5x)

Complete the square. Remember to balance the equation by adding the same constants to each side

7-12.50=-2(x^{2} -5x+2.5^{2})

-5.50=-2(x^{2} -5x+2.5^{2})

Divide both sides by -2

2.75=(x^{2} -5x+2.5^{2})

Rewrite as perfect squares

2.75=(x-2.5)^{2}

Taking the square roots of both sides (square root property of equality)

x-2.5=(+/-)\sqrt{2.75}

Remember that

\sqrt{2.75}=\sqrt{\frac{11}{4}}= \frac{\sqrt{11}}{2}

x-2.5=(+/-)\frac{\sqrt{11}}{2}

x=2.5(+/-)\frac{\sqrt{11}}{2}

x=2.5+\frac{\sqrt{11}}{2}=\frac{5+\sqrt{11}}{2}

x=2.5-\frac{\sqrt{11}}{2}=\frac{5-\sqrt{11}}{2}

<u>the answer is</u>

The solutions are

x=\frac{5+\sqrt{11}}{2}

x=\frac{5-\sqrt{11}}{2}


5 0
2 years ago
Read 2 more answers
The cost of 5 scarves is $91.25. What is the unit price? The unit price is $ per scarf.
NikAS [45]

Answer:

18.25

Step-by-step explanation:

just divide the total cost of scarves by the number of scarves:

91.25 / 5 = 18.25

6 0
2 years ago
In ΔGHI, the measure of ∠I=90°, IG = 6.8 feet, and HI = 2.6 feet. Find the measure of ∠G to the nearest degree.
Luda [366]

Answer:

\angle G=20.8\approx 21^{\circ}

Step-by-step explanation:

Given: In ΔGHI, \angle I=90°, IG = 6.8 feet, and HI = 2.6 feet

To find: \angle G

Solution:

Trigonometry defines relationship between the sides and angles of the triangle.

For any angle \theta,

tan\theta = side opposite to \theta/side adjacent to \theta

In ΔGHI,

tan G=\frac{HI}{GI}

Put HI=2.6  \,\,feet\,,\,GI=6.8\,\,feet

So,

tan G=\frac{2.6}{6.8}=0.38

Therefore, \angle G=20.8\approx 21^{\circ}

Download pptx
3 0
2 years ago
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