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denpristay [2]
2 years ago
12

A golf player is trying to make a hole-in-one on the miniature golf green shown.

Mathematics
1 answer:
Flauer [41]2 years ago
5 0

The path of the ball is an illustration of absolute equation.

The equation of the path is: y =- \frac{12}7|x- 4.8| + 6.4

The given parameters are:

(h,k) = (4.8,6.4) --- the vertex (i.e. the point where the ball hits the wall)

(x_1,y_1) = (2,1.6)

(x_2,y_2) = (7.6,1.6)

An absolute function is represented as:

y = a |x - h| + k

Substitute (h,k) = (4.8,6.4)

y = a |x - 4.8| + 6.4

Substitute (x_1,y_1) = (2,1.6) for x and y

1.6 = a|2 - 4.8| + 6.4

1.6 = a|- 2.8| + 6.4

Remove absolute bracket

1.6 = 2.8a + 6.4

Collect like terms

2.8a = -6.4+1.6

2.8a =- 4.8

Solve for a

a =-\frac{4.8}{2.8}

Simplify

a =-\frac{12}{7}

Substitute a =-\frac{12}{7} in y = a |x - 4.8| + 6.4

y =- \frac{12}7|x- 4.8| + 6.4

Hence, the equation of the path is: y =- \frac{12}7|x- 4.8| + 6.4

See attachment for graph that models the path

Read more about absolute equations at:

brainly.com/question/2166748

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Answer:

B) 4.07

Step-by-step explanation:

First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:

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39.31             36.69                38.99             40.04

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Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63

Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):

SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12

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Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom

SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974

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For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom

Now we can establish the hypothesis for the test:

H0: μ1 = μ2 = μ3 = μ4

The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:

H1: the grade of gasoline does makes a difference

We will use the F statistic to test the hypothesis, which is calculated like follows:

F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19

We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:

Fc = 4.07

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