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Lyrx [107]
3 years ago
11

The Scatter plot below shows the linear trend of the number of golf carts a company sold the month of February and a line of bes

t fit representing this trend.
IT IS A DIFFERENT QUESTION FROM REPAIRING GOLF CARTS PLEASE READ


A. Write a function that models the number of golf carts sold as a function of the number of days in the month of February

B. What is the meaning of the slope as a rate of change for this line of best fit.

Mathematics
1 answer:
Andrei [34K]3 years ago
4 0

Answer:

Part A) y=-5.625x+90

Part B) see the explanation

Step-by-step explanation:

Part A) Write a function that models the number of golf carts sold as a function of the number of days in the month of February

Let

x ----> the number of days

y ----> the number of golf carts sold

we know that

The linear equation in slope intercept form is equal to

y=mx+b

where

m is the slope or unit rate

b is the y-intercept or initial value

In this problem

The y-intercept is the point (0,90)

so

b=90

Find the slope m

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

take two points from the graph

(0,90) and (16,0)

substitute the values in the formula

m=\frac{0-90}{16-0}

m=\frac{-90}{16}

simplify

m=-\frac{45}{8}  ---> is negative because is a decreasing function

therefore

The linear equation is equal to

y=-\frac{45}{8}x+90

Part B) What is the meaning of the slope as a rate of change for this line of best fit

The slope is m=-\frac{45}{8}\ golf\ carts\ sold/days

That means ----> every 8 days 45 less cars are sold

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SOVA2 [1]

Any arithmetic operation on polynomials except division* will result in a polynomial. Appropriate choices are ...

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E. (4/8)x . . . . . but not if you mean 4/(8x)

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7 0
3 years ago
Read 2 more answers
The desired percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is 5.5. To test whether the true average
Reil [10]

Answer:

a)  Null hypothesis : H₀ : μ = 5.5

 Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

        |t| = |-3.33| = 3.33

c) P - value lies between in these intervals

0.001 < P < 0.005

Step-by-step explanation:

<u><em>Step( i )</em></u>:-

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the  percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

<u><em> Null hypothesis : H₀ : μ = 5.5</em></u>

<u><em>  Alternative Hypothesis : H₁ : μ < 5.5</em></u>

 Level of significance ∝ = 0.01

<u><em>Step(ii)</em></u>:-

 The test statistic

                              t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

                             t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }

On calculation , we get

                            t = -3.33

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<u><em>Step(iii)</em></u>:-

<u><em>P - value</em></u>

<u><em>The degrees of freedom γ = n-1 = 16-1 =15</em></u>

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

<u>Condition(i)</u>

P - value < ∝ then reject H₀

<u>Condition(ii)</u>

P - value > ∝ then Accept H₀

we observe that  0.001 < P < 0.005

P- value < 0.01

we rejected  H₀

<em>(or)</em>

The tabulated value  = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

<u><em>Conclusion</em></u>:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

<u><em></em></u>

             

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