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Sveta_85 [38]
2 years ago
8

Classic Golf Inc. manages five courses in the Jacksonville, Florida, area. The director of golf wishes to study the number of ro

unds of golf played per weekday at the five courses. He gathered the following sample information. Day Rounds Monday 120 Tuesday 105 Wednesday 100 Thursday 150 Friday 150 At the 0.01 significance level, is there a difference in the number of rounds played by day of the week?
(Round your answers to 3 decimal places.)
Mathematics
1 answer:
pochemuha2 years ago
8 0

Answer:

After statistical analysis we find that the critical value with respect to a level of significance of 0.01 is greater, we deduce that if there is a difference in the number of rounds played per day in the week

Step-by-step explanation:

120+105+100+150+150=625/5=125= Expected frecuency

(observed - expected)^2/expected

Monday. 120 2

Tuesday. 105. 3.2

Wednesday. 100. 5

Thursday. 150. 5

Friday 150. 5

The total sum will give us the statistical chi-square test that is 20.2

according to the previous listing we look for the degrees of freedom:

5-1=4

Critical value = 3.747

After statistical analysis we find that the critical value with respect to a level of significance of 0.01 is greater, we deduce that if there is a difference in the number of rounds played per day in the week

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Answer:

Step-by-step explanation:

Given the equation  4x²+ 49y² = 196

a) Differentiating implicitly with respect to y, we have;

8x + 98y\frac{dy}{dx} = 0\\98y\frac{dy}{dx}  = -8x\\49y\frac{dy}{dx}  = -4x\\\frac{dy}{dx} = \frac{-4x}{49y}

b)  To solve the equation explicitly for y and differentiate to get dy/dx in terms of x,

First let is make y the subject of the formula from the equation;

If 4x²+ 49y² = 196

49y² = 196 - 4x²

y^{2} =  \frac{196}{49}  - \frac{4x^{2} }{49} \\y = \sqrt{\frac{196}{49}  - \frac{4x^{2} }{49} \\} \\

Differentiating y with respect to x using the chain rule;

Let u=  \frac{196}{49}  - \frac{4x^{2} }{49}

y =  \sqrt{u} \\y =u^{1/2} \\

\frac{dy}{dx}  = \frac{dy}{du} * \frac{du}{dx}

\frac{dy}{du} = \frac{1}{2}u^{-1/2} \\

\frac{du}{dx} =  0 - \frac{8x}{49} \\\frac{du}{dx} =\frac{-8x}{49} \\\frac{dy}{dx} = \frac{1}{2} ( \frac{196}{49}  - \frac{4x^{2} }{49})^{-1/2} *  \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} (  \frac{196-4x^{2} }{49})^{-1/2} *  \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} ( \sqrt{ \frac{49}{196-4x^{2} })} *  \frac{-8x}{49}\\\frac{dy}{dx} = \frac{1}{2} *{ \frac{7}\sqrt {196-4x^{2} }} *  \frac{-8x}{49}\\

\frac{dy}{dx} = \frac{-4x}{7\sqrt{196-4x^{2} } }

c) From the solution of the implicit differentiation in (a)

\frac{dy}{dx} = \frac{-4x}{49y}

Substituting y = \sqrt{\frac{196}{49}  - \frac{4x^{2} }{49} \\ into the equation to confirm the answer of (b) can be shown as follows

\frac{dy}{dx} = \frac{-4x}{49\sqrt{\frac{196-4x^{2} }{49} } }\\\frac{dy}{dx}  =  \frac{-4x}{49\sqrt{196-4x^{2}}/7} }\\\\\frac{dy}{dx}  = \frac{-4x}{7\sqrt{196-4x^{2}}}

This shows that the answer in a and b are consistent.

6 0
3 years ago
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