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sergeinik [125]
3 years ago
15

Sales of a new line of athletic footwear are crucial to the success of a newly formed company, Fleet Shoes. Fleet wishes to esti

mate the average weekly sales of the new footwear within $200 with 95% reliability. The initial sales indicate the standard deviation of the weekly sales figures to be approximately $1,500. How many weeks of data must be sampled for Fleet to get the information is desires?
Mathematics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

The number of weeks required = 216 weeks

Step-by-step explanation:

Given that:

Margin of Error E = 200

Confidence interval = 95% = 0.95

Level of SIgnificance  = 1 - C.I

= 1 - 0.95

= 0.05

Standard deviation = 1500

The Critical value for Z :

Z_{\alpha/2} =Z_{0.05/2}  \\ \\  = Z_{0.025} =  1.96

The number of weeks( i.e the sample size (n) ) required is :

n = (\dfrac{Z_{\alpha/2} \times \sigma}{E})^2

n = (\dfrac{1.96 \times 1500}{200})^2

n = (14.7)^2

n = 216.09

n ≅ 216 weeks

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Chech the picture below.

if the person kicked it from the ground, that means its initial height is 0.

it reaches its maximum height at the y-coordinate of its vertex, and it will hit the ground when y = 0, as you see in the picture.

\bf ~~~~~~\textit{initial velocity}
\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}
\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{8}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{0}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+8t+0

now let's find the y-coordinate of its vertex

\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+8}t\stackrel{\stackrel{c}{\downarrow }}{+0}
\qquad \qquad 
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~  c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(\qquad ,~~0-\cfrac{8^2}{4(-16)}  \right)\implies (\quad ,~0+1)\implies (\quad ,~1)

when will it hit the ground?

\bf \stackrel{h(t)}{0}=-16t^2+8t\implies 16t^2-8t=0\implies 8t(2t-1)=0
\\\\\\
8t=0\implies \stackrel{\textit{0 seconds when it took off first}}{t=0}
\\\\\\
2t-1=0\implies 2t=1\implies \stackrel{\textit{half a second later it  when it came back down}}{t=\cfrac{1}{2}}

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