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I am Lyosha [343]
3 years ago
15

A random sample is drawn from a population with mean μ = 66 and standard deviation σ = 5.5. use table 1.

Mathematics
1 answer:
creativ13 [48]3 years ago
7 0
<span>A random sample is drawn from a population with mean μ = 66 and standard deviation σ = 5.5. use table 1.
a. is the sampling distribution of the sample mean with n = 16 and n = 36 normally distributed? yes, both the sample means will have a normal distribution. no, both the sample means will not have a normal distribution. no, only the sample mean with n = 16 will have a normal distribution. no, only the sample mean with n = 36 will have a normal distribution.
b. can you use the standard normal distribution to calculate the probability that the sample mean falls between 66 and 68 for both sample sizes? yes, for both the sample sizes, standard normal distribution could be used. no, for both the sample sizes, standard normal distribution could not be used. no, only for the sample size with n = 16, standard normal distribution could be used. no, only for the sample size with n = 36, standard normal distribution could be used.
c. calculate the probability that the sample mean falls between 66 and 68 for n = 36. (round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.)</span>
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tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

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\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
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vazorg [7]

Answer:

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Step-by-step explanation:

assuming that the desktop screen is a rectangle, then it's diagonal will form two right triangles.

by the Pythagorean theorem

{a}^{2}  +  {b}^{2}  =  {c}^{2}  \\  a = 14 \: and \: b = 18 \\ c  = the \: hypoteneuse

then

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3 years ago
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