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ololo11 [35]
3 years ago
15

A sample of 92 observations is taken from a process (an infinite population). the sampling distribution of x̄ is approximately n

ormal because
Mathematics
1 answer:
Sindrei [870]3 years ago
6 0
It is normal based on the Central Limit Theorem. According to the theorem, an appropriately big sample (infinite) size from a population with a limited level of variance, the average of all samples from the same populace will be roughly equivalent to the mean of the population.
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Answer:

-\frac{10}{3}ft/s

Step-by-step explanation:

We are given that

Height of man=5 foot

\frac{dy}{dt}=-10ft/s

Height of street light=20ft

We have to find the rate of change of the length of his shadow when he is 25 ft form the street light.

ABE and CDE are similar triangle because all right triangles are similar.

\frac{20}{5}=\frac{x+y}{x}

4=\frac{x+y}{x}

4x=x+y

4x-x=y

3x=y

3\frac{dx}{dt}=\frac{dy}{dt}

\frac{dx}{dt}=\frac{1}{3}(-10)=-\frac{10}{3}ft/s=-\frac{10}{3}ft/s

Hence, the rate of change of the length of his shadow when he is 25 ft from the street light=-\frac{10}{3}ft/s

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3 years ago
Help with algebra problem
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andrezito [222]

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36/10? = 18/5 = 3.6 Are u sure that's the question?
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Find the expression that represents the length of the hypotenuse of a right triangle whose legs measure m^2-n^2 and 2mn
My name is Ann [436]
\bf c^2=a^2+b^2\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
------\\
a=m^2-n^2\\
b=2mn
\end{cases}\\\\\\ c^2=(m^2-n^2)^2+(\underline{2mn})^2
\\\\\\
c^2=m^4-2m^2n^2+n^4+\underline{2^2m^2n^2}
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c^2=m^4-2m^2n^2+n^4+{4m^2n^2}
\\\\\\
c^2=m^4+2m^2n^2+n^4\impliedby \textit{perfect square trinomial}
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c^2=(m^2+n^2)^2\implies c=\sqrt{(m^2+n^2)^2}\implies c=m^2+n^2
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