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tester [92]
3 years ago
15

Two studies were completed in Georgia. One study in northern Georgia involved 3,000 patients; 84% of them experienced flulike sy

mptoms during the month of December. The other study, in southern Georgia, involved 2,000 patients; 75% of them experienced flulike symptoms during the same month. Which study has the smallest margin of error for a 99% confidence interval?
The northern Georgia study with a margin of error of 2.5%.
The southern Georgia study with a margin of error of 2.5%.
The northern Georgia study with a margin of error of 1.7%.
The southern Georgia study with a margin of error of 1.7%.
Mathematics
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer:

The northern Georgia study with a margin of error of 1.7%.

All else being equal, the study with the larger sample size has the smaller margin of error. So the result makes sense (as 0.84 is pretty close to 0.75)

Step-by-step explanation:

The best and most correct answer among the choices provided by the question is the third choice "The northern Georgia study with a margin of error of 1.7%."

All else being equal, the study with the larger sample size has the smaller margin of error. So the result makes sense (as 0.84 is pretty close to 0.75)

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MrMuchimi
It would be 1/64 I hope this helps with your question.
8 0
3 years ago
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Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
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3 years ago
Jensen is building a snow fort. Each block in the fort weighs about 1 kilogram. Jensen hopes to make about 40 blocks for the for
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Answer: 13,333 snowflakes

Step-by-step explanation:

For this exercise let be "x" represents the number of snowflakes that will be in the fort.

According to the information given in the exercise, the weight of one block is 1 kilogram. Knowing that the fort must have 40 blocks, the total weight is:

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Therefore, through this procedure you get the following result:

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Therefore, the there will be 13,333 snowflakes in the fort.

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Now, multiply the units by the numbers in the ratio.
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