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True [87]
4 years ago
13

a bottle contains 4 fluid ounces of medicine. About how many milliliters of medicine are in the bottle?

Mathematics
1 answer:
yarga [219]4 years ago
4 0
There is 118.2941184 milliliters in four fluid ounces
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Need rn. worth 30.
Likurg_2 [28]

Considering the situation described, the confidence interval is given by:

(0.35, 0.51).

<h3>What is a confidence interval?</h3>

It is given by the <u>estimate plus/minus the margin of error</u>.

For this problem, we have that:

  • The estimate is of 0.43.
  • The margin of error is half the range of the observations, hence M = (0.48 - 0.32)/2 = 0.08.

Then, the bounds of the interval are given by:

  • 0.43 - 0.08 = 0.35.
  • 0.43 + 0.08 = 0.51.

More can be learned about confidence intervals at brainly.com/question/25890103

#SPJ1

4 0
2 years ago
Felicia has $6.40 in dimes and quarters. She has 34 coins in all. How many of each type of coin does Felicia have?
jonny [76]

Answer:

you would use 20 quarters and 14 dimes

Step-by-step explanation:

25*20=500

10*14=140

5.00+1.40=6.40

4 0
3 years ago
What is 5p to the second power plus 7p to the second power
Sauron [17]
The answer is 74. I'm 100% sure.
3 0
3 years ago
If the graphed line goes up from left to right, then the slope is positive.
irga5000 [103]

Answer:

a) True

Step-by-step explanation:

The left side is where the corner or 0 is, so if a line goes up towards the right, the answer would be positive.

7 0
3 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
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