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vova2212 [387]
2 years ago
14

Find the surface area of a hemisphere that has a volume of 486π

title="cm^{3}" alt="cm^{3}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Xelga [282]2 years ago
6 0

\textit{volume of a hemisphere}\\ V=\cfrac{4\pi r^3}{3}\cdot \cfrac{1}{2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=486\pi \end{cases}\implies 486\pi =\cfrac{4\pi r^3}{6} \implies 6(486\pi )=4\pi r^3 \\\\\\ \cfrac{6(486\pi )}{4\pi }=r^3\implies 729=r^3\implies \sqrt[3]{729}=r\implies \boxed{9=r} \\\\[-0.35em] ~\dotfill

\textit{surface area of a hemisphere}\\\\ SA=4\pi r^2\cdot \cfrac{1}{2}\implies \stackrel{\textit{we know that r = 9}}{SA=2\pi (9)^2}\implies SA=162\pi \implies SA\approx 508.94

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Alenkasestr [34]

The better statistics that can be used to compare the scores of the two teams are: <u>D. Median and IQR​.</u>

  • The diagram given shows two box plots representing the data of the two team's scores.

<u><em>Box plots</em></u><u><em> displays the </em></u><u><em>5-number summary</em></u><u><em>, namely:</em></u>

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The Interquartile Range (IQR) of the data can also be calculated by finding the difference between the First Quartile (Q_1), and the Third Quartile (Q_3).

The Interquartile Range (IQR) helps us to determine the variability of the data set.

The median that is gotten from the box pot measures the center tendency of the data.

Both median and IQR are better statistics that can be easily gotten from box plots to compare the team scores.

Therefore, the better statistics to use for comparing the scores of the teams are: <u>D. Median and IQR​.</u>

Learn more here:

brainly.com/question/15800880

4 0
2 years ago
Doughnuts are sold in bags and cartons.
irga5000 [103]

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Answer:

22.\frac{}{2} %

Step-by-step explanation:

\frac{Amount ( part)}{Base (whole) } =\frac{percent}{100}

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Which of the following equations represents a line that passes through the points
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y+15= -3(x-4)

or

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5 0
3 years ago
Read 2 more answers
☆15 POINTS AND MARKED BRAINLIEST IF CORRECT☆<br>look at the image above to view the question!​
swat32

Answer:

3125 bacteria.

Step-by-step explanation:

We can write an exponential function to represent the situation.

We know that the current population is 100,000.

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The standard exponential function is given by:

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So, we can say that t = -5. The negative represent the number of days that has passed.

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We know that the bacterial population had been doubling for 5 days. Let A represent the initial population. So, our function is:

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After 5 days, we reach the 100,000 population. So, when t = 5, P(t) = 100000:

100000=A(2)^5

And solving for A, we acquire:

\displaystyle A=\frac{100000}{2^5}=3125

So, our function in terms of the original day is:

P (t) = 3125 (2)^t

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3 years ago
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