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aev [14]
3 years ago
10

Average daily temperatures in Tucson,Arizona in December 67,57,52,51,64,58,67,58,55,59,66,50,57,62,58,50,58,50,60,63

Mathematics
2 answers:
valkas [14]3 years ago
4 0
Add all the numbers then divide by the number of numbers there are. 51.1
Virty [35]3 years ago
3 0
The average temperature is 58 degrees
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You need to add the data set

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4 years ago
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A prop for the theater club’s play is constructed as a cone topped with a half-sphere. What is the volume of the prop? Round you
prisoha [69]

The volume of the prop is calculated to be 1,875.6 cubic inches.

Step-by-step explanation:

Step 1:

The prop consists of a cone and a half-sphere on top. We will have to calculate the volumes of the cone and the half-sphere separately and then add them to obtain the total volume.

Step 2:

The volume of a cone is determined by multiplying \frac{1}{3} with π, the square of the radius (r²) and height (h). Here we substitute π as 3.14. The radius is 8 inches and the height is 12 inches.

The volume of the cone: \frac{1}{3} \pi r^{2} h = \frac{1}{3} (3.14) (8^{2}) (12)= 803.84 cubic inches.

Step 3:

The area of a half-sphere is half of a full sphere. The volume of a sphere is given by multiplying \frac{4}{3} with π and the cube of the radius (r³).

Here the radius is 8 inches. We take π as 3.14.

The volume of a full sphere= \frac{4}{3} \pi r^{3} = \frac{4}{3} (3.14) (8^{3} ) = 2,143.573 cubic inches.

The volume of the half-sphere= \frac{2,143.573}{2} = 1,071.7865 cubic inches.

Step 4:

The total volume = The volume of the cone + The volume of the half sphere,

The total volume =  803.84+1,071.7865 = 1,875.6265 cubic inches.

Rounding this off, we get the volume of the prop as 1,875.6 cubic inches.

6 0
3 years ago
Can you help me please?
morpeh [17]

\bf 3\sqrt{2}-\sqrt{50}+3-\sqrt{8}+\sqrt{288}-\sqrt{9}~~ \begin{cases} 50=2\cdot 5\cdot 5\\ \qquad 2\cdot 5^2\\ 8=2\cdot 2\cdot 2\\ \qquad 2^2\cdot 2\\ 288=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\\ \qquad 2^4\cdot 2\cdot 3^2\\ \qquad 2 (2^2)^2 3^2\\ \qquad 2(2^2 3)^2\\ \qquad 2(12)^2\\ 9=3^2 \end{cases}


\bf 3\sqrt{2}-\sqrt{2\cdot 5^2}+3-\sqrt{2^2\cdot 2}+\sqrt{2\cdot 12^2}-\sqrt{3^2} \\\\\\ 3\sqrt{2}-5\sqrt{2}+3-2\sqrt{2}+12\sqrt{2}-3 \\\\\\ 3\sqrt{2}-5\sqrt{2}-2\sqrt{2}+12\sqrt{2}+3-3\implies \stackrel{\textit{adding like-terms}}{8\sqrt{2}}

7 0
3 years ago
How can i do this? Plz help me
FinnZ [79.3K]
Write the area I guess ex. 14 = 6t + 6
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What's bigger 3.05kg or 3.56
Julli [10]

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3.56

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