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Bezzdna [24]
3 years ago
6

Help me plz i don't even know if the first one is correct

Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0

Here's what i got:

Answer:

A)

<u><em>2.2 * 10^7</em></u>

B)

<u><em>0.072 = 7% are female</em></u>

C)

0.10455 * 22,000,000 = C

<u><em>0.10455 * 22,000,000 = 2,300,100</em></u>


D)

<u><em>0.0195 = 2% of the living veterans served in both. </em></u>





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You can work no more than 60 hours each week at your two jobs. Dog walking pays $7 per hour and your sales job at Computers &amp
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Answer:

No

Step-by-step explanation:

You cannot work negative 3 hours, it's impossible. One possible solution would be to work 50 hours a week at dog walking for 7 dollars an hour, and 10 hours a week at Computers & More, Inc. for 12 dollars a week. This would give you 350 dollars from dog walking, and 120 dollars from Computers & More, Inc. This would be a total of 470 dollars.

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3 years ago
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A sector of a circle has a central angle of 60°. find the area of the sector if the radius of the circle is 6 mi.
professor190 [17]
The entire circle has a central angle of 360°.
Because the radius is 6 mi, the area of the circle is
π(6)² = 36π mi².

The sector has a central angle of 60°, therefore it occupies 60/360 = 1/6 of the area of the circle.
The area of the sector is
(1/6)*36π = 6π mi² = 18.85 mi²

Answer: 6π mi²  or  18.85 mi².

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He perimeter of the figure is given. Find the length of the indicated side.
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3 0
3 years ago
Please solve the problem with steps
Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}

The first one: (\frac{1}{2})^{n-1} is a geometric sequence of first term (a_1) "1" and common ratio (r) " \frac{1}{2} ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2

The second one: \frac{1}{6^{n-1}} is a geometric sequence of first term "1", and common ratio (r) " \frac{1}{6} ". Again, since the common ratio is smaller than one, we can find its infinite sum:

Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{6} } =\frac{1}{\frac{5}{6} } =\frac{6}{5}

now we simply combine the results making sure we do the indicated difference: Infinite total sum= 2-\frac{6}{5} =\frac{10-6}{5} =\frac{4}{5}

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3 years ago
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0.000701 to a scientific notation
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7.01×10^-4. udufhfhf

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