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Ilya [14]
3 years ago
9

What are some sources of new mathematical symbols that have been widely used?

Mathematics
1 answer:
Svet_ta [14]3 years ago
7 0
The sum of three cubes
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brainly.com/question/9513977
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Can anybody give real answers please? Thanks.
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Electronics Unlimited sells TVs. There are 110 TVs on display in the showroom, and each TV is turned on to a random channel from
Lynna [10]

Answer:

  • <u><em>Option B. There will be close to 40 TVs but probably not exactly 40 TVs not showing a sports channel.</em></u>

Explanation:

There are a total of <em>7 sports channels</em> and 4 non-sports channels o<em>ut of 11 channels.</em>

The probability that one <em>TV will be not be showing a sports channel</em>, P(not S), is:

  • P(not S) = number of non-sports channels / number of channels.

  • P(not S) = 4/11.

The <em>best prediction</em> on <em>how many TVs will not be showing a sports channel </em>is, the expected value, which is equal to the number of TVs mulitplied by P(not S):

  • P(not S) = 110 × 4/11 = 40.

Since this is a random variable, the expected value is not the exact number of TV but just a probability.

Hence, the answer is the option <em>B: There will be close to 40 TVs but probably not exactly 40 TVs not showing a sports channel.</em>

3 0
3 years ago
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The answer to these please!
igomit [66]
\sqrt[4]{(4a^{2})^{6}} = \sqrt[4]{4096a^{12}} = 8a^{3}

\sqrt[7]{(b^{\frac{5}{3}})^{8}} = \sqrt[7]{b^\frac{40}{3}} = b^{\frac{40}{21}} = \sqrt[21]{b^{40}} = \sqrt[21]{b^{21} * b^{19}} = \sqrt[21]{b^{21}} \sqrt[21]{b^{19}} = b\sqrt[21]{b^{19}}

\sqrt[3]{\frac{c^{9}}{c^{3}}} = \sqrt[3]{c^{6}} = c^{2}

\sqrt[3]{27d^{6} * 8d^{-4}} = \sqrt[3]{27d^{6} * \frac{8}{d^{4}}} = \sqrt[3]{216d^{2}} = 6\sqrt[3]{d^{2}} = 6d^{\frac{2}{3}}
4 0
3 years ago
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Esther. Vida and Clair were asked to consider two different cash flows: GH¢1000 that they
Leona [35]

Answer:

Esther is right. Take the cash today

Step-by-step explanation:

The value of money tends to decrease due to inflation. With GH¢1000 now, she can invest and work to earn more herself.

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