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wolverine [178]
2 years ago
5

Anyone knows how to do this please tell me how?

Mathematics
2 answers:
yulyashka [42]2 years ago
8 0

Step-by-step explanation:

0.000672, 6.72×10⁵, 67.2×10‐⁴, 672×10⁴

Marizza181 [45]2 years ago
3 0
<h2>Scientific Notation</h2>

Scientific notation is a way of writing numbers that looks like the following:

a\times10^n

  • <em>a</em> is any number between 1 and 10
  • <em>n</em> is any integer

For instance, we can write 2930 in scientific notation.

  1. First, move the decimal so that the number is between 1 and 10:
    ⇒ 2.93
  2. Notice how we moved the decimal 3 places <em>to the left.</em> This means <em>n</em> will be 3. (Note that when we move to the <em>right</em>, <em>n</em> would be negative.)
    ⇒ 2.93 × 10³

<h2>Solving the Question</h2>

First, write all the given values in scientific notation:

  1. 6.72\times10^5      ⇒ 6.72\times10^5
  2. 67.2\times 10^{-4}    ⇒ 6.72\times 10^{-3}
  3. 672\times 10^4       ⇒ 6.72\times 10^6
  4. 0.000672        ⇒ 6.72\times10^{-4}

Now, let's compare the values of <em>n,</em> and organize the numbers from least to greatest:

6.72\times10^{-4}

6.72\times 10^{-3}

6.72\times10^5

6.72\times 10^6

Finally, rewrite all the numbers as their given format:

0.000672

67.2\times 10^{-4}

6.72\times10^5

672\times 10^4

<h2>Answer</h2>

0.000672

67.2\times 10^{-4}

6.72\times10^5

672\times 10^4

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Answer:

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Step-by-step explanation:

<u>Vertical Throw</u>

It refers to a situation where an object is thrown verticaly upwards with some inicial speed v_o and let in free air (no friction) until it completes its movement up and finally returns to the very same point of lauch. The only acting force is gravity

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This is a second-degree equation which will be solved with the formula

\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\displaystyle x=\frac{-90\pm \sqrt{90^2-4(-16)(-120)}}{2(-16)}

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We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

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the probability for the first two footballers to be born on two different days is thus

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Similarly, the third footballer can be born on any day, except d_1 and d_2:

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so, the probability for the first three footballers to be born on three different days is

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The probability of all 22 footballers being born on 22 different days is thus

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1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

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