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Anna35 [415]
2 years ago
13

A coin lands on heads 300 times. The relative frequency of heads is 0.6. How many times was the coin flipped?

Mathematics
1 answer:
Ksenya-84 [330]2 years ago
6 0

Answer: Number of times the coin is flipped =500

Step-by-step explanation:

Relative frequency of head = \dfrac{\text{Number of times heads occur}}{\text{Total number of times coin is flipped}}

Let x= Number of times the coin is flipped, then

0.6=\dfrac{300}{x}\\\\\Rightarrow\ x=\dfrac{300}{0.6}\\\\\Rightarrow\ x=\dfrac{3000}{6}\\\\\Rightarrow\ x=500

Hence, Number of times the coin is flipped =500

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An arithmetic sequence has this recursive formula: (a^1 =8, a^n= a^n-1 -6
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Answer:

a_n = 8 + (n - 1) (-6)

Step-by-step explanation:

Given

a_1 = 8

Recursive: a_{n} = a_{n-1} - 6

Required

Determine the formula

Substitute 2 for n to determine a_2

a_{2} = a_{2-1} - 6

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Next is to determine the common difference, d;

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The nth term of an arithmetic sequence is calculated as

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Answer:

Step-by-step explanation:

CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Subtract row 3 from row 1:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]

Subtract row 3 from 7 times row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]

Divide row 3 by 17:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 2 of row 3 from row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 6 of row 2 from row 1:

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]

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