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2 years ago

A fair die is rolled 10 times. What is the average number of even number outcomes?

Mathematics

2 answers:

ankoles [38]2 years ago

7 0

Answer:

3/6=1/2 so one half

Step-by-step explanation:

3/6 × 10 =30/60=3/6=1/2

Hatshy [7]2 years ago

3 0

Answer:

Average number of even number outcomes =5

Step-by-step explanation:

Probability = number of possible outcome / sample space

A fair die has sides labeled 1,2,3,4,5,6.

Therefore sample space = 6

Odd numbers = 1,3,5

Possible outcome of odd numbers = 3

Even numbers = 2,4,6

Possible outcome of even numbers = 3

Probability of even numbers = possible outcome of even numbers / sample space

Probability of even numbers = 3/6 = 1/2.

If the die is rolled 10times

Total number of outcome = 10

Average number of even number outcomes = probability of even numbers * total number of outcome

= 1/2 x 10

= 5

Average number of even number outcomes =5

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Which of the following matrices is the solution matrix for the given system of equations? x + 5y = 11 x - y = 5

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<h2>The required solution is x = 6 and y = 11 </h2>

Step-by-step explanation:

Given system of equations are

x+5y = 11 and x-y =5

$A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]$ $X=\left[\begin{array}{c}x\\y\end{array}\right]$

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$adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]$

$A= \left|\begin{array}{cc}1&5\\1&-1\end{array}\right|=-6$

∴ $A^{-1} =\frac{adj A}{|A|}$

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$A^{-1} ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}$

$X =A^{-1}\times B$

⇒ $\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]$

⇒ $\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c} {6}\\ {11} \end{array}\right]}$

∴ x= 6 and y = 11

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