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aleksandrvk [35]
3 years ago
13

What is the determinant of the coefficient matrix of the system

Mathematics
1 answer:
vovikov84 [41]3 years ago
8 0

The coefficient matrix is

\begin{bmatrix}-1&-1&2\\3&2&-1\\4&4&-8\end{bmatrix}

Its determinant is then

\det\begin{bmatrix}-1&-1&2\\3&2&-1\\4&4&-8\end{bmatrix}=-4\det\begin{bmatrix}-1&-1&2\\3&2&-1\\-1&-1&2\end{bmatrix}

which is equal to 0.

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\sqrt{3Answer:

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take 45 degree as reference angle

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3 years ago
A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?
seraphim [82]

Answer:

P\left(E\right)=\frac{55}{1024}

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

nCr=\frac{n!}{r!\left(n-r\right)!}

Plug in n = 12 and r = 9

       =\frac{12!}{9!\left(12-9\right)!}

       =\frac{12!}{9!\cdot \:3!}

       =\frac{12\cdot \:11\cdot \:10}{3!}            ∵ \frac{12!}{9!}=12\cdot \:11\cdot \:10

       =\frac{1320}{6}                   ∵ 3!\:=\:3\times 2\times 1=6

       =220

Thus, the probability will be:

P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}

         =\frac{220}{4096}

         =\frac{55}{1024}

Thus, the probability of the coin landing tails up exactly nine times will be:

P\left(E\right)=\frac{55}{1024}

4 0
3 years ago
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