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netineya [11]
4 years ago
9

The probability that a lab specimen contains high levels of contamination is 0.12. A group of 4 independent samples are checked.

Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that none contain high levels of contamination
Mathematics
1 answer:
sergiy2304 [10]4 years ago
5 0

Answer:

0.5996 is the probability that none contain high levels of contamination.

Step-by-step explanation:

We are given the following information:

We treat  lab specimen containing high levels of contamination as a success.

P( lab specimen containing high levels of contamination) = 0.12

Then the number of lab specimens follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 4

We have to find the probability that none of the lab specimen consist of high level of contamination.

We have to evaluate:

P(x = 0)\\= \binom{4}{0}(0.12)^0(1-0.12)^{(4-0)}\\= 0.5996

0.5996 is the probability that none contain high levels of contamination.

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Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0
Arada [10]

Answer:

b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0

Step-by-step explanation:

The given matrix addition is

B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}

We need to find the elements of matrix B.

Let B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}

Substitute the value of matrix.

\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}

After addition of two matrix we get

\begin{bmatrix}b_{11}+15&b_{12}-7&b_{13}+4\\ b_{21}+0&b_{22}+1&b_{23}+2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}

On equating both sides.

b_{11}+15=1\Rightarrow b_{11}=-14

b_{12}-7=2\Rightarrow b_{12}=9

b_{13}+4=12\Rightarrow b_{13}=8

b_{21}+0=4\Rightarrow b_{21}=4

b_{22}+1=0\Rightarrow b_{22}=-1

b_{23}+2=2\Rightarrow b_{23}=0

Therefore, the elements of matrix B are b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0.

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3 years ago
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