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Andre45 [30]
4 years ago
11

A new surgery is successful 85% of the time. If the results of 6 such surgeries are randomly sampled, what is the probability th

at fewer than 4 of them are successful
Mathematics
1 answer:
Leno4ka [110]4 years ago
8 0

The probability that fewer than 4 of them are successful are 0.9527

<u>Explanation:</u>

Given:

number of trials, n = 6

p(successful) = 0.85

P(x >= 4) = ?

The problem can be solved using binomial probability formula.

 P(x >= 4) = 1 - binomcdf(6,0.85,4) = 1 - 0.0473

               = 0.9527

Therefore, the probability that fewer than 4 of them are successful are 0.9527

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X + y + w = b
White raven [17]

Answer:

(a) a=6 and b≠\frac{11}{4}

(b)a≠6

(c) a=6 and b=\frac{11}{4}

Step-by-step explanation:

writing equation in agumented matrix form

\begin{bmatrix}1 &1 & 0 &1 &b\\ 2 &3 & 1 &5 &6\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}

now R_{2} =R_{2}-2\times R_{1}

\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}

now R_{4} =R_{4}-2\times R_{2}

\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 0 & 0 &a-6 &4b-11\end{bmatrix}

a) now for inconsistent

rank of augamented matrix ≠ rank of matrix

for that  a=6 and b≠\frac{11}{4}

b) for consistent w/ a unique solution

rank of augamented matrix = rank of matrix

  a≠6

c) consistent w/ infinitely-many sol'ns

  rank of augamented matrix = rank of matrix < no. of variable

for that condition

 a=6 and b=[tex]\frac{11}{4}

then rank become 3 which is less than variable which is 4.

4 0
4 years ago
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