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Vaselesa [24]
3 years ago
10

A blind taste test will be conducted with 9 volunteers to determine whether people can taste a difference between bottled water

and tap water. Each participant will taste the water from two different glasses and then identify which glass he or she thinks contains the tap water. Assuming that people cannot taste a difference between bottled water, what is the probability that at least 8 of the 9 people will correctly identify the tap water?
Mathematics
1 answer:
Aloiza [94]3 years ago
6 0

Answer:

The answer is \frac{1}{256}

Step-by-step explanation:

If we assume that people cannot taste a difference between bottled water, then the probability of identifying tap water is 0.5

Thus, P(identify tap water)=0.5

The probability that at least 8 of the 9 people identify the tap water correctly is the sum of the probabilities

  • 8 of 9 people identified correctly or
  • 9 of 9 people identified correctly

Since P(identify tap water)=0.5 each probabilities are the same and equal to

0.5^9 =\frac{1}{512}

So we have 2*\frac{1}{512} = \frac{1}{256}

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An electronics company produces​ transistors, resistors, and computer chips. Each transistor requires 3 units of​ copper, 1 unit
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Answer:

An electronics company can be produce 350 transistors and 340 computer chips, they can´t produce resistors.

Step-by-step explanation:

1. We will name the variables for transistors, resistors and the computer chips.

a = Transistors

b= Resistors

c = Computer chips

2. We propose three linear equations, one for the copper, one for the zinc and one for the glass.

\left \{ {{3a+3b+2c=1730} \atop {a+2b+c=690}}\atop {2a+b+2c=1380}} \right.

3. We write the matrix form as Ax=d

A=\left(\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right)

x=\left(\begin{array}{ccc}a\\b\\c\end{array}\right)

A=\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)

With this formula the solution of x is x=\frac{d}{A} or x=A^{-1}d

4. We will find the inverse matrix A^{-1} using the formula:

A^{-1} = \frac{1}{detA} (C_{A})^{T}

a. det A

det A=\left[\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right] =3*(4-1)-3*(2-2)+2*(1-4)=9-0-6=3

b. (C_{A})^{T}

C_{A}=\left(\begin{array}{ccc}4-1&.(2-2)&1-4\\-(6-2)&6-4&-(3-6)\\3-4&-(3-2)&6-3\end{array}\right)

C_{A}=\left(\begin{array}{ccc}3&.0&-3\\-4&2&3\\-1&-1&3\end{array}\right)

(C_{A}) ^T=\left(\begin{array}{ccc}3&0&-3\\-4&2&3\\-1&-1&3\end{array}\right)^T

(C_{A}) ^T=\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)

c.A^{-1}

A^{-1}=\frac{1}{3} \left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)

5. As x=\frac{d}{A} or x=A^{-1}d, the solution of x is:

x=\frac{1}{3}\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)

x=\frac{1}{3}\left(\begin{array}{ccc}(3*1730)+(-4*690)+(-1*1380)\\(0*1730)+(2*690)+(-1*1380)\\(-3*1730)+(3*690)+(3*1380)\end{array}\right)

x=\frac{1}{3}\left(\begin{array}{ccc}1050\\0\\1020)\end{array}\right)

X=\left[\begin{array}{ccc}350\\0\\340\end{array}\right]

<u><em>Therefore:</em></u>

<u><em>a= 350 Transistors</em></u>

<u><em>b=0 Resistors</em></u>

<u><em>c=340 Computer chips</em></u>

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