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Alja [10]
2 years ago
8

Write an EQUATION with an undefined slope:

Mathematics
1 answer:
Phantasy [73]2 years ago
3 0

Answer:

Example of an equation with an undefined slope: x = 2

Step-by-step explanation:

<h2>Definitions</h2>

The standard form of linear equations with an <u>undefined slope</u> is <em>x</em> = <em>a</em>, whose graph represents a vertical line. The value of a in the standard form is the x-intercept, (<em>a</em>, 0).  

The slope is the ratio of the vertical change in y-values to the horizontal change in x-values.

\displaystyle\mathsf{Slope(m) =\:\frac{\triangle y}{\triangle x}\:=\frac{y_2 - y_1}{x_2 - x_1}}

The slope of a vertical line is undefined because if we were to solve its slope, the denominator will be zero. As we know, division by zero is an undefined operation.  

<h2>Example:</h2>

For example, suppose that we have the following points (2, 5) (2, 10).

Let (x₁, y₁) = (2, 5)

     (x₂, y₂) = (2, 10)

Substitute these values into the slope formula:

\displaystyle\mathsf{Slope(m) =\:\frac{\triangle y}{\triangle x}\:=\frac{y_2 - y_1}{x_2 - x_1}\:=\:\frac{10\:-\:5}{2\:-\:2}\:=\frac{5}{0}}

Dividing the numerator, 5, by the denominator, 0, will have an undefined quotient.

Thus, the equation of the <u>vertical line</u> will be: x = 2, where <em>a</em> = 2.

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A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade
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Part a

The probability that assembly is replaced the first day is 0.7069.

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The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

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The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

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The probability distribution of the hyper geometric is,

P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}

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P(x\geq1)=1-P(x

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly  n= 5

Number of blades in an assembly dull is M  = 10.

The probability mass function is,

P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

P(x\geq 1)=1- P(x

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly  N = 48

Number of blades selected at random from the assembly  N = 5

Number of blades in an assembly dull is  M = 10

From the information,

The probability that assembly is replaced (P)  is 0.7069.

The probability that assembly is not replaced is (Q)  is,

q=1-p\\= 1-0.7069= 0.2931

The geometric probability mass function is,

P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)

The probability that assembly is replaced no replaced until the third day of evaluation is,

P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly   N = 48

Number of blades selected at random from the assembly  n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 2.

P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 6.

P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

P(x\geq 1)=1- P(x

 

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

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