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IrinaK [193]
3 years ago
9

A factory has three machines that all manufacture the same engine part. From long historical records the company has determined

that the first machine produces 4% defectives, the second machine produces 6% defectives, and the third machine produces 1% defectives. Suppose also that the first machine produces 30%, the second machine 20%, and the third machine 50% of the engine parts. An engine part is selected at random. If the part is defective, what is the probability that was produced by the first machine?
Mathematics
1 answer:
Alex Ar [27]3 years ago
3 0

Answer:

Step-by-step explanation:

P(D/M₁) = .04 ( probability of defect from a 1 st machine )

P(D/M₂) = .06

P(D/M₃) = .01

P( M₁ )= .3 ( probability of manufacture from 1 st machine )

P(M₂) = .2

P(M₃)= .5

P( M₁/D) ( Probability of manufacture from 1 st machine , given it is defective )

= P( M₁ )xP(D/M₁) / [P( M₁ )xP(D/M₁) +P( M₂ )xP(D/M₂)+P( M₃ )xP(D/M₃)]

= .3 x .06 / .3 x .06 + .2 x .06 + .5 x .01

= .018 / .035

= 18/35

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Teo spins the spinner 120 times. He expects to land on one particular color 30 times. What color is it?
olasank [31]

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Red

Step-by-step explanation:

P = 30/120 = 1/4

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The color is red.

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4 years ago
Find the values of k so that each remainder is three. <br> 10. (x^2+ 5x + 7) = (x + k)
Goryan [66]

Answer:

k=1\text{ or } k=4

Step-by-step explanation:

We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a <em>binomial</em> in the form (x - a), then our remainder will be P(a).

We are dividing:

(x^2+5x+7)\div(x+k)

So, a polynomial by a binomial factor.

Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.

We want our remainder to be 3. So, P(a)=P(-k)=3.

Therefore:

(-k)^2+5(-k)+7=3

Simplify:

k^2-5k+7=3

Solve for <em>k</em>. Subtract 3 from both sides:

k^2-5k+4=0

Factor:

(k-1)(k-4)=0

Zero Product Property:

k-1=0\text{ or } k-4=0

Solve:

k=1\text{ or } k=4

So, either of the two expressions:

(x^2+5x+7)\div(x+1)\text{ or } (x^2+5x+7)\div(x+4)

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3 years ago
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abruzzese [7]

Answer:

The answer is <em>1</em>.

Step-by-step explanation:

Given the expression:

|z-6|-|z-5|,\ if\ z

To find:

The expression without absolute value.

Solution:

First of all, let us learn about the absolute value function:

y = f(x) = |x| =\left \{ {{x\ if\ x>0} \atop {-x\ if\ x

i.e. value is x if x is positive

value is -x if x is negative

Here the given expression contains two absolute value functions:

|z-6| and |z-5|

Using the definition of absolute value function as per above definition.

|z-5| =\left \{ {{(z-5)\ if\ z>5} \atop {-(z-5)\ if\ z

|z-6| =\left \{ {{(z-6)\ if\ z>6} \atop {-(z-6)\ if\ z

Now, it is given that z < 5 that means z will also be lesser than 6 i.e. z < 6

So, given expression |z-6|-|z-5|,\ if\ z will be equivalent to :

-(z-6) - (-(z-5))\\\Rightarrow -z+6 +z-5 = \bold{1}

So, the expression is equivalent to <em>1</em>.

6 0
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Graph the hyperbola using the transverse axis, vertices, and co-vertices:
Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

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The transverse axis is calculated as:

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Evaluate the quotient

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This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

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This means that

(h, k) = 0

And

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The co-vertices are

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So, we have:

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See attachment for the graph of the hyperbola

Read more about hyperbola at:

brainly.com/question/26250569

#SPJ1

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