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borishaifa [10]
3 years ago
7

True or false If a system produced a true statement by the elimination method. It has no solution?

Mathematics
1 answer:
MariettaO [177]3 years ago
5 0
Slide brainliest. Lol Jkjk. True
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11/20 as a simplified fraction
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Fraction already reduced simplified to lowest terms gcf ( 11/20 ) =1 numerator and denominator are coprime number they no common prime factors.
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Help please!! I’ve gotten this wrong twice and if I get it wrong again then I’m going to fail.
serious [3.7K]

Answer: Choice A (yes it is a function; one range value exists for each domain value)

Put another way, each x corresponds to exactly one output only. We do not have any repeat x values. Any input you specify, there is only one output. If for example we had the two points (3,5) and (3,7) then the input x = 3 leads to multiple outputs y = 5 and y = 7 at the same time. This example is a non-function because of this. In this case, we don't have such repeated x values so that is why we have a function.

Try graphing out the four points given. You'll notice you cannot draw a vertical line through more than one point. Therefore, this graph passes the vertical line test. The vertical line test is to see if it's possible to draw a vertical line through more than one point on the graph. If so, then the relation fails to be a function.

8 0
3 years ago
Which ordered pair is a solution to the equation?
mezya [45]

Answer:

Step-by-step explanation:

4(4) = -7(-3) - 5

16 = 21 - 5

16 = 16

answer is D

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3 years ago
What time is this I can’t read clocks ?
tensa zangetsu [6.8K]

Answer:

12:40

Step-by-step explanation:

Brainliest!

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3 years ago
g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa
adoni [48]

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is  \sigma =  0.0133

Step-by-step explanation:

From the question we are told that

   The upper specification is  USL  =  1.68 \ mm

    The lower specification is  LSL  = 1.52  \  mm

     The sample mean is  \mu =  1.6 \  mm

     The standard deviation is  \sigma =  0.03 \ mm

Generally the capability index in mathematically represented as

             Cpk  =  min[ \frac{USL -  \mu }{ 3 *  \sigma }  ,  \frac{\mu - LSL }{ 3 *  \sigma } ]

Now what min means is that the value of  CPk is the minimum between the value is the bracket

          substituting value given in the question

           Cpk  =  min[ \frac{1.68 -  1.6 }{ 3 *  0.03 }  ,  \frac{1.60 -  1.52 }{ 3 *  0.03} ]

=>      Cpk  =  min[ 0.88 , 0.88  ]

So

         Cpk  = 0.88

Now from the question we are asked to evaluated the value of  standard deviation that will produce a  capability index of 2

Now let assuming that

         \frac{\mu - LSL  }{ 3 *  \sigma } =  2

So

         \frac{ 1.60 -  1.52  }{ 3 *  \sigma } =  2

=>    0.08 = 6 \sigma

=>     \sigma =  0.0133

So

        \frac{ 1.68  - 1.60 }{ 3 *  0.0133 }

=>      2

Hence

      Cpk  =  min[ 2, 2 ]

So

    Cpk  = 2

So    \sigma =  0.0133 is  the value of standard deviation required

3 0
3 years ago
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