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Mamont248 [21]
2 years ago
10

F(x)=5-1/5x What are the x and y intercepts?

Mathematics
1 answer:
Bogdan [553]2 years ago
3 0

Answer:

X intercepts 1/25=0

Y intercepts none

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HELPPPPP ASAPPPP!!!!!!
In-s [12.5K]

Hey there!!

Let's take the number as x and y

Given,

The sum of x and y = 12

One number is 2 more than the other

Let's take y = x + 2

Equations

x + y = 12

As we know y = x + 2 , plug the value in

x + x + 2 = 12

2x + 2 = 12

Subtract 2 on both sides

2x = 10

Divide by 5 on both sides

x = 5

y = 2 + x

y = 7

The numbers are 5 and 7

Hope my answer helps!

3 0
3 years ago
Read 2 more answers
When The Product of 15 and 40 is divided by the sum of 15 and 45 what is the quotient?
BaLLatris [955]

Answer:

10

Step-by-step explanation:

(15 x 40) / (15+40)

solve inside the parenthesis:

600/60

divide:

10

6 0
3 years ago
Find the value of X below
Paul [167]

Answer: x= 80

Step-by-step explanation:

it’s an isoceles triangle so bottom 2 angles are 50

50+50= 100

180-100= x

7 0
2 years ago
Estimate the unit rate to the nearest hundredth $2.35 12 Grade AA eggs.<br>​
Wewaii [24]

Answer:

2.4

Step-by-step explanation:

6 0
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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