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kondor19780726 [428]
3 years ago
6

2а +3Ь — 5 b = a – 5

Mathematics
1 answer:
schepotkina [342]3 years ago
8 0

Answer:

a= -2b

Step-by-step explanation:

math.

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A summer camp is organizing a hike and needs to buy granola bars for the campers. The granola bars come in small boxes and large
Irina18 [472]

Answer:

The Answer is: There are 8 small boxes and 9 large boxes. See explanation below for variables and variable definitions.

Step-by-step explanation:

Let s = small boxes. Let b = large boxes.

s + b = 17

You can solve for s:

s = 17 - b

You can solve for b:

b = 17 - s

10 times the number of small boxes plus 24 times the number of large boxes is equal to 296 granola bars.

10s + 24b = 296

Substitute:

10(17 - b) + 24b = 296

170 - 10b + 24b = 296

14b = 296 - 170

14b =  126

b = 126 / 14 = 9 large boxes

Find the number of small boxes, s:

s = 17 - b = 17 - 9 = 8 small boxes

There are 8 small boxes and 9 large boxes.

Proof:

10(8) + 24(9) = 296

80 + 216 = 296

296 = 296

4 0
3 years ago
Consider this expression.
USPshnik [31]

Answer:

-16

Step-by-step explanation:

m^2 + n^2

-5^2 + 3^2 = -16

5 0
1 year ago
What is the solution of the system of equations that contains the equation - 3x - 9y = 18 and the equation of the line shown on
Diano4ka-milaya [45]

Answer:

No solution

Step-by-step explanation:

7 0
2 years ago
Pls help there are 5 pics with multiple questions inside of them pls thank you
Bogdan [553]

Answer:

12. The second one

13. The first one

14. The last one

15. The first one

16. 30 degrees

Step-by-step explanation:

7 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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