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Mashutka [201]
2 years ago
10

$84.12 for 12 lbs of chocolate ; $35.05 for p lbs of chocolate

Mathematics
1 answer:
olga_2 [115]2 years ago
3 0

Answer:

$35,05 for 5 lbs of chocolate

Step-by-step explanation:

8412:12=701

7,01 for 1

3505:701=5

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3. Formulate 900 lbs of a 35% CP ration using Soybean Meal (46% CP) and Oats (12% CP). How many pounds of soybean meal will be u
Anarel [89]

Answer:

The amount of Soyabean = 608.82 pounds.

The amount of Oats  = 900-608.82 = 291.18 pounds.

Step-by-step explanation:

Given that the Soybean Meal has 46% CP and Oats has 12% CP.

The total amount of ration is 900 lbs of 35% CP.

Let x pounds of Soyabean has been used, so,

the amount of Oats = 900-x lbs.

The CP amount in 900 lbs mixture will be equal to the sum of CP amounts in x lbs Soyabean and 900-x lbs Oats, i.e

35% of 900 = 46% of x + 12% of (900-x)

\Rightarrow \frac{35}{100}\times 900=\frac{46}{100}\times x+\frac{12}{100}\times (900-x)

\Rightarrow  35\times 900 = 46x + 12\times 900 - 12 x

\Rightarrow 34x= 900(35-12)

\Rightarrow x = (900\times 23)/34

\Rightarrow x = 608.82 lbs.

So, the amount of Soyabean = 608.82 pounds.

The amount of Oats  = 900-608.82 = 291.18 pounds.

5 0
2 years ago
Can anyone evaluate this problem for me .
AURORKA [14]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
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Express each of the following in standard form 0.0 0 0 0 0 0 0 0 0 2 2 1​
Anastasy [175]
answer= 2.21 x 10^-10
3 0
3 years ago
Please help me ASAP please
lianna [129]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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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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