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Aleksandr [31]
3 years ago
6

The weight of an elephant is 1000 times the weight of a cat. If the elephant weighs 14,000 pounds how many pounds does the cat w

eigh
Mathematics
2 answers:
ohaa [14]3 years ago
5 0
To solve this problem, all you have to do is 14,000/1,000, which is 14, because the elephant weighs 14,000 lbs and the cat is 1/1000 of that weight. The cat weighs 14 pounds.
Ad libitum [116K]3 years ago
4 0
Weight of the elephant= 1,000x
x= weight of the cat
14,000=1,000x
x=14 pounds is what the cat weighs. 
Check: 14*1,000= 14,000
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If town A had 35 people, 12 left, 13 joined, 39 joined, 25 got killed, how many people were there left?
Makovka662 [10]

Answer:

50

Step-by-step explanation:

35 - 12 + 13 + 39 - 25 = 50

6 0
3 years ago
Suppose KLMN ~ PQRS. Find the value of x.
arlik [135]

Complete question is attached.

Answer:

B) 8

Step-by-step explanation:

In this question, KLMN ~ PQRS. This means they are similar shapes.

From the image, we could denote that:

KL ~ PQ

LM ~ QR

Given that:

KL = 12

PQ = 30

QR = 20

LM = x

To find x, let's use tge expression:

\frac{KL}{PQ} = \frac{LM}{QR}

=  \frac{12}{30} = \frac{x}{20}

Cross multiplying, we have:

12 * 20 = 30x

240 = 30x

To find x, let's divide both sides by 30.

\frac{240}{30} = \frac{30x}{30}

8 = x

x = 8

The correct option is option B.

3 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

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3 years ago
5 76/10 x 1.75 i need help
alekssr [168]

Answer:

22.05

Step-by-step explanation:

7 0
3 years ago
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Answer: 1) The value of the 3 in option A is 10x the value of the 3 in option B

2) Option A has the 3 in the ones places, Option B has the 3 in the tenths place

Step-by-step explanation:

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3 years ago
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