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Tema [17]
3 years ago
12

Clarisse decided to do the 52 a week Money Saving Challenge. In the first week of January, she saved Php 5.00. In the second wee

k of January, she saved Php 10.00. In the third week of January, she saved Php 15.00. If the increment was the same for the whole year,
-How much did she save on the 52nd week?

-How much was her total savings after 52 weeks?
Mathematics
1 answer:
Brrunno [24]3 years ago
4 0

Answer:

1378

Step-by-step explanation:

once you’ve decided on your New Year’s resolutions, the best way to make sure you actually follow through is by developing a plan for how to accomplish your new goals.

For anyone with “saving money” at the top of their list in 2020, the 52-week money challenge is a simple and effective way to stay on track. And at the end of the year, you’ll have $1,378 extra dollars to bulk up your emergency savings or put towards next year’s vacation.

Take some time now to determine what, exactly, you want to save for so you’ll be more motivated to stay the course throughout the year. Once your savings goals are set, jump in and start saving with the 52-week money challenge.

What is the 52-week money challenge?

Using the 52-week money challenge, you should deposit an increasing amount of money each week for one year.

Match each week’s savings amount with the number of the week in your challenge. In other words, you’ll save $1 the first week, $2 the second week, $3 the third week, and so on until you put away $52 in week 52. You can also get creative with the amount you save each week if you’d like to save more over the year or you don’t have enough extra cash available for the regular challenge.

If you stick to this challenge throughout the entire year, you’ll save a total of $1,378.

WEEK NUMBER WEEKLY DEPOSIT TOTAL SAVINGS

Week 1 $1 $1

Week 2 $2 $3

Week 3 $3 $6

Week 4 $4 $10

Week 5 $5 $15

Week 6 $6 $21

Week 7 $7 $28

Week 8 $8 $36

Week 9 $9 $45

Week 10 $10 $55

Week 11 $11 $66

Week 12 $12 $78

Week 13 $13 $91

Week 14 $14 $105

Week 15 $15 $120

Week 16 $16 $136

Week 17 $17 $153

Week 18 $18 $171

Week 19 $19 $190

Week 20 $20 $210

Week 21 $21 $231

Week 22 $22 $253

Week 23 $23 $276

Week 24 $24 $300

Week 25 $25 $325

Week 26 $26 $351

Week 27 $27 $378

Week 28 $28 $406

Week 29 $29 $435

Week 30 $30 $465

Week 31 $31 $496

Week 32 $32 $528

Week 33 $33 $561

Week 34 $34 $595

Week 35 $35 $630

Week 36 $36 $666

Week 37 $37 $703

Week 38 $38 $741

Week 39 $39 $780

Week 40 $40 $820

Week 41 $41 $861

Week 42 $42 $903

Week 43 $43 $946

Week 44 $44 $990

Week 45 $45 $1,035

Week 46 $46 $1,081

Week 47 $47 $1,128

Week 48 $48 $1,176

Week 49 $49 $1,225

Week 50 $50 $1,275

Week 51 $51 $1,326

Week 52 $52 $1,378

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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 ]

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    Cpk  = 2

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

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Answer:

The pvalue of the test is 0.0124 < 0.1, which means that there is sufficient evidence at the 0.1 level to support the testing firm's claim.

Step-by-step explanation:

An automobile manufacturer has given its van a 59.5 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating:

At the null hypothesis, we test if the mean is the same, that is:

H_0: \mu = 59.5

At the alternate hypothesis, we test that it is different, that is:

H_a: \mu \neq 59.5

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

59.5 is tested at the null hypothesis:

This means that \mu = 59.5

After testing 250 vans, they found a mean MPG of 59.2. Assume the population standard deviation is known to be 1.9.

This means that n = 250, X = 59.2, \sigma = 1.9

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{59.2 - 59.5}{\frac{1.9}{\sqrt{250}}}

z = -2.5

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean that differs from 59.5 by at least 0.3, which is P(|Z|>-2.5), which is 2 multiplied by the pvalue of Z = -2.5.

Looking at the z-table, Z = -2.5 has a pvalue of 0.0062

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The pvalue of the test is 0.0124 < 0.1, which means that there is sufficient evidence at the 0.1 level to support the testing firm's claim.

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