# Rounding off significant figures worksheet: Rounding Significant Figures

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## Rounding to Significant Figures Worksheet

1. Three different people measure the length of a room. The measurements given by them are 7.8 meters, 8 meters and 7.76 meters. Which is the most appropriate measurement ?

2. Round 6.7845 to 3 significant figures.

3. Round 0.0360578 to 4 significant figures.

4. Round 10.86 to 2 significant figures.

5. Evaluate the following multiplication and round your answer to the approximate number of significant digits.

8.0000 x 0.004

6. Evaluate the following division and round your answer to the approximate number of significant digits.

30,000 ÷ 3.004

7. Evaluate the following multiplication and round your answer to the approximate number of significant digits.

6.14 x 30.5 x 500

8. Evaluate the following division and round your answer to the approximate number of significant digits.

0.009 ÷ 7

9. Using the approximation 1 in. = 2.54 cm, convert 24.25 inches to centimeters.

10. Using the approximation 1 kg = 2.2 pounds, convert 125 pounds to kilograms.

11. Evaluate :

3.78 + 2.7 + 4.203

12. Evaluate :

2.56 — 1.235

This can be determined by the number of “significant figures”, in the measurement.

7.8 meters has 2 significant figures

8 meters has 1 significant figure

7.76 meters has 3 significant figures

The measurement of 7.76 meters is more precise than 7.8 meters or 8 meters because it has more significant digits.

All the digits in 6.7845 are non-zero digits.

6.7845 has 5 significant figures

If you want to round 6.7845 to 3 significant figures, look at the digit after the third significant figure.

In 6.7845, the digit after the third significant figure is 4.

6.7845

Since the digit after the third significant figure is 4, keep the previous digit 8 same.

6.7845 to 3 significant figures = 6.78

0. 0360578 has 6 significant figures

If you want to round 0.0360578 to 4 significant figures, look at the digit after the fourth significant figure.

In 0.0360578, the digit after the fourth significant figure is 7.

0.0360578

Since the digit after the fourth significant figure is 7, increase the previous digit 5 by 1.

0.0360578 to 4 significant figures = 0.03606

10.86 has 4 significant figures

If you want to round 10.86 to 2 significant figures, look at the digit after the second significant figure.

In 10.86, the digit after the second significant figure is 8.

10.86

Since the digit after the second significant figure is 8, increase the previous digit 0 by 1.

10.86 to 2 significant figures = 11

8.0000 x 0.004 = 0.032

8.0000 has 5 significant figures

0.004 has 1 significant figure

We have the least number of significant figures in 0.004, that is 1 significant figure.

So the final answer will have 1 significant figure.

30,000 ÷ 3.004 = 9986.68442………

30,000 has 5 significant figures

3.004 has 4 significant figures

We have the least number of significant figures in 3.004, that is 4 significant figures.

So the final answer will have 4 significant figures.

6.14 x 30.5 x 500 = 93,635

6.14 has 3 significant figures

30.5 has 3 significant figures

500 has 3 significant figures

In all the three numbers, we have the number of significant figures, that is 3 significant figures.

So the final answer will have 3 significant figures.

0.009 ÷ 7 = 0.012857………

0.009 has 1 significant figure

7 has 1 significant figure

In both the numbers, we have the same number of significant figures, that is 1 significant figure.

So the final answer will have 1 significant figure.

24.25 x 2.54 = 61.595

24.25 has 4 significant figures

2.54 has 3 significant figures

We have the least number of significant figures in 2.54, that is 3 significant figures.

So the final answer will have 3 significant figures.

125 ÷ 2.2 = 56.818181………

125 has 3 significant figures

2.2 has 2 significant figures

We have the least number of significant figures in 2.2, that is 2 significant figures.

So the final answer will have 2 significant figures, making it no more precise than the least precise.

3.78 + 2.7 + 4.203 = 10.683

3.78 has 2 decimal places.

2.7 has 1 decimal place.

4.203 has 3 decimal places.

The fewest decimal places is 1.

The final answer is 10. 7.

2.56 — 1.235 = 1.325

2.56 has 2 decimal places.

1.325 has 3 decimal places.

The fewest decimal places is 2.

Note :

It is not necessary to count the number of significant figures when doing addition or subtraction.

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## 4.6: Significant Figures and Rounding

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• Page ID
3555
• Stephen Lower
• Simon Fraser University
##### Learning Objectives
• Give an example of a measurement whose number of significant digits is clearly too great, and explain why.
• State the purpose of rounding off, and describe the information that must be known to do it properly.
• Round off a number to a specified number of significant digits.
• Explain how to round off a number whose second-most-significant digit is 9.
• Carry out a simple calculation that involves two or more observed quantities, and express the result in the appropriate number of significant figures.

The numerical values we deal with in science (and in many other aspects of life) represent measurements whose values are never known exactly. Our pocket-calculators or computers don’t know this; they treat the numbers we punch into them as «pure» mathematical entities, with the result that the operations of arithmetic frequently yield answers that are physically ridiculous even though mathematically correct. The purpose of this unit is to help you understand why this happens, and to show you what to do about it.

### Digits: Significant and otherwise

Consider the two statements shown below:

• «The population of our city is 157,872. «
• «The number of registered voters as of Jan 1 was 27,833.

Which of these would you be justified in dismissing immediately? Certainly not the second one, because it probably comes from a database which contains one record for each voter, so the number is found simply by counting the number of records. The first statement cannot possibly be correct. Even if a city’s population could be defined in a precise way (Permanent residents? Warm bodies?), how can we account for the minute-by minute changes that occur as people are born and die, or move in and move away?

What is the difference between the two population numbers stated above? The first one expresses a quantity that cannot be known exactly — that is, it carries with it a degree of uncertainty. It is quite possible that the last census yielded precisely 157,872 records, and that this might be the “population of the city” for legal purposes, but it is surely not the “true” population. To better reflect this fact, one might list the population (in an atlas, for example) as 157,900 or even 158,000. These two quantities have been rounded off to four and three significant figures, respectively, and the have the following meanings:

• 157900 (the significant digits are underlined here) implies that the population is believed to be within the range of about 157850 to about 157950. In other words, the population is 157900±50. The “plus-or-minus 50” appended to this number means that we consider the absolute uncertainty of the population measurement to be 50 – (–50) = 100. We can also say that the relative uncertainty is 100/157900, which we can also express as 1 part in 1579, or 1/1579 = 0.000633, or about 0.06 percent.
• The value 158000 implies that the population is likely between about 157500 and 158500, or 158000±500. The absolute uncertainty of 1000 translates into a relative uncertainty of 1000/158000 or 1 part in 158, or about 0.6 percent.

Which of these two values we would report as “the population” will depend on the degree of confidence we have in the original census figure; if the census was completed last week, we might round to four significant digits, but if it was a year or so ago, rounding to three places might be a more prudent choice. In a case such as this, there is no really objective way of choosing between the two alternatives.

This illustrates an important point: the concept of significant digits has less to do with mathematics than with our confidence in a measurement. This confidence can often be expressed numerically (for example, the height of a liquid in a measuring tube can be read to ±0.05 cm), but when it cannot, as in our population example, we must depend on our personal experience and judgment.

So, what is a significant digit? According to the usual definition, it is all the numerals in a measured quantity (counting from the left) whose values are considered as known exactly, plus one more whose value could be one more or one less:

• In “157900” (four significant digits), the left most three digits are known exactly, but the fourth digit, “9” could well be “8” if the “true value” is within the implied range of 157850 to 157950.
• In “158000” (three significant digits), the left most two digits are known exactly, while the third digit could be either “7” or “8” if the true value is within the implied range of 157500 to 158500.

Although rounding off always leads to the loss of numeric information, what we are getting rid of can be considered to be “numeric noise” that does not contribute to the quality of the measurement. The purpose in rounding off is to avoid expressing a value to a greater degree of precision than is consistent with the uncertainty in the measurement.

##### Implied Uncertainty

If you know that a balance is accurate to within 0.1 mg, say, then the uncertainty in any measurement of mass carried out on this balance will be ±0.1 mg. Suppose, however, that you are simply told that an object has a length of 0.42 cm, with no indication of its precision. In this case, all you have to go on is the number of digits contained in the data. Thus the quantity “0. 42 cm” is specified to 0.01 unit in 0 42, or one part in 42 . The implied relative uncertainty in this figure is 1/42, or about 2%. The precision of any numeric answer calculated from this value is therefore limited to about the same amount.

### Rounding Error

It is important to understand that the number of significant digits in a value provides only a rough indication of its precision, and that information is lost when rounding off occurs. Suppose, for example, that we measure the weight of an object as 3.28 g on a balance believed to be accurate to within ±0.05 gram. The resulting value of 3.28±.05 gram tells us that the true weight of the object could be anywhere between 3.23 g and 3.33 g. The absolute uncertainty here is 0.1 g (±0.05 g), and the relative uncertainty is 1 part in 32.8, or about 3 percent.

How many significant digits should there be in the reported measurement? Since only the left most “3” in “3.28” is certain, you would probably elect to round the value to 3. 3 g. So far, so good. But what is someone else supposed to make of this figure when they see it in your report? The value “3.3 g” suggests an implied uncertainty of 3.3±0.05 g, meaning that the true value is likely between 3.25 g and 3.35 g. This range is 0.02 g below that associated with the original measurement, and so rounding off has introduced a bias of this amount into the result. Since this is less than half of the ±0.05 g uncertainty in the weighing, it is not a very serious matter in itself. However, if several values that were rounded in this way are combined in a calculation, the rounding-off errors could become significant.

### Rules for Rounding

The standard rules for rounding off are well known. Before we set them out, let us agree on what to call the various components of a numeric value.

• The most significant digit is the left most digit (not counting any leading zeros which function only as placeholders and are never significant digits. )
• If you are rounding off to n significant digits, then the least significant digit is the nth digit from the most significant digit. The least significant digit can be a zero.
• The first non-significant digit is the n+1th digit.

Rounding-off rules

• If the first non-significant digit is less than 5, then the least significant digit remains unchanged.
• If the first non-significant digit is greater than 5, the least significant digit is incremented by 1.
• If the first non-significant digit is 5, the least significant digit can either be incremented or left unchanged (see below!)
• All non-significant digits are removed.

Students are sometimes told to increment the least significant digit by 1 if it is odd, and to leave it unchanged if it is even. One wonders if this reflects some idea that even numbers are somehow “better” than odd ones! (The ancient superstition is just the opposite, that only the odd numbers are «lucky». )

In fact, you could do it equally the other way around, incrementing only the even numbers. If you are only rounding a single number, it doesn’t really matter what you do. However, when you are rounding a series of numbers that will be used in a calculation, if you treated each first nonsignificant 5 in the same way, you would be over- or understating the value of the rounded number, thus accumulating round-off error. Since there are equal numbers of even and odd digits, incrementing only the one kind will keep this kind of error from building up. You could do just as well, of course, by flipping a coin!

Table \(\PageIndex{1}\): Examples of rounding-off

number to round

number of significant digits

result

comment

34. 216 3 34.2 First non-significant digit (1) is less than 5, so number is simply truncated.
2.252 2 2.2 or 2.3 First non-significant digit is 5, so least sig. digit can either remain unchanged or be incremented.
39.99 3 40.0 Crossing «decimal boundary», so all numbers change.
85,381 3 85,400 The two zeros are just placeholders
0. 04597 3 0.0460 The two leading zeros are not significant digits.

#### Rounding up the Nines

Suppose that an object is found to have a weight of 3.98 ± 0.05 g. This would place its true weight somewhere in the range of 3.93 g to 4.03 g. In judging how to round this number, you count the number of digits in “3.98” that are known exactly, and you find none! Since the “4” is the left most digit whose value is uncertain, this would imply that the result should be rounded to one significant figure and reported simply as 4 g. An alternative would be to bend the rule and round off to two significant digits, yielding 4.0 g. How can you decide what to do? In a case such as this, you should look at the implied uncertainties in the two values, and compare them with the uncertainty associated with the original measurement.

Table \(\PageIndex{2}\)

rounded value

implied max

implied min

absolute uncertainty

relative uncertainty

3.98 g 3.985 g 3.975 g ±.005 g or 0.01 g 1 in 400, or 0.25%
4 g 4. 5 g 3.5 g ±.5 g or 1 g 1 in 4, 25%
4.0 g 4.05 g 3.95 g ±.05 g or 0.1 g 1 in 40, 2.5%

Clearly, rounding off to two digits is the only reasonable course in this example. Observed values should be rounded off to the number of digits that most accurately conveys the uncertainty in the measurement.

• Usually, this means rounding off to the number of significant digits in in the quantity; that is, the number of digits (counting from the left) that are known exactly, plus one more.
• When this cannot be applied (as in the example above when addition of subtraction of the absolute uncertainty bridges a power of ten), then we round in such a way that the relative implied uncertainty in the result is as close as possible to that of the observed value.

### Rounding the Results of Calculations

When carrying out calculations that involve multiple steps, you should avoid doing any rounding until you obtain the final result. Suppose you use your calculator to work out the area of a rectangle:

rounded value

relative implied uncertainty

1.58 1 part in 158, or 0.6%
1.6 1 part in 16, or 6 %
##### Note

Your calculator is of course correct as far as the pure numbers go, but you would be wrong to write down «1.57676 cm2» as the answer. Two possible options for rounding off the calculator answer are shown at the right.

It is clear that neither option is entirely satisfactory; rounding to 3 significant digits overstates the precision of the answer, whereas following the rule and rounding to the two digits in «. 42″ has the effect of throwing away some precision. In this case, it could be argued that rounding to three digits is justified because the implied relative uncertainty in the answer, 0.6%, is more consistent with those of the two factors.

The «rules» for rounding off are generally useful, convenient guidelines, but they do not always yield the most desirable result. When in doubt, it is better to rely on relative implied uncertainties.

In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise that the least precise number used to get the answer. When adding or subtracting, we go by the number of decimal places (i.e., the number of digits on the right side of the decimal point) rather than by the number of significant digits. Identify the quantity having the smallest number of decimal places, and use this number to set the number of decimal places in the answer.

#### Multiplication and Division

The result must contain the same number of significant figures as in the value having the least number of significant figures.

#### Logarithms and antilogarithms

If a number is expressed in the form a × 10b («scientific notation») with the additional restriction that the coefficient a is no less than 1 and less than 10, the number is in its normalized form. Express the base-10 logarithm of a value using the same number of significant figures as is present in the normalized form of that value. Similarly, for antilogarithms (numbers expressed as powers of 10), use the same number of significant figures as are in that power.

##### Examples \(\PageIndex{1}\)

The following examples will illustrate the most common problems you are likely to encounter in rounding off the results of calculations. They deserve your careful study!

calculator result

rounded remarks
1. 6 Rounding to two significant figures yields an implied uncertainty of 1/16 or 6%, three times greater than that in the least-precisely known factor. This is a good illustration of how rounding can lead to the loss of information.
1.9E6 The «3.1» factor is specified to 1 part in 31, or 3%. In the answer 1.9, the value is expressed to 1 part in 19, or 5%. These precisions are comparable, so the rounding-off rule has given us a reasonable result.

A certain book has a thickness of 117 mm; find the height of a stack of 24 identical books:

2810 mm The “24” and the “1” are exact, so the only uncertain value is the thickness of each book, given to 3 significant digits. The trailing zero in the answer is only a placeholder.
10.4 In addition or subtraction, look for the term having the smallest number of decimal places, and round off the answer to the same number of places.

23 cm see below

The last of the examples shown above represents the very common operation of converting one unit into another. There is a certain amount of ambiguity here; if we take «9 in» to mean a distance in the range 8.5 to 9.5 inches, then the implied uncertainty is ±0. 5 in, which is 1 part in 18, or about ± 6%. The relative uncertainty in the answer must be the same, since all the values are multiplied by the same factor, 2.54 cm/in. In this case we are justified in writing the answer to two significant digits, yielding an uncertainty of about ±1 cm; if we had used the answer «20 cm» (one significant digit), its implied uncertainty would be ±5 cm, or ±25%.

When the appropriate number of significant digits is in question, calculating the relative uncertainty can help you decide.

This page titled 4.6: Significant Figures and Rounding is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Stephen Lower via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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The values ​​\u200b\u200bgiven in mathematical tables are sometimes exact, but more often approximate, representing the results of rounding off exact values, and contain errors that do not exceed half the units of the digit of the last digit. If the value is not taken directly from the table, but found by interpolation, the error may be larger, but in the vast majority of cases it does not exceed the last digit of one digit.
When calculating using tables, as with any calculation, the following rules must be observed:

• It is necessary to distinguish which data are accurate and which are approximate. Approximate data must be rounded, keeping only reliable figures in them and no more than one not quite reliable.
• When writing integer approximate numbers, zeros placed instead of unknown digits should be avoided.
• When adding and subtracting approximate numbers, the result should retain as many decimal places as there are in the approximate datum with the fewest decimal places.
• Note.
«Decimal places» numbers are those digits that are located to the right of the fractional sign.

• When multiplying and dividing, the result should retain as many significant digits as the approximation with the fewest significant digits has.
• Note.
«Significant digits» of a number are all its digits, except for zeros located to the left of its first non-zero digit.

• When squaring and cubeping, as a result, as many significant digits should be stored as the approximate number being raised to a power has.
• Note.
The last digit of the square and especially the cube is less reliable than the last digit of the base.

• When extracting square and cube roots, as a result, take as many significant digits as the radical (approximate) number has.
• Note.
The last digit of the square and especially the cube root is more reliable than the last digit of the root.

• When calculating intermediate results, you should take one digit more than the previous rules recommend.
• Note.
In the final result, this «spare digit» is discarded. It is recommended to highlight it.

• If some data has more decimal places (for steps I) or more significant figures (for steps II and III) than others, then they must first be rounded, keeping only one extra digit.
• If the data can be taken with arbitrary precision, then to obtain a result with k digits, the data should be taken with the number of digits that gives, according to rules 3 — 6, k + 1 digit in the result.
• When calculating by means of logarithms the value of an expression that does not contain addition and subtraction operations, you should calculate the number of significant digits in the approximate datum that has the smallest number of significant digits, and take the logarithm table with the number of decimal places greater than 1. In the final result, the last significant digit is discarded.
• When applying these rules, it should be remembered that they by no means guarantee the exact last digit of the result. This last figure may have an error, reaching even several units in some cases, but small values ​​of this error are more likely than large ones.

Other notes on algebra and geometry

Useful information?

## Computational methods

Laboratory work No. 1

Error estimation methods

I. Job description

Topic : Methods for estimating errors in approximate values.

Task 1. Rounding exact numbers to three significant figures, determine the absolute and relative errors of the obtained approximate numbers.

Given:

Find:

Solution:

— approximate value of the number A

Absolute error:

Relative error:

Task 2. Determine the absolute error of approximate numbers by their relative error.

Given:

Find:

Solution:

Absolute error:

When measuring length with an accuracy of 5 m, km is obtained, and when determining another length with an accuracy of 0.5 cm, meters are obtained. Which measurement is better in terms of quality?

Given: Km, M, M, Sm

Compare: and

Solution: So, according to the 1st measurement, the result is Km = M to within M ( — the absolute error of the value ).

Then the relative error: %

According to the 2nd measurement, the result is Km with an accuracy of Cm =M ( — absolute error of the value ).

Then the relative error: %

Since , the measurement can be considered better in quality than .

Answer: quality measurement is better than .

Task 4. a) Determine the number of correct characters in the number if its limiting absolute error is known true in the narrow sense, if the absolute error of this number does not exceed half the unit digit of the least significant digit, counting from left to right.

Absolute error: , so significant figures 8 and 4 of 0.00842 are correct in a narrow sense.

Answer: the number X has two correct digits in the narrow sense (8 and 4), that is,

B) Determine the number of correct characters in the number if its limiting relative error is known.

Given: %

Find:

Solution:

Absolute error limit:

First significant digit 1 of A 9 only0141 is correct in a narrow sense.

Answer: the number A has one correct digit in the narrow sense (1), that is,

believing that they have all the written signs are correct;

B) knowing that

Compare the errors and draw the necessary conclusions.

Given: , ,

Find:

Solution:

A):

If we assume that all the recorded characters are correct in the narrow sense, then the absolute error:

Maximum absolute error:

Then the maximum relative error:

%

:

that all the written signs are correct in the narrow sense, then the absolute error:

Limit absolute error:

Then the limit relative error:

%

:

If we assume that all the recorded characters are correct in the narrow sense, then the absolute error:

Limit absolute error:

Then limit relative error: 90 003

%

B) Let (interrupt write the number on the 7th decimal place and consider the resulting number as the exact value of the number).

Then the absolute error of the first representation of the number : .

Relative error: %

Absolute error of the second number representation: .

Relative error: %

Absolute error of the third number representation: %.

Relative error: %

Conclusions:

1) You can see that , that is ;

, that is ;

, i.e.

In other words, for three numbers their “true” relative error is limited by the limiting relative error determined from the condition that the signs of the numbers are correct. Moreover, for each number, the two estimates differ by less than an order of magnitude. Hence, the assumption about the correctness of all signs of the numbers Justified by .

2) Comparison of the relative errors of numbers:

shows

That the numbers are listed

In order of increasing accuracy of the representation of the number,

That is, more precisely, more precisely.

b)

Task 6. Find the sum of approximate numbers , , assuming that all signs in them are correct, i.e. that the absolute error of each term does not exceed half the unit of the least significant digit of this term. Determine the absolute and relative errors of the sum.

Given: , ,

Find:

Solution:

0003

The number with the largest absolute error.

2) The rest of the numbers are rounded, keeping one spare decimal place compared to the previously selected least accurate term:

, absolute rounding error

, absolute rounding error

3) Add all these numbers, taking into account all the saved signs:

4) The result is rounded by one sign (formally):

, the absolute rounding error is

5) The total absolute error of the sum will be added from three components:

B) the absolute value of the sum of rounding errors of terms;

C) the final rounding error of the result.

is the absolute error of the sum.

% — relative sum error.

Task 7. Find the maximum absolute and relative errors in calculating the volume of a right circular cylinder, if the values ​​of its height and radius of the base have all the correct signs.

Given: ,

Find:

Solution:

,

Take

The number with the largest absolute error.

Round the number R, keeping one spare decimal place in comparison with the previously selected least exact term:

, absolute rounding error (rounding is not required)

3) Received the result is rounded, keeping as many significant digits as there are correct digits in the number H , that is, 2 significant digits:

;

Absolute rounding error

4) The total absolute error of the product will be added from two terms:

A) the maximum absolute error of the product before its rounding;

B) the final rounding error of the product.

The absolute error of the product before rounding is calculated on the basis of the previously found relative error of the product of the rounded factors:

%.

Total absolute error

Now let’s move on to the desired volume.

(Here the result is rounded up to three significant figures).

— limiting absolute error of the volume.

% — limiting relative error of volume.

Task 8. Give an example of loss of accuracy when subtracting two close numbers.

Solution:

Let and be two close numbers; we assume that they have the same number of decimal places.

We consider that all signs are in numbers and are correct in a narrow sense. Then absolute errors:

Relative errors:

%

%

result error:

Relative result error: %

When subtracting two close numbers and relative error increased by 3 orders of magnitude!

Lab #2

Gauss method

I. Description of work

Topic : Solving a system of linear inhomogeneous algebraic equations by the Gauss method (single division scheme).

Job. Solve a system of three equations in three unknowns with an accuracy of the required unknowns up to .

Conduct intermediate calculations with two spare characters.

,

Solution:

We write the initial data and all calculation results in Table 1.

Forward run

1. We write the coefficients of this system in three rows and four columns of section 1 of Table 1.

2. We sum up all the coefficients in a row and write the sum in a column (control column), for example.

3. Divide all the numbers in the first line by and write the results in the 4th line of section 1.

4. Calculate and check if calculations are carried out with 6 or more decimal places, then the numbers should not differ by more than one last digit:

We write the results in the first two lines of the section:

6. We do the check. The sum of the elements of each line should not differ from more than 1-2 units of the last digit. Note that ,

,

,

0003

8. We check:

9. By the formulas we calculate:

We write the results in 1 line section 3.

10. Checking:

,

11. Divide all elements of line 1 of section 3 are on and the results are written in the next (second) line of this section.

12. Checking:

Reverse move

1. In section 4, write the units

2. Write down.

3. To calculate and use only section lines containing 1.

4. Calculate by the formula: .

5. Calculate by the formula:

.

6. Similarly, we carry out the reverse run in the control system. We write down,

we calculate and with the replacement of and with and, respectively:

We do the usual line-by-line check — it should be, with an accuracy of 1-2 units of the last digit.

Valid:

Fill in Table 1 with the calculation results:

Table 1

 Times Del 1 1 2 3 2 2 3 3 3 4 1 1 1 1 1

We round the obtained solution up to , according to the requirement of the problem:

The final check of the accuracy of the obtained solution of the system will be performed by substituting this solution into the system.

## By alexxlab

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