A sample of size 25 is drawn from a normal population with a population standard deviation of 100. Suppose the mean of the sample is x(bar) = 35. Recall that z0.025=1.96. A 95% confidence interval for the population mean is equal to

The standard form of the parabola with the equation 2x² - 8x - 3y + 11 = 0 is y = 2/3x² - 8/3x + 11/3

Given that

2x² - 8x - 3y + 11 = 0

To put the given equation in standard form, we need to isolate the y term on one side of the equation.

Here's how to do it:

2x² - 8x - 3y + 11 = 0

Isolate 3y

This gives

3y = 2x² - 8x + 11

Divide through by 3

y = 2/3x² - 8/3x + 11/3

Hence, the standard form is y = 2/3x² - 8/3x + 11/3

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Answer:

(a) $ 67.1

(b) $98.60

(c) $ 1.26

Step-by-step explanation:

Given equation is
[tex]y\:=\:-1.26x\:+\:98.6[/tex]

where x is the temperature in °F and y is the heating cost in $

To find heating cost for a specific temperature just substitute the value of x (temperature) in the equation

Part (a)
For x = 25 we get
y = -1.26(25)+ 98.6 = $ 67.1

Part (b)

For x = 0 we get
y = -1.26(0) + 98.6 = 0 + 98.6 = $98.60

Part (c)
For a 1 degree increase in temperature, the predicted decrease is the slope of the line
Slope of line = coefficient of x = -1.26

So decrease in cost for 1 degree increase in temperature = $1.26

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

[tex]y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{7}}x+9\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ \cfrac{-2}{7}} ~\hfill \stackrel{reciprocal}{\cfrac{7}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{7}{-2} \implies \cfrac{7}{ 2 }}}[/tex]

so we're really looking for the equation of a line whose slope is 7/2 and it passes through (4 , -6)

[tex](\stackrel{x_1}{4}~,~\stackrel{y_1}{-6})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{7}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{ \cfrac{7}{2}}(x-\stackrel{x_1}{4}) \implies y +6= \cfrac{7}{2} (x -4) \\\\\\ y+6=\cfrac{7}{2}x-14\implies {\Large \begin{array}{llll} y=\cfrac{7}{2}x-20 \end{array}}[/tex]

Answer:- Hence, the  combination will maximize total utility is [tex]12.[/tex]

Step-By-Step-Explanation:-

What is cost?

Cost is the amount of money you spent to make the product or service. Price is what you will charge for it.

Here given that,

The cups and plates budget is [tex]$18 (3 plates x $6 each + 6 cups x $3 each)[/tex]

Then the utility for the combination is [tex]12 (3 plates x 4 utility each + 6 cups x 2 utility each).[/tex]

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Can you post the image in your question.

The probability that a person chosen at random at age 40 or older has only completed high school is 0.2912.

Probability is the chance or likelihood that something will occur. It is quantified using numbers, ranging from 0 (no chance) to 1 (certainty). It also helps us to draw conclusions about the chances of something happening based on the available data.

The total number of people who have only completed high school

=(5394)

the total number of people in the age group 40 and over =(18521)

By dividing both the data, we get

5394/18521= 0.2912

From the given data in the table, we can calculate the probability that a person chosen at random at age 40 or older has only completed high school.

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The given family of functions is y = c1ex + c2e−x which is the general solution of the differential equation y'' − y = 0 on the indicated interval which is (−∞, ∞).

       Now, we are required to find a member of the family that is a solution to the initial-value problem which is

                             y(0) = 0 and y′(0) = 5.

     The differential equation is y'' − y = 0

      The characteristic equation is r2 − 1 = 0r2 = 1r1 = 1 and r2 = −1

     The general solution of the differential equation is y = c1ex + c2e−x

      Let us solve for the constants by using the given initial conditions:

                    At x = 0,y(0) = c1e0 + c2e0 = 0 + 0 = 0y(0) = 0

                    means c1 + c2 = 0or c1 = -c2At x = 0, y′(0) = c1ex |x=0 + c2e−x |x=0(d/dx)(c1ex + c2e−x) |x=0y′(0) = c1 - c2 = 5c1 - c2 = 5c1 - (-c1) = 5c1 + c1 = 5c1 = 5/2c1 = 5/2

                    Let's replace c1 = 5/2 in c1 = -c2, c2 = -5/2

                The solution of the initial-value problem y = (5/2)ex − (5/2)e−x is a member of the family y = c1ex + c2e−x that is a solution of the initial-value problem y(0) = 0 and y′(0) = 5.

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The value of the pooled sample variance is 25 when the first sample has a sample variance of 12 and a second sample has a sample variance of 22.

If a first sample has a sample variance of 12 and a second sample has a sample variance of 22, then the possible values of the pooled sample variance are given by the formula below:

Formula:

pooled sample variance = [(n₁ - 1) s₁² + (n₂ - 1) s₂²] / (n₁ + n₂ - 2)

Where s₁ and s₂ are the sample standard deviations of the first and second samples,

n₁ and n₂ are the sample sizes of the first and second samples, respectively.

Thus, substituting the given values into the formula above, we have pooled sample variance:

= [(n₁ - 1) s₁² + (n₂ - 1) s₂²] / (n₁ + n₂ - 2)

= [(n₁ - 1) 12 + (n₂ - 1) 22] / (n₁ + n₂ - 2)

Checking each of the answer options:

If pooled sample variance is 1, then:

(n₁ - 1) 12 + (n₂ - 1) 22

= (n₁ + n₂ - 2)(1)

= 12n₁ + 22n₂ - 34

= (12n₁ - 12) + (22n₂ - 22)

= 12(n₁ - 1) + 22(n₂ - 1)

The expression on the right-hand side of the equation is a sum of multiples of 12 and 22, and therefore, the expression itself will be a multiple of the greatest common divisor of 12 and 22, which is 2.

Since 34 is not a multiple of 2, the equation cannot be true if the pooled sample variance is 1.

Thus, 1 is not a possible value of the pooled sample variance.

If pooled sample variance is 10, then:

(n₁ - 1) 12 + (n₂ - 1) 22

= (n₁ + n₂ - 2)(10)

= 12n₁ + 22n₂ - 34

= (12n₁ - 12) + (22n₂ - 22)

= 12(n₁ - 1) + 22(n₂ - 1)

The expression on the right-hand side of the equation is a sum of multiples of 12 and 22, and therefore, the expression itself will be a multiple of the greatest common divisor of 12 and 22, which is 2.

Since 34 is not a multiple of 2, the equation cannot be true if the pooled sample variance is 10.

Thus, 10 is not a possible value of the pooled sample variance.

If pooled sample variance is 16, then:

(n₁ - 1) 12 + (n₂ - 1) 22

= (n₁ + n₂ - 2)(16)

= 12n₁ + 22n₂ - 34

= (12n₁ - 12) + (22n₂ - 22)

= 12(n₁ - 1) + 22(n₂ - 1)

The expression on the right-hand side of the equation is a sum of multiples of 12 and 22, and therefore, the expression itself will be a multiple of the greatest common divisor of 12 and 22, which is 2.

Since 34 is not a multiple of 2, the equation cannot be true if the pooled sample variance is 16.

Thus, 16 is not a possible value of the pooled sample variance.

If pooled sample variance is 25, then:

(n₁ - 1) 12 + (n₂ - 1) 22

= (n₁ + n₂ - 2)(25)

= 12n₁ + 22n₂ - 34

= (12n₁ - 12) + (22n₂ - 22)

= 12(n₁ - 1) + 22(n₂ - 1)

The expression on the right-hand side of the equation is a sum of multiples of 12 and 22, and therefore, the expression itself will be a multiple of the greatest common divisor of 12 and 22, which is 2.

Since 46 is a multiple of 2, the equation can be true if the pooled sample variance is 25.

Thus, 25 is a possible value of the pooled sample variance.

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Explanation:

2n is greater than 20, so 2n > 20. This solves to n > 10 after dividing both sides by 2. That inequality is the same as 10 < n.

5n is less than 60. It leads to 5n < 60. That solves to n < 12 after dividing both sides by 5.

To summarize:

Combine 10 < n with n < 12 to write a compound inequality:

10 < n < 12

The value n is between 10 and 12. We exclude each endpoint.

The only possible whole number that works here is n = 11

The Dow Jones average after 4 more weeks of the same rate of drop would be 11,167.

What is average?

Average, also known as mean, is a measure of central tendency that represents the typical or common value in a set of data. It is calculated by adding up all the values in a data set and then dividing the sum by the total number of values.

To calculate the drop in the Dow Jones average, we subtract the initial value from the final value:

Drop = Final Value - Initial Value

Drop = 12,503 - 12,837

Drop = -334

So the Dow Jones average dropped by 334 points in one week.

If the price continued to drop at the same rate for 4 more weeks, then the total drop after 5 weeks would be:

Total Drop = 5 x Drop

Total Drop = 5 x (-334)

Total Drop = -1670

To calculate the Dow Jones average after 4 more weeks, we need to subtract the total drop from the initial value:

New Dow Jones Average = Initial Value - Total Drop

New Dow Jones Average = 12,837 - 1,670

New Dow Jones Average = 11,167

Therefore, the Dow Jones average after 4 more weeks of the same rate of drop would be 11,167.

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The answers of the question number 3 and 4 are given below respectively.

A graph is a visual representation of data or mathematical relationships.

It typically involves plotting points on a coordinate plane and connecting them with lines or curves.

Graphs can help to illustrate patterns, trends, and relationships in data and make it easier to understand complex information.

3.Assuming that the sale of bracelets and necklaces are represented by the variables B and N respectively, the total sale would be represented by the mathematical statement:

Total sale = B + N

Graph of Total Sale = B + N

The x-axis represents the number of bracelets sold (B), and the y-axis represents the number of necklaces sold (N).

4.To represent Nimfa's goal of achieving a total sale of at least Php 15,000, we can use the inequality:

Total sale ≥ Php 15,000

or, using the expression from the previous question:

2B + N ≥ Php 15,000

Graph of Total Sale ≥ Php 15,000

The shaded area above the plane represents all the combinations of sales of bracelets and necklaces that result in a total sale of at least Php 15,000. Any point above the plane is a valid combination of sales that meet Nimfa's goal.

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3. The x-axis represents the number of bracelets sold (B), and the y-axis represents the number of necklaces sold (N).

4. The shaded area above the plane represents all the combinations of sales of bracelets and necklaces that result in a total sale of at least Php 15,000.

A graph is a visual representation of data or mathematical relationships.

It typically involves plotting points on a coordinate plane and connecting them with lines or curves.

Graphs can help to illustrate patterns, trends, and relationships in data and make it easier to understand complex information.

3.Assuming that the sale of bracelets and necklaces are represented by the variables B and N respectively, the total sale would be represented by the mathematical statement:

Total sale = B + N

Graph of Total Sale = B + N

4.To represent Nimfa's goal of achieving a total sale of at least Php 15,000, we can use the inequality:

Total sale ≥ Php 15,000

or, using the expression from the previous question:

2B + N ≥ Php 15,000

Graph of Total Sale ≥ Php 15,000

The shaded area above the plane represents all the combinations of sales of bracelets and necklaces that result in a total sale of at least Php 15,000. Any point above the plane is a valid combination of sales that meet Nimfa's goal.

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The Central Limit Theorem (CLT) states that as the sample size n increases, the sample mean approaches a normal distribution with a mean of μ and a standard deviation of σ/√n.

Therefore, when the number of houses in a population is unknown and a random sample of 48 houses is drawn from it, the mean for the sample sum distribution can be calculated using the CLT as follows:Mean for the sample sum distribution = nμ = 48 * 1670 = 80,160. The standard deviation for the sample sum distribution is given by:σ/√n = 140/√48 ≈ 20.20. Therefore, the sample sum distribution has a mean of 80,160 and a standard deviation of 20.20.To verify this, a histogram can be plotted in Excel using the following steps:Enter the data for the square footage of the 48 houses in column A of the Excel worksheet.Highlight column B and enter the formula =SUM(A1:A48) to sum the data in column A and store the result in column B.Highlight column C and enter the formula =B1/48 to calculate the mean for the sample sum distribution and store it in column C.Highlight column D and enter the formula =140/SQRT(48) to calculate the standard deviation for the sample sum distribution and store it in column D.Highlight columns B, C, and D, and select the Insert tab.Click on the Histogram icon under the Charts group.Select the Histogram chart type, and click OK to generate the histogram.The histogram should show a bell-shaped curve with a mean of 80,160 and a standard deviation of 20.20, indicating that the sample sum distribution is approximately normal.

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The grocer must use 45 pounds of each kind of candy

A grocer has two kinds of candy
The first type of candy sells for 90 cents per pound

The second type of candy sells for 40 cents per pound

The number of pounds needed to make each kind of candy if he uses 50 pounds can be calculated as follows

The first step is to multiply 50 by 85%
50(85/100)

= 50(0.85)

= 42.5

Let y represent the unknown

y(90/100) + (50-y)(40/100)

0.9y + 0.4(50-y)

0.9y + 20-0.4y

0.9y-0.4y+20

0.5y+20(equate this value with 42.5)

0.5y+20= 42.5

collect the like terms

0.5y= 42.5-20

0.5y= 22.5

y= 22.5/0.5

y= 45

Hence the grocer needs 45 pounds for each kind of candy

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Answer:

Step-by-step explanation:

To find the product

Its gonna be

12g^4h^5

the area of the shaded region is approximately 37.68 cm².

The area of a circle is the amount of space that is enclosed by the circle in a two-dimensional plane. It is defined as the product of the square of the radius (r) of the circle and the mathematical constant pi (π), which is approximately equal to 3.14159.

The area of the larger circle is:

A1 = π(4 cm)² = 16π cm²

The area of the smaller circle is:

A2 = π(2 cm)² = 4π cm²

To find the area of the shaded region, we subtract A2 from A1:

A1 - A2 = 16π cm²- 4π cm² = 12π cm²

To round the final answer to the nearest hundredth, we can use the approximation π ≈ 3.14:

12π cm² ≈ 12 × 3.14 cm² ≈ 37.68 cm²

Therefore, the area of the shaded region is approximately 37.68 square centimeters.

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Answer: 28cm²

Step-by-step explanation: To calculate the area of the trapezium, we can use the formula:

Area = (1/2) x (sum of parallel sides) x (height)

In this case, the two parallel sides are 6 cm and 8 cm, and the height is 4 cm.

Plugging in the values, we get:

Area = (1/2) x (6 cm + 8 cm) x 4 cm

Area = (1/2) x 14 cm x 4 cm

Area = 28 cm²

The value οf a = 1/2  and b = 19/4 in the equatiοn x² - x + 5 = (x-a)² + b.

The part οf mathematics in which letters and οther general symbοls are used tο represent numbers and quantities in fοrmulae and equatiοn is called algebra.

x² - x + 5 = (x-a)² + b

x² - x + 5 = x² + a² - 2xa + b

-x + 5 = -2xa + a² + b

By matching cοrrespοnding terms,

2a = 1          and        a²+ b = 5

a= 1/2          and        a²+ b = 5

Substituting value οf "a"

(1/2)² + b = 5

1/4 + b = 5

b = 5- 1/4

b = 19/4

Thus, a = 1/2  and b = 19/4.

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Answer:

Step-by-step explanation:

100 students all together.

50+70=120....so 20 must like both.

In mathematics, a logarithm is a mathematical operation that determines how many times a given number (known as the base) must be multiplied by itself to produce another given number.

Logarithms are used to simplify complex calculations involving exponents and to convert between exponential and logarithmic expressions.

The logarithm of a number x to a given base b is represented as logb(x). For example, log10(100) = 2 because 10 multiplied by itself twice equals 100.

Logarithms have a wide range of applications in various fields, including mathematics, physics, engineering, finance, and computer science. They are particularly useful in scientific calculations involving large numbers, where working with exponents can become cumbersome. Logarithms are also used in the study of exponential growth and decay, as well as in the analysis of data sets with a wide range of values.

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The function is constant from approximately x = -4 to x = -3 and from approximately x = -1 to x = 1. So, the constant intervals are (-4, -3) and (-1, 1)

A function, which produces one output from a single input, is an illustration of a rule. The picture was obtained from Alex Federspiel. The equation y=x2 serves as an example of this.

a) We can see that the function is increasing from approximately x = -3 to x = -1 and from approximately x = 1 to x = 2.5. So, the increasing intervals are (-3,-1) and (1, 2.5)

(b) We can see that the function is decreasing from approximately x = -2 to x = -0.5 and from approximately x = 3 to x = 4. So, the decreasing intervals are (-2, -0.5) and (3, 4)

(c) We can see that the function is constant from approximately x = -4 to x = -3 and from approximately x = -1 to x = 1. So, the constant intervals are (-4, -3) and (-1, 1)

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Answer: The other leg is 7 m, and the hypotenuse is 7√2 m.

Step-by-step explanation:

This is just a rule that in all cases, the two legs are equal and the hypotenuse is equal to the length of a leg times the square root of 2.

Hope this helps :)

Answer: cab

Step-by-step explanation:

you start with the last point and move up

According to the given information Ling has enough string to produce at least 20 bracelets.

Numbers serve the purpose of counting, measuring, organising, indexing, and other purposes. Natural numbers, whole numbers, rational and irrational numbers, integers, actual values, complex numbers, even and odd numbers, and so on are all distinct forms of numbers.

A number is really a numerical value that is used to express quantity. As a consequence, a number seems to be a mathematical idea that may be used to count, measure, and name objects. As little more than a result, numbers are the foundation of mathematics. Here is one butterfly, and here are four butterflies.

Each bracelet requires 7 1/2 inches of string. So, the total length of string required to make 20 bracelets is:

20 × 7 1/2 = 150 inches

Therefore, Ling needs to buy at least 150 inches of string.

We can write the inequality to represent this situation as:

length of string ≥ 150

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Answer:

Width = 17 inches

Length = 43 inches

Step-by-step explanation:

  Let the width of the rectangle = x inches

              Length = 2*width +9

                          = (2x + 9) cm

    [tex]\boxed{\bf Perimeter \ of \ rectangle = 2* (length + width)}[/tex]

        2* (length + width) = 120 inches

         2* (2x + 9 + x)       = 120

                     2* (3x + 9) = 120

Use Distributive property,

                 2 * 3x + 2*9  = 120

                         6x + 18   = 120

Subtract 18 from both sides,

                                   6x = 120 - 18

                                   6x = 102

Divide both sides by 6,

                                     x = 102 ÷ 6

                                     x = 17

Width = 17 inches

Length = 2*17 +9

           = 34 + 9

           = 43 inches

Answer:

[tex] |x - 300| \leqslant 10[/tex]

Answer:  0.11

Step-by-step explanation:

5/45=0.11

Answer:

  3. 79.9 mm²

  4. 6.4 in

  5. 177.5 mi²

  6. 60.3°

Step-by-step explanation:

Given various quadrilaterals and their dimensions, you want to find missing dimensions.

In all cases, one or more area formulas and trig relations are involved. The trig relations are summarized by the mnemonic SOH CAH TOA. The relevant relation for these problems is ...

  Tan = Opposite/Adjacent

It is also useful to know that 1/tan(x) = tan(90°-x).

The formula for the area of a trapezoid is ...

  A = 1/2(b1 +b2)h

The relevant formula here for the area of a parallelogram is ...

  A = bs·sin(α) . . . . . where α is the angle between sides of length b and s

Using the area formula above, we find the area to be ...

  A = (21 mm)(9 mm)·sin(155°) ≈ 79.9 mm²

The area is about 79.9 square mm.

The given figure shows two unknowns. We can write equations for these using the area formula and using a trig relation.

If we draw a vertical line through the vertex of the marked angle, the base of the triangle to the right of it is (b2-4). The acute angle at the top of that right triangle is (121°-90°) = 31°. The tangent relation tells us ...

  tan(31°) = (b2 -4)/h   ⇒   h = (b2 -4)/tan(31°)

Using the area formula we have ...

  A = 1/2(b2 +4)h

and substituting for A and h, we get ...

  20.8 = 1/2(b2 +4)(b2 -4)/tan(31°)

  2·tan(31°)·20.8 = (b2 +4)(b2 -4) = (b2)² -16 . . . . . multiply by 2tan(31°)

  (b2)² = 2·tan(31°)·20.8 +16 . . . . . . . add 16

  b2 = √(2·tan(31°)·20.8 +16) ≈ 6.4 . . . . . . take the square root

Base 2 of the trapezoid is about 6.4 inches.

To find the area, we need to know the height of the trapezoid. To find the height we can solve a triangle problem.

If we draw a diagonal line parallel to the right side through the left end of the top base, we divide the figure into a triangle on the left and a parallelogram on the right. The triangle has a base width of 10 mi, and base angles of 60° and 75°.

Drawing a vertical line through the top vertex of this triangle divides it into two right triangles of height h. The top angle is divided into two angles, one being 90°-60° = 30°, and the other being 90°-75° = 15°. The bases of these right triangles are now ...

and their sum is 10 mi.

The height h can now be found to be ...

  h·tan(30°) +h·tan(15°) = 10

  h = 10/(tan(30°) +tan(15°))

Back to our formula for the area of the trapezoid, we find it to be ...

  A = 1/2(b1 +b2)h = 1/2(20 +10)(10/(tan(30°) +tan(15°)) ≈ 177.5

The area of the trapezoid is about 177.5 square miles.

The final formula we used for problem 5 can be used for problem 6 by changing the dimensions appropriately.

  A = 1/2(b1 +b2)(b1 -b2)/(tan(90-x) +tan(90-x))

  112 = 1/2(20+12)(20-12)/(2·tan(90-x)) = (20² -12²)·tan(x)/4

  tan(x) = 4·112/(20² -12²)

  x = arctan(4·112/(20² -12²)) = arctan(7/4) ≈ 60.3°

Angle x° in the trapezoid is about 60.3°.

__

Additional comment

There is no set "step by step" for solving problems like these. In general, you work from what you know toward what you don't know. You make use of area and trig relations as required to create equations you can solve for the missing values. There are generally a number of ways you can go at these.

A nice scientific calculator has been used in the attachment for showing the calculations. A graphing calculator can be useful for solving any system of equations you might write.

The second attachment shows a graphing calculator solution to problem 5, where we let y = area, and x = the portion of the bottom base that is to the left of the top base. Area/15 represents the height of the trapezoid. This solution also gives an area of 177.5 square miles.

If the shape is made up of one half-circle attached to an equilateral triangle with side lengths 20 inches then, the perimeter of the shape is 91.4 inches.

To find the perimeter of the shape, we need to know the length of the curved boundary (the circumference of the half-circle) and the length of the straight boundary (the perimeter of the equilateral triangle).

The radius of the half-circle is half the length of the side of the equilateral triangle, which is 10 inches. Therefore, the circumference of the half-circle is:

C = πr = π(10) = 31.4 inches.

The perimeter of the equilateral triangle is 3 times the length of one side, which is 20 inches. Therefore, the perimeter of the triangle is:

P = 3s = 3(20) = 60 inches

Finally, the perimeter of the entire shape is the sum of the lengths of the curved and straight boundaries:

Perimeter = C + P = 31.4 + 60 = 91.4 inches.

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Complete Question

This shape is made up of one half-circle attached to a square with side lengths 11 inches. Find the perimeter of the shape.

You can use 3.14 as an approximation for π. help i don't know it.

hmmm what we do is firstly make the recurring part a variable, then we multiply it such that, the recurring digits move over from the decimal point to the left, so we'd multiply it by some power of 10, in this case power of 3, because we have three digits to move, 246, so let's do all that

[tex]0.\overline{246}\hspace{5em}x=0.\overline{246}\hspace{5em} \begin{array}{llll} 1000x&=&246.\overline{246}\\\\ &&246+0.\overline{246}\\\\ &&246+x \end{array} \\\\[-0.35em] ~\dotfill\\\\ 1000x=246+x\implies 999x=246\implies x=\cfrac{246}{999}\implies x=\cfrac{82}{333}[/tex]