A new L.E.D. light bulb has an expected life time of 25000 hours. Your guess for the probability that it will last more than 3 years is closest to: (Assume life times follow the exponential distribution) (A) 100% (B) 99% (C) 53% (D) 35% (E) 0%

The two bases B1 and B2 span the same subspace of R3.

Let B1 be the basis {(1,2,3),(2,−1,2)}={u1,u2} and B2 be the basis {(0,5,4),(1,−8,−5)}={v1,v2}. We want to show that these two bases span the same subspace of R3.

To do this, we need to show that any vector in the subspace spanned by B1 can also be written as a linear combination of vectors in B2, and vice versa.

Let x be a vector in the subspace spanned by B1. Then x = a*u1 + b*u2 for some scalars a and b. Substituting the values of u1 and u2, we get:

x = a*(1,2,3) + b*(2,−1,2)

= (a+2b, 2a-b, 3a+2b)

Now we want to express x as a linear combination of v1 and v2:

x = c*v1 + d*v2

= c*(0,5,4) + d*(1,−8,−5)

= (d, 5c-8d, 4c-5d)

Setting the two expressions for x equal to each other, we get the following system of equations:

a+2b = d

2a-b = 5c-8d

3a+2b = 4c-5d

Solving this system of equations, we can find values of c and d that satisfy the equations. This shows that any vector in the subspace spanned by B1 can also be written as a linear combination of vectors in B2.

Similarly, we can show that any vector in the subspace spanned by B2 can also be written as a linear combination of vectors in B1.

Therefore, the two bases B1 and B2 span the same subspace of R3.

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The average cost of a call depends on the initial period, subsequent period, and unit cost for each subsequent period. By using the formula derived above, we can calculate the average cost of a call for different values of I, S, and p.

The average cost of a call can be calculated using the formula: Average Cost = r + p(S-1)/(S-1+1/e^S), where r is the cost for the initial period, p is the unit cost for each subsequent period, and S is the length of each subsequent period.

If the cost of a call is $1 per minute and the average call length is one minute, then the average cost of a call is just $1. However, if the call is rounded up to the next minute (I = S = 1), then the average cost can be calculated using the formula derived above: Average Cost = r + p(S-1)/(S-1+1/e^S) = $1 + $1(1-1)/(1-1+1/e^1) = $1 + $1(0)/(0+1/e) = $1.

If I = S = 0.1 (i.e., rounded up to the next 6 seconds), then the average cost can be calculated using the formula derived above: Average Cost = r + p(S-1)/(S-1+1/e^S) = $1 + $1(0.1-1)/(0.1-1+1/e^0.1) = $1 + $1(-0.9)/(-0.9+1/e^0.1) = $1.07.

Therefore, the average cost of a call depends on the initial period, subsequent period, and unit cost for each subsequent period. By using the formula derived above, we can calculate the average cost of a call for different values of I, S, and p.

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Answer:

2. 29 3. 9:11, [tex]\frac{9}{11}[/tex], 55 4(a) 10.7

Step-by-step explanation:

2. 29

3. 9:11

   [tex]\frac{9}{11}[/tex]

   [tex]\frac{11}{20} *100%\\= 55%[/tex]

4(a)10.7

Isabel had 10 kilogram of oranges when she started. She ate 2 kilogram of the oranges, leaving her with 8 kilogram. This means that she has 8/10 kilogram of oranges left.

It is important to understand how to calculate fractions in order to answer questions like this. In this case, the question was asking how much of a kilogram remains after Isabel ate 2 kilogram. To answer this, we had to figure out what fraction of the 10 kilogram Isabel ate, which was 2/10. We then subtracted this fraction from 1 in order to figure out how much of a kilogram remains, which was 8/10.

This kind of calculation is useful when trying to figure out how much of something remains after a certain amount has been taken away. It can be used in a wide variety of contexts, such as when trying to figure out how much of a product or item remains after some has been sold or used. Understanding how to calculate fractions can be very helpful in these kinds of scenarios.

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Answer: option D is correct

Step-by-step explanation

Angle makes Quadrant III is 60°

To find : terminal side angle .

solution : we know that sum of total angle

is 360° and half of it in 180°

here the possible angle of an

that has terminal side  in quadrant III

and makes a 60 degree with the x axis,

=) 120° + 60° =240°

                                          Hence the correct option is in D) 240°

A \times B \times C = \left[\begin{array}{ll} -40 & -60 \\ -68 & 70 \\ 6 & 0 \end{array}\right]

To find the product of the matrices A, B and C, we can use the following equation:



$$A \times B \times C = \left[\begin{array}{lll} (A \times B)_{11} & (A \times B)_{12} & (A \times B)_{13} \\ (A \times B)_{21} & (A \times B)_{22} & (A \times B)_{23} \\ (A \times B)_{31} & (A \times B)_{32} & (A \times B)_{33} \end{array}\right] \times C = \left[\begin{array}{ll} (A \times B \times C)_{11} & (A \times B \times C)_{12} \\ (A \times B \times C)_{21} & (A \times B \times C)_{22} \\ (A \times B \times C)_{31} & (A \times B \times C)_{32} \end{array}\right]$$



To find each element of the product, we use the following equation:



$$(A \times B \times C)_{ij} = \sum_{k=1}^{3} A_{ik} \times B_{kj} \times C_{ij}$$



Where $i$ and $j$ represent the row and column numbers respectively. For example, to find the element $(A \times B \times C)_{11}$, we have:



$$(A \times B \times C)_{11} = \sum_{k=1}^{3} A_{1k} \times B_{k1} \times C_{11} = (-5 \times -8 \times -4) + (1 \times 7 \times -4) + (-7 \times 5 \times -4) = -40$$



Therefore, the product of A, B and C is:



$$A \times B \times C = \left[\begin{array}{ll} -40 & -60 \\ -68 & 70 \\ 6 & 0 \end{array}\right]$$

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The temperature at 7 AM is approximately 87 degrees.

The outside temperature over a day can be modeled as a sinusoidal function. In this case, we know that the high temperature of 98 degrees occurs at 6 PM and the average temperature for the day is 85 degrees. To find the temperature at 7 AM, we can use the equation:

T(t) = A sin(B(t - C)) + D

where T(t) is the temperature at time t, A is the amplitude, B is the frequency, C is the horizontal shift, and D is the vertical shift.

We know that the average temperature is 85 degrees, so D = 85. We also know that the high temperature of 98 degrees occurs at 6 PM, so A = (98 - 85)/2 = 6.5 and C = 6.

Now we can plug in the values and solve for the temperature at 7 AM:

T(7) = 6.5 sin(B(7 - 6)) + 85

Since we want to find the temperature to the nearest degree, we can round the answer to the nearest whole number:

T(7) = 6.5 sin(B) + 85 ≈ 87 degrees

Therefore, the temperature at 7 AM is approximately 87 degrees.

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By assuming and solving with the rectangle measurements, we get the equation, 2x^2 + 6x - 33 = 0.

Follow the steps below to reach the answer:

A region of 11 cm2 divides the two rectangles A and B.

Rectangle A has a length of x cm.

Rectangle B has a length of (x+3) cm.

Form an equation in x and demonstrate that it simplifies when given the fact that rectangle A's width is 2 cm bigger than rectangle B's width:

2x^2 + 6x - 33 = 0

Rectangle A's length and area being x and 11, respectively, results in the following for rectangle A's width: 11 / x

Rectangle B's length is equal to x + 3 and its area is equal to 11, hence its width is as follows: 11 / (x + 3)

Given that width of the rectangle, A is 2 cm Larger than the width of rectangle B, we obtain: 11/x =  (11/ ( x + 3 ) ) + 2

In addition, based on the situation, we get

11/x - 11/ (x+3) = 2 cm

Let this be equation (1)

The main challenge at hand is to change this equation into the quadratic equation's standard form.

It is obtained by multiplying both sides of equation (1) by x*(x+3). As a result, you will get,

11*(x+3) - 11x = 2x*(x+3)

11x + 33 - 11x = 2x^2 + 6x

2x^2 + 6x - 33 = 0.

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The solution for x is 1/(3^16).

To solve for x, we need to convert the logarithmic equation to exponential form. The general formula for converting a logarithmic equation to an exponential equation is:

log_b(x) = y => b^y = x

In this case, the base is 3, the exponent is -2 - 6 - 8 + (1/9) - (1/9), and x is the value we are trying to find. So, we can write the exponential equation as:

3^(-2 - 6 - 8 + (1/9) - (1/9)) = x

Simplifying the exponent gives us:

3^(-16) = x

Now, we can solve for x by taking the inverse of both sides:

x = 1/(3^16)

Therefore, the solution for x is 1/(3^16).

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The numerator is \((25x^{2}+15x+9)\) and the denominator is \(2(x+2)\).

To simplify the given rational expression, we need to factor the numerator and denominator and then cancel out any common factors.

First, let's factor the numerator:
\(500x^{3}-108=4(125x^{3}-27)=4(5x-3)(25x^{2}+15x+9)\)

Next, let's factor the denominator:
\(40x^{2}+56x-48=8(5x^{2}+7x-6)=8(5x-3)(x+2)\)

Now, we can cancel out the common factor of \(4(5x-3)\) from the numerator and denominator:
\(\frac{4(5x-3)(25x^{2}+15x+9)}{8(5x-3)(x+2)}=\frac{(25x^{2}+15x+9)}{2(x+2)}\)

Therefore, the simplified rational expression is \(\frac{(25x^{2}+15x+9)}{2(x+2)}\).

In this simplified form, the numerator is \((25x^{2}+15x+9)\) and the denominator is \(2(x+2)\).

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The distance across the lake is 120 feet.

Right angled triangle are those triangle for which one of the angle is 90 degrees.

Given is a right angled triangle.

Pythagoras theorem states that,

Square of the hypotenuse of a right angled triangle is equal to the sum of the squares of the two legs of the triangle.

The distance across the lake is one of the leg of the right triangle.

Length of hypotenuse = 208 feet + 104 feet

                                     = 312 feet

Length of other leg = 192 feet + 96 feet

                                = 288 feet

Distance across the lake = √(312² - 288²)

                                         = √14,400

                                         = 120 feet

Hence 120 feet is the distance across the lake.

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Answer:

x ∈ { ( 6 + [tex]\sqrt{23}[/tex] * i) , { ( 6 - [tex]\sqrt{23}[/tex] * i) }

Step-by-step explanation:

Given: [tex]x^2-12x+59=0[/tex]

First, collect like terms. [tex]x^2[/tex] and -12x, 59 and 0:

[tex]x^2-12x=0-59[/tex]

+59 is changed to -59 when transferred. Then subtract:

( x - 12 ) x = -59

Finally divide both sides by -59:

x ∈ { ( 6 + [tex]\sqrt{23}[/tex] * i) , { ( 6 - [tex]\sqrt{23}[/tex] * i) }

Answer:

10 cm

Step-by-step explanation:

The volume of a cylinder is:

V = πr²h

Plug in the known values:

1884 cm³ = 3.14 × r² × 6 cm

Solve for r²:

r² = (1884 cm³)/(3.14 × 6 cm)

r² = 100 cm²

Solve for r:

r = √(100 cm²)

r = 10 cm

The correct answer is d. happens when A is not in [-pi/2, pi/2]. The inverse sine function, sin^-1, or arcsin, is the function that reverses the sine function.

It is defined for values in the range [-1, 1] and has a range of [-pi/2, pi/2]. This means that if A is not in the range [-pi/2, pi/2], then sin^-1 (sin A) will not equal A.

For example, if A = pi, then sin A = 0, but sin^-1 (0) = 0, not pi. This is because pi is not in the range [-pi/2, pi/2], so the inverse sine function cannot return it as an answer.

Therefore, The correct answer is d. happens when A is not in [-pi/2, pi/2].

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The water level after 4 weeks will be 29.2 feet.

The water level of the lake will increase by 10% each week during the winter rainstorms. This means that each week, the water level will increase by 0.10 × 20 = <<0.10*20=2>>2 feet.

After the first week, the water level will be 20 + 2 = <<20+2=22>>22 feet.
After the second week, the water level will be 22 + 2 = <<22+2=24>>24 feet.
After the third week, the water level will be 24 + 2 = <<24+2=26>>26 feet.
And so on.

We can use the formula A = P(1 + r)^n to find the water level after n weeks, where A is the final amount, P is the initial amount, r is the rate of increase, and n is the number of weeks.

In this case, P = 20, r = 0.10, and n is the number of weeks.

So, the water level after n weeks will be:

A = 20(1 + 0.10)^n

We can plug in different values of n to find the water level after a certain number of weeks.

For example, to find the water level after 4 weeks, we can plug in n = 4:

A = 20(1 + 0.10)^4 = 29.2 feet

So, the water level after 4 weeks will be 29.2 feet.

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To derive the properties of R2 for the model m Ji = Σβ,tij + u; j=1, we need to first understand what R2 represents. R2, also known as the coefficient of determination, is a statistical measure that represents the proportion of the variance in the dependent variable that is explained by the independent variable(s) in a regression model.
In the given model, the dependent variable is m Ji and the independent variable is tij. The coefficient β represents the slope of the regression line and u represents the error term.

To derive the properties of R2, we can use the formula:
R2 = 1 - (SSres/SStot)
Where SSres is the sum of squared residuals and SStot is the total sum of squares.
SSres = Σ(m Ji - ŷi)2
SStot = Σ(m Ji - ȳ)2
Where ŷi is the predicted value of m Ji and ȳ is the mean of m Ji.

By substituting the values of SSres and SStot into the formula for R2, we can derive the properties of R2 for the given model.
R2 = 1 - (Σ(m Ji - ŷi)2/Σ(m Ji - ȳ)2)

The properties of R2 for the given model are:
1) R2 is always between 0 and 1. A value of 0 indicates that the independent variable(s) do not explain any of the variance in the dependent variable, while a value of 1 indicates that the independent variable(s) explain all of the variance in the dependent variable.
2) R2 is a measure of the strength of the relationship between the independent variable(s) and the dependent variable. The closer R2 is to 1, the stronger the relationship.
3) R2 is affected by the number of independent variables in the model. The more independent variables there are, the higher the R2 value will be.
4) R2 does not indicate causation. It only measures the strength of the relationship between the independent variable(s) and the dependent variable.

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Answer:

2

Step-by-step explanation:

Using the scale factor, we found the volume of the larger cylinder as 70 m³.

One of the most fundamental curvilinear geometric shapes, a cylinder has historically been a three-dimensional solid. It is regarded as a prism with a circle as its base in basic geometry. One of the fundamental three-dimensional shapes in geometry is the cylinder, which has two distant, parallel circular bases. At a predetermined distance from the centre, a curved surface connects the two circular bases. The axis of the cylinder is the line segment connecting the centres of two circular bases. The height of the cylinder is defined as the distance between the two circular bases.

Given,

The scale factor of the cylinders = 5:2

The scaling factor indicates how much a figure has increased or decreased from its initial value.

The volume of the smaller cylinder = 28 m³

We are asked to find the volume of the larger cylinder.

let the volume of the larger cylinder be x.

x / 28 = 5/2

x  =(5/2) × 28 = 70

Therefore using the scale factor, we found the volume of the larger cylinder as 70 m³.

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The equation (1 + 4/a+4/a²)/ (1- 8/a - 20/a² simplifies to  -(a-10)/(a+2) . (C)

The given expression can be simplified by using the distributive law of multiplication over addition and the inverse law of multiplication to simplify the denominator.

By applying these laws, the given expression becomes: (1 + 4/a+4/a²)/ (1- 8/a - 20/a² = (1 + 4/a+4/a²)/ (a² - 8a - 20) = (a+2)/(a² - 8a - 20). By further simplifying it, we get (a+2)/(a-10). Hence, the answer is C. -(a-10)/(a+2).

In order to simplify the given expression, the distributive law of multiplication over addition is used by multiplying the numerator and denominator separately with the denominator's terms.

The inverse law of multiplication is then used to simplify the denominator. This gives us the simplified expression of (a+2)/(a-10).

By substituting the values of a in the expression, we can verify the answer. If a = 5, then the expression simplifies to

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Complete question:

[tex]\[ \frac{1+\frac{4}{a}+\frac{4}{a^{2}}}{1-\frac{8}{a}-\frac{20}{a^{2}}} \] a. \( \frac{a+2}{a-10} \) b. \( \frac{a+2}{a^{2}} \) c. \( -\frac{a-10}{a+2} \) d. \( \frac{4}{a-10} \)[/tex]

(1 + 4/a+4/a²)/ (1- 8/a - 20/a²) = ?

A. (a+2)/(a-10) b. (a+2)/a² C. -(a-10)/(a+2) D. 4/(a-10)

The domain of \(f(x)=\sin^{-1}x\) is \([-1, 1]\) and the range is \([-π/2, π/2]\).

This is because the inverse sine function, \(sin^{-1}x\), is defined only for values of x between -1 and 1, resulting in a domain of \([-1, 1]\). The range of \(sin^{-1}x\) is the set of all possible output values, which are angles between \(-π/2\) and \(π/2\).



Therefore, the domain of \(f(x)=\sin^{-1}x\) is \(\boxed{[-1, 1]}\) and the range is \(\boxed{[-π/2, π/2]}\).

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$$x \frac{dy}{dx}\sin \left(\frac{y}{x}\right)=y \sin \left(\frac{y}{x}\right)-x$$

QUESTION 1:

Using the substitution $u = xy$, we have the following equation:

$$(x^2u+xu-u)dx + (x^2u-2x^2)dy = 0$$

Multiplying both sides by $\frac{dy}{dx}$, we get:

$$(x^2u+xu-u)\frac{dy}{dx}dx + (x^2u-2x^2)\frac{dy}{dx}dy = 0$$

Simplifying, we have:

$$(x^2u+xu-u)\frac{dy}{dx} + (x^2u-2x^2)dy = 0$$

Rearranging, we get:

$$x\frac{dy}{dx}(u \sin \left(\frac{y}{x}\right)) + y\sin \left(\frac{y}{x}\right) - x = 0$$

Which simplifies to:

$$x \frac{dy}{dx}\sin \left(\frac{y}{x}\right)=y \sin \left(\frac{y}{x}\right)-x$$

QUESTION 2:

The solution to the given differential equation is:

$$x \frac{dy}{dx}\sin \left(\frac{y}{x}\right)=y \sin \left(\frac{y}{x}\right)-x$$

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The probability that the product operates is 0.6055.

The probability can be meant as the chance something will be happened or not. The probability that the product operates is the probability that all 47 integrated circuits are not defective. Since the probability that any integrated circuit is defective is 0.01, the probability that it is not defective is 1 - 0.01 = 0.99. Since the integrated circuits are independent, the probability that all 47 are not defective is the product of the probability that each one is not defective. This is given by:

P(product operates) = 0.99⁴⁷ ≈ 0.6055

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Applying Trigonometric ratios, the missing sides, expressed as radicals in simplest form are:

26. u = 3√2; v = 3

27. a = 3/2; b = 3/2

28. x = 7√2; y = 7

29. u = 5√2; v = 5√2

Radicals are mathematical expressions that involve the square root or nth root of a number or a variable.

We can express each of the missing sides in radicals using the appropriate Trigonometry ratio in each case as follows:

26. Use sine ratio to find u:

sin 45 = 3/u

u = 3/sin 45

u = 3 / 1/√2 [sin 45 = 1/√2]

u = 3 * √2/1

u = 3√2

Use tangent ratio to find v:

tan 45 = 3/v

v = 3/tan 45

v = 3/1 [tan 45 = 1]

v = 3

27. Use sine ratio to find a:

sin 45 = a / 3√2

a = 3√2 * sin 45

a = 3√2 * 1/√2 [sin 45 = 1/√2]

a = 3/2

Use cosine ratio to find b:

cos 45 = b / 3√2

b = 3√2 * cos 45

b = 3√2 * 1/√2 [cos 45 = 1/√2]

b = 3/2

28. Use cosine ratio to find x:

cos 45 = 7 / x

x = 7/ cos 45

x = 7 / 1/√2 [cos 45 = 1/√2]

x = 7 * √2/1

x = 7√2

Use tangent ratio to find y:

tan 45 = y/7

y = 7 * tan 45

y = 7 * 1 [tan 45 = 1]

y = 7

29. Use sine ratio to find u:

sin 45 = u/10

u = 10 * sin 45

u = 10 * 1/√2 [sin 45 = 1/√2]

u = 10/√2

Rationalize

u = 5√2

Use cosine ratio to find v:

cos 45 = v/10

v = 10 * cos 45

v = 10 * 1/√2 [sin 45 = 1/√2]

v = 10/√2

Rationalize

v = 5√2

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For the given equation of an ellipse, Center: (1,3), Vertices: (6,3) and (-4,3), Co-Vertices: (1,5) and (1,1), and Foci: (5.58,3) and (-3.58,3).

The given equation is in the standard form of an ellipse with center at (h,k) with a horizontal major axis.

To find the center, we can simply look at the values of h and k in the equation. In this case, h = 1 and k = 3, so the center is (1,3).

To find the vertices, we need to find the values of a and b, which are the semi-major and semi-minor axes, respectively. In this case, a^2 = 25 and b^2 = 4, so a = 5 and b = 2.

The vertices are located at a distance of a units from the center along the major axis. Since the major axis is horizontal, the vertices are (1 + 5, 3) and (1 - 5, 3), or (6,3) and (-4,3).

The co-vertices are located at a distance of b units from the center along the minor axis. Since the minor axis is vertical, the co-vertices are (1, 3 + 2) and (1, 3 - 2), or (1,5) and (1,1).

To find the foci, we need to find the value of c, which is related to a and b by the equation c^2 = a^2 - b^2. In this case, c^2 = 25 - 4 = 21, so c ≈ 4.58.

The foci are located at a distance of c units from the center along the major axis. Since the major axis is horizontal, the foci are (1 + 4.58, 3) and (1 - 4.58, 3), or (5.58,3) and (-3.58,3).

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Answer:

The answer is 2/15

Step-by-step explanation:

2/5 × 1/3

The minima and minima of the function y=x^4-(32/3)x^3-2x^2+2x in the interval [0,3] are:
Maxima: x=2

Minima: None

The minima and maxima of a function are the lowest and highest points on the function within a given interval. To find these points, we need to take the derivative of the function and set it equal to zero to find the critical points. The critical points are where the function changes direction, and are potential minima or maxima. We can then use a conditional statement to determine if the critical points are within the given interval and if they are minima or maxima.

The derivative of the function is:
y'=4x^3-3(32/3)x^2-4x+2

Setting the derivative equal to zero, we get:
4x^3-32x^2-4x+2=0

We can use the Rational Root Theorem to find the potential rational roots of this equation. The potential rational roots are ±1, ±2, ±1/2, and ±1/4. Using synthetic division, we find that x=2 is a root. This gives us the factor (x-2), and we can use synthetic division again to find the other factors. The factored form of the equation is:
(x-2)(4x^2-12x+1)=0

Using the quadratic formula, we can find the other two roots:
x=3±√(3^2-4(4)(1))/2(4)
x=3±√(9-16)/8
x=3±√(-7)/8
x=3±i√7/8

The only real root is x=2, so this is the only critical point. We can use a conditional statement to determine if this critical point is within the given interval and if it is a minima or maxima. The critical point x=2 is within the interval [0,3], so we need to determine if it is a minima or maxima. We can do this by taking the second derivative of the function and plugging in the critical point:
y''=12x^2-6(32/3)x-4
y''(2)=12(2^2)-6(32/3)(2)-4
y''(2)=48-64-4
y''(2)=-20

Since the second derivative is negative at the critical point, this means that the critical point is a maxima. Therefore, the maxima of the function is at x=2.

In conclusion, the minima and maxima of the function y=x^4-(32/3)x^3-2x^2+2x in the interval [0,3] are:
- Maxima: x=2
- Minima: None

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v = -29/5 is the simplest form of the (4)/(v+5)=5.

To solve for v, we need to isolate the variable on one side of the equation. We can do this by following these steps:

Step 1: Multiply both sides of the equation by (v+5) to get rid of the fraction. This gives us:
-(4) = 5(v+5)

Step 2: Distribute the 5 on the right side of the equation:
-(4) = 5v + 25

Step 3: Subtract 25 from both sides of the equation:
-29 = 5v

Step 4: Divide both sides of the equation by 5 to solve for v:
v = -29/5

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Answer:

Step-by-step explanation:

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The graph of this equation will be a straight line with a slope of 1/2 and a y-intercept of -5/6.

To solve the given linear equation, we can use the method of isolating the variable on one side of the equation. Here are the steps to solve the equation:

Step 1: Start with the given equation: -3x + 6y = -5

Step 2: Isolate the variable on one side of the equation. We can do this by adding 3x to both sides of the equation:

-3x + 6y + 3x = -5 + 3x

Simplifying the equation gives us:

6y = 3x - 5

Step 3: Now, we can isolate the y variable by dividing both sides of the equation by 6:

(6y)/6 = (3x - 5)/6

Simplifying the equation gives us:

y = (3/6)x - (5/6)

Step 4: Simplify the fractions by reducing them:

y = (1/2)x - (5/6)

This is the final solution of the given linear equation. The graph of this equation will be a straight line with a slope of 1/2 and a y-intercept of -5/6.

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The time taken to empty one tank is 1000 seconds or 16.67 minutes. Therefore, the total time required to empty all three tanks one after another is 50 minutes (16.67 minutes x 3).

The volume of water that can be emptied through the hole at the bottom of each tank per second is 0.2 liters. Therefore, the time taken to empty 200 liters of water from one tank is:

200 liters ÷ 0.2 liters per second = 1000 seconds

Converting the time to minutes:

1000 seconds ÷ 60 seconds per minute = 16.67 minutes

So, it takes 16.67 minutes to empty one tank. Multiplying this by three gives us the total time required to empty all three tanks:

16.67 minutes x 3 = 50 minutes

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