let's firstly convert the mixed fractions to improper fractions and them sum them up.
[tex]\stackrel{mixed}{7\frac{5}{12}}\implies \cfrac{7\cdot 12+5}{12}\implies \stackrel{improper}{\cfrac{89}{12}}~\hfill \stackrel{mixed}{11\frac{2}{3}} \implies \cfrac{11\cdot 3+2}{3} \implies \stackrel{improper}{\cfrac{35}{3}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{89}{12}+\cfrac{35}{3}\implies \cfrac{(1)89~~ + ~~(4)35}{\underset{\textit{using this LCD}}{12}}\implies \cfrac{89+140}{12}\implies \cfrac{229}{12}\implies {\Large \begin{array}{llll} 19\frac{1}{12} \end{array}}[/tex]
Isabel has (9)/(10) kilogram of oranges. She ate (2)/(10) kilogram. How much of a kilogram remains
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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The water level of a lake was 20 feet and increases 10% each week during winter rainstorms
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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Mary has 7.2 yards of cloth. She uses 3.25 yards. She claims that she has 4.5 yards of cloth left. Do you agree?
How do u do this question
yes or no. Mary has 3.95,4.5, or 10.45 yards left.
Answer:
Step-by-step explanation:
jkhnb and then efhrfehnb c3454k3
Find the missing side links, leave your answers as radicals in simplest form
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
What are Radicals?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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E. 4(x - 1)^2 + 25(y - 3)^2 = 100 Center: ____ Vertices: ____ Co-Vertices: ____
Foci: ____
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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Fill in each blank so that the resulting statement is true. The domain of \( f(x)=\sin ^{-1} x \) is and the range is The domain of \( f(x)=\sin ^{-1} x \) is and the range is
Fill in each blank so t
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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16) if a cylinder has a volume of 1884 cubic centimeters and a height of six cm, explain how to find the radius of the base circle. (Hint: work backwards and remember that r-squared means r x r )
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
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 98 degrees occurs at 6 PM and the average temperature for the day is 85 degrees. Find the temperature, to the nearest degree, at 7 AM. degrees
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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\[ \frac{1+\frac{4}{a}+\frac{4}{a^{2}}}{1-\frac{8}{a}-\frac{20}{a^{2}}} \] a. \( \frac{a+2}{a-10} \) b. \( \frac{a+2}{a^{2}} \) c. \( -\frac{a-10}{a+2} \) d. \( \frac{4}{a-10} \)
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)
QUESTION 1 \[ \left(x^{2} y+x y-y\right) d x+\left(x^{2} y-2 x^{2}\right) d y=0 \] QUESTION 2 \[ x \frac{d y}{d x} \sin \left(\frac{y}{x}\right)=y \sin \left(\frac{y}{x}\right)-x \text { using the sub
$$x \frac{dy}{dx}\sin \left(\frac{y}{x}\right)=y \sin \left(\frac{y}{x}\right)-x$$
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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Knowledge Check Solve for v. -(4)/(v+5)=5 Simplify your answer as much as possible.
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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sin
^-1 (sin A) ≠ A
A. implies that A is not in the domain
b. requires that A = 0
c. is not possible because arcsin reverses sin
d. happens when A is not in [-pi/2, pi/2]
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 scale factor of two similar cylinders is 5:2. The volume of the smaller cylinder is 28 m3. What is the volume of the larger cylinder?
70 m3
175 m3
350 m3
700 m3
437.5 m3
Using the scale factor, we found the volume of the larger cylinder as 70 m³.
What is a cylinder?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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57. FO correct to two decimal places. (c) (d) Rectangle A X Rectangle B x+3 Two rectangles, A and B, each have an area of 11 cm². The length of rectangle A is x cm. The length of rectangle B is (x+3) cm. (a) Find, in terms of x, an expression for the width of i. Rectangle A Rectangle B (b) Given that the width of rectangle A is 2 cm greater than the width of rectangle B, form an equation in x and show that it simplifies to 2x² +6x-33=0. Solve the equation 2x² +6x-33=0, giving both answers correct to 2 decimal places. Hence find the width of rectangle B. Two docimal places N98/2/2b J99/2/7 N99/2/2c
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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What is a possible measure of an angle that has a terminal side in Quadrant III and makes a 60º angle with the x-axis?
A. 120°
B. 210°
C. 150°
D. 240°
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°
Solve for x in the equation x squared minus 12 x + 59 = 0.
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) }
Consider the telephone tariff (charge) schedule by the telephone company: a. First examine the cost of an average phone call with the following charge schedule: each call is charged r for the initial period of length / (or fraction thereof). For call length exceeding !, you are charged incrementally for each subsequent period of length S(or fraction thereof), at a unit cost of p. What is the average cost of a call? (Note that it is a standard practice in the industry to assume that call duration is exponentially distributed.) b. If the cost of a call is $1 per minute, and your call is always pro-rated (as fraction of a minute) and the average call length is one minute, then the average cost of a call is just $1. If your call is rounded up to the next minute (I = S = 1), use the formula derived above to find the average cost. (Are you surprised?) c. If I = S = 0.1 (i.e., rounded up to the next 6 seconds), compute the average cost again.
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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\[ \begin{array}{c} A=\left[\begin{array}{lll} -5 & 1 & -7 \end{array}\right] \\ B=\left[\begin{array}{llll} -8 & 7 & 5 & -5 \end{array}\right] \\ C=\left[\begin{array}{ll} -4 & -2 \end{array}\right]
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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4. A surveying team needs to measure the distance across the lake. They make the
measurements shown along the ground. What is the distance across the lake?
The distance across the lake is 120 feet.
What is Right Angled Triangle?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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Solve for x by converting the logarithmic equation to exponential form. log_(3)(x)=-2 -6 -8 (1)/(9) -(1)/(9)
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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lowing linear equation. -3x+6y=-5 aph. Any lines or curves will be drawn once all required points are plotted. Enable
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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2) Consider the model m Ji = Σβ,tij + u; j=1 where there is no constant in the equation. Derive the properties of R2 for this model.
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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What is the hcf of 124
Answer:
2
Step-by-step explanation:
I need help as soon as possible!
Answer:
The answer is 2/15
Step-by-step explanation:
2/5 × 1/3
Please i need this by tnight
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
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}. Show that these two bases span the same subspace of R3
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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There are 3 tanks each filled with 200 liters of water. There is a hole at the bottom of each tank through which water can be let out at a constant rate of 0. 2 liters per second. A person wants to empty the tanks one after another which means the hole of the next tank will be opened only after the previous tank is empty. Not including the time taken to open the holes of the tanks, what is the total time, in minutes, required to empty all three tanks?
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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The school packs one lunch based on each of these choices. If Dr. Higgins wants a "Turkey, Water and Cookie" or a " Turkey, Water and Brownie" (I could eat either lunch) - what is the probability that Dr. Higgins randomly picks up one or the other of his Favorite Lunches?
The probability of selecting either lunch is 50%, as both lunches are equally likely to be chosen. This is because both lunches have the same ingredients, with the only difference being the dessert item.
What is probability?Probability is a measure of the likelihood of a certain event or outcome occurring. It is expressed as a number between 0 and 1, where 0 indicates that the event or outcome is impossible and 1 indicates that the event or outcome is certain to occur. Probability is an important concept in mathematics and statistics, and it is widely used in fields such as finance, science, engineering, and gaming.
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Simplify the following rational expression: \( \frac{500 x^{3}-108}{40 x^{2}+56 x-48} \) Leave the numerator and denominator in your final answer in fac torm.
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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Progress An electronic product contains 47 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the integrated circuits are independent. The product operates only if there are no defective integrated circuits. What is the probability that the product operates? Round your answer to four decimal places (e.g. 98.7654). The probability is
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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