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Find x given sample of 266 people with 71% being affected

Find x given sample of 266 people with 71% being affected

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71 103 105 109 124 104 116 97 99 ... x x n s i 45 Sample Standard Deviation 1 ( )2 i ... Find the correlation coefficient X 10 15 20 30 40
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41 CFR Ch. 101 (7-1-00 Edition) Federal Property Management ...


These two publications must be used together to determine the latest version of any given ... Code users may find the text of ... A List of CFR Sections Affected ...
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261-270 - SECTION 5.4 SAMPLING DISTRIBUTIONS AND THE CENTRAL ...


Recall that you can use technology Interpretation So, about 71% of such ... the given sample mean ... 64, find the probability of a sample mean being less ...
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Introduction Math 10 - professormo.com


71 103 105 109 124 104 116 97 99 ... x x n s i 45 Sample Standard Deviation 1 ( )2 5 ... Find the correlation coefficient X 10 15 20 30 40
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Mark Scheme (Results) Summer 2014


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Calculate the linear correlation coefficient for given value


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Find x given sample of 266 people with 71% being affected

??????????? yu want 71% av 266 ????????? =0.71*266 =188.86 maebee round tu 189?????
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The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m.

The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m. A) What is the temperature of the greenhouse at 1:00 a.m. to the nearest Fahrenheit? (B) Find the average temperature between 7:00 p.m. and 7:00 a.m. to the nearest tenth of a degree Fahrenheit. (C) When the temperature of the greenhouse drops below 80 degreeso Fahrenheit, a heating system will automatically be turned on a maintain the temperature at minimum of 80 degrees Fahrenheit. At what values of the to the nearest tenth is the heating system turned on? (D) The cost of heating the greenhouse is $0.25 per hour for each degree. What is the total cost to the nearest dollar to heat the greenhouse from 7:00 p.m. and 7:00 a.m.?   The equation is: f(t) = 96 – 20sin(t/4), 0 <= t <= 12 (A)  At 1.00 a.m. t = 6 f(6) = 96 – 20.sin(6/4) = 96 – 20*0.99749 = 96 – 19.9499 f(6) = 76 ⁰F (B)  The average temperature would need to be worked out by sampling the temperature at different times throughout the night. Divide the temperature range into N equal intervals, giving N+1 sampling points. We would then have T1 = f(δt), T2 = f(2δt), T3 = f(3δt), ... ,Tn = f(nδt) Where δt = range/N = 12/N, and n = 0..N Giving Tn = 96 – 20.sin((12n/N)/4) = 96 – 20.sin(3n/N) Then Tav = (1/N)*sum(Tn, n = 0 .. N) i.e. Tav = (1/N)*sum(96 – 20.sin(3n/N), n = 0 .. N Tav = 96 – 20. (1/N)*sum(sin(3n/N), n = 0 .. N I used Maple to evaluate the above summation. The results are tabulated as follows.                                  Average Temperature Num Intervals            3       4        6       10       20      50     100    200 Tav over the range 83.39 83.01 82.78 82.69 82.69 82.71 82.72 82.73 As can be seen from the table the temperature is averaging out at:  Tav = 82.7 ⁰F (C)  T = f(t) = 96 – 20sin(t/4), 0 <= t <= 12 At T = 80 ⁰F,            96 – 20sin(t/4) = 80 20.sin(t/4) = 96 – 80 = 16 sin(t/4) = 0.8 t/4 = 0.927295 t = 3.70918 t = 3.7 (to nearest tenth) (D)  The temperature will (normally) drop to 80 ⁰F after t = 3.7 hours and rise again to 80 ⁰F when t = 12 – 3.7 = 8.3 hours. Heating system is turned on for 8.3 – 3.7 = 4.6 hours Cost of heating is 4.6*80*0.25 = 4.6*20 = 92 Cost = $92  
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given: cos theta = 3/5 theta is in Q4 FIND cos2theta, sin2theta,tan2theta

cos(2theta)=2cos^2(theta)-1; sin(2theta)=2sin(theta)cos(theta); tan(2theta)=sin(2theta)/cos(2theta). These are trigonometric identities, true for all values of theta. In Q4, draw a triangle ABC with the following coordinates for the vertices: B(0,0), A(3,0), C(0,-4). In this triangle, AB=3, BC=-4, AC=sqrt(3^2+4^2)=5 (hypotenuse) using Pythagoras' theorem. Angle BAC=theta. So cos(theta)=AB/AC=3/5; sin(theta)=BC/AC=-4/5, and sin(2theta)=2sin(theta)cos(theta)=2*3/5*(-4/5)=-24/25, cos(2theta)=2(3/5)^2-1=18/25-1=(18-25)/25=-7/25; tan(2theta)=-24/25÷(-7/25)=24/7. Draw a triangle DEF with E(0,0),  D(-7,0), F(0,-24) in Q3. Angle EDF=2theta, sin(2theta)=EF/DF=-24/25, cos(theta)=DE/DF=-7/25, tan(2theta)=EF/DE=-24/-7=24/7.
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one in four adults claims to have no trouble sleeping at night. you randomly select five adults and ask them if they have any trouble sleeping at night

  Probability p=1/4=0.25; 1-p=0.75 the probability of having trouble sleeping. Mean=5*0.25=1.25 average of people claiming to have no trouble sleeping. Variance=5*0.25*0.75=0.9375, so standard deviation is sqrt(0.9375)=0.97 approx. In the small sample the mean and standard deviation are respectively 1.25 and 0.97, which means that the expected results of a survey of 5 people lie in the range 1.25-0.97=0.28 and 1.25+0.97=2.22, i.e., between 0.28 and 2.22. Binomial distribution is based on the expansion of (0.25+0.75)^5 which has coefficients 1 5 10 10 5 1 (Pascal). The coefficients are applied to 0.25^(5-r)0.75^r, where r=0 to 5: for the purposes of comparison and for drawing a histogram, the values shown are multiples of 2^-10 (1/1024): All 5 claiming to have no trouble sleeping: 1 (r=0) 4 claiming and 1 not claiming: 5*(0.25^4)(0.75)=15 (r=1) 3 claiming and 2 not claiming: 90 (r=2) 2 claiming and 3 not claiming: 270 1 claiming and 4 not claiming: 405 All not claiming: 243 The magnitude of these values can be applied to the length of blocks in a histogram. The peak is clearly (5) above. The average is 1024/5=204.8.    
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A standard Hyperbola has its vertices ar 5,7 and 3,-1.Find its equation and its asymptotes

this equation expands toThe vertices share no common coordinate so the hyperbola cannot be orientated in the usual x-y form. The hyperbola must therefore be skew (tilted axis) and the line joining the vertices is sloping. This line is given by the equation y=((7+1)/(5-3))x+c where c is to be found. y=4x+c, and -1=12+c, plugging in (3,-1). c=-13. Therefore the axis of the hyperbola is y=4x-13. The centre or origin of the hyperbola lies midway between them at ((1/2)(5+3),(1/2)(7-1))=(4,3) and the equation of the conjugate axis is given by y=-x/4+k where k is found by plugging in the midpoint: 3=-4/4+k, so k=4 and y=4-x/4 is the equation of the conjugate axis. Let's define the axes of the hyperbola as X and Y instead of x and y. We can place the centre of the hyperbola at (0,0) in the X-Y plane so that in this plane the equation of the hyperbola is Y^2/A^2-X^2/B^2=1. The vertices are found by finding Y when X=0, so (0,A) and (0,-A) are the vertices. The asymptotes are given by (Y/A+X/B)(Y/A-X/B)=0. So Y=AX/B and Y=-AX/B are their equations. The distance between the vertices is 2A so we can work this out from the x-y coordinates of the vertices using Pythagoras' Theorem: 2A=sqrt((7+1)^2+(5-3)^2)=sqrt(68). A=sqrt(68)/2, A^2=68/4=17. Since the hyperbola is rectangular, B=A, and the equation of the hyperbola is Y^2-X^2=17. The asymptotes then become Y=X and Y=-X. Now we have to transform X-Y to x-y, so that the equation can be expressed in the x-y plane. We use geometry and trigonometry  for this. Let O'(4,3) be the location of the origin of the X-Y plane. Let P(x,y) be any point. In the X-Y plane this is P'(X,Y). Join O'P'=O'P. This line can be represented in both reference frames. O'P^2=X^2+Y^2=(x-4)^2+(y-3)^2 by Pythagoras. The gradient of the Y axis with respect to the x-y frame is 4, and we can represent this as tan(w) so w=tan^-1(4) or tan(w)=4; sin(w)=4/sqrt(17); cos(w)=1/sqrt(17). The angle z that O'P makes with the line y=3 is given by tan(z)=(y-3)/(x-4). sin(z)=(y-3)/O'P; cos(z)=(x-4)/O'P. Let v=90-w. sin(v+z)=Y/O'P; cos(v+z)=X/O'P. v+z=90-w+z=90-(w-z), so sin(v+z)=cos(w-z)=cos(w)cos(z)+sin(w)sin(z)=(x-4)/(O'Psqrt(17))+4(y-3)/(O'Psqrt(17))=(x+4y-16)/(O'Psqrt(17)). cos(v+z)=sin(w-z)=sin(w)cos(z)-cos(w)sin(z)=4(x-4)/(O'Psqrt(17))-(y-3)/(O'Psqrt(17))=(4x-y-13)/(O'Psqrt(17)). Y=O'P*sin(v+z)=(x+4y-16)/sqrt(17); X=O'P*cos(v+z)=(4x-y-13)/sqrt(17). So P maps to the x-y plane using these transformations. Y^2-X^2=17 becomes (x+4y-16)^2-(4x-y-13)^2=289, the equation of the hyperbola. This expands to x^2+8x(y-4)+16y^2-128y+256-(16x^2-8x(y+13)+y^2+26y+169)=289; -15x^2+16xy+15y^2+72x-154y-202=0. The asymptotes are x+4y-16+4x-y-13=0; 5x+3y-29=0 and x+4y-16-4x+y+13=0; 5y-3x-3=0.
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Discuss at least three applications of graph theory in the field of computer sciences?

Graphs are among the most ubiquitous models of both natural and human-made structures. They can be used to model many types of relations and process dynamics in physical, biological[1] and social systems. Many problems of practical interest can be represented by graphs. In computer science, graphs are used to represent networks of communication, data organization, computational devices, the flow of computation, etc. One practical example: The link structure of a website could be represented by a directed graph. The vertices are the web pages available at the website and a directed edge from page A to page B exists if and only if A contains a link to B. A similar approach can be taken to problems in travel, biology, computer chip design, and many other fields. The development of algorithms to handle graphs is therefore of major interest in computer science. There, the transformation of graphs is often formalized and represented by graph rewrite systems. They are either directly used or properties of the rewrite systems (e.g. confluence) are studied. Complementary to graph transformation systems focussing on rule-based in-memory manipulation of graphs are graph databases geared towards transaction-safe, persistent storing and querying of graph-structured data. Graph-theoretic methods, in various forms, have proven particularly useful in linguistics, since natural language often lends itself well to discrete structure. Traditionally, syntax and compositional semantics follow tree-based structures, whose expressive power lies in the Principle of Compositionality, modeled in a hierarchical graph. More contemporary approaches such as Head-driven phrase structure grammar (HPSG) model syntactic constructions via the unification of typed feature structures, which are directed acyclic graphs. Within lexical semantics, especially as applied to computers, modeling word meaning is easier when a given word is understood in terms of related words; semantic networks are therefore important in computational linguistics. Still other methods in phonology (e.g. Optimality Theory, which uses lattice graphs) and morphology (e.g. finite-state morphology, using finite-state transducers) are common in the analysis of language as a graph. Indeed, the usefulness of this area of mathematics to linguistics has borne organizations such as TextGraphs, as well as various 'Net' projects, such as WordNet, VerbNet, and others. Graph theory is also used to study molecules in chemistry and physics. In condensed matter physics, the three dimensional structure of complicated simulated atomic structures can be studied quantitatively by gathering statistics on graph-theoretic properties related to the topology of the atoms. For example, Franzblau's shortest-path (SP) rings. In chemistry a graph makes a natural model for a molecule, where vertices represent atoms and edges bonds. This approach is especially used in computer processing of molecular structures, ranging from chemical editors to database searching. In statistical physics, graphs can represent local connections between interacting parts of a system, as well as the dynamics of a physical process on such systems. Graph theory is also widely used in sociology as a way, for example, to measure actors' prestige or to explore diffusion mechanisms, notably through the use of social network analysis software. Under the umbrella of Social Network graphs there are many different types of graphs: Starting with the Acquaintanceship and Friendship Graphs, these graphs are useful for representing whether n people know each other. next there is the influence graph. This graph is used to model whether certain people can influence the behavior of others. Finally there's a collaboration graph which models whether two people work together in a particular way. The measure of an actors' prestige mentioned above is an example of this, other popular examples include the Erdős number and Six Degrees Of Separation Likewise, graph theory is useful in biology and conservation efforts where a vertex can represent regions where certain species exist (or habitats) and the edges represent migration paths, or movement between the regions. This information is important when looking at breeding patterns or tracking the spread of disease, parasites or how changes to the movement can affect other species. In mathematics, graphs are useful in geometry and certain parts of topology, e.g. Knot Theory. Algebraic graph theory has close links with group theory. A graph structure can be extended by assigning a weight to each edge of the graph. Graphs with weights, or weighted graphs, are used to represent structures in which pairwise connections have some numerical values. For example if a graph represents a road network, the weights could represent the length of each road. A digraph with weighted edges in the context of graph theory is called a network. Network analysis have many practical applications, for example, to model and analyze traffic networks. Applications of network analysis split broadly into three categories: First, analysis to determine structural properties of a network, such as the distribution of vertex degrees and the diameter of the graph. A vast number of graph measures exist, and the production of useful ones for various domains remains an active area of research. Second, analysis to find a measurable quantity within the network, for example, for a transportation network, the level of vehicular flow within any portion of it. Third, analysis of dynamical properties of networks.
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find the center of mass of the solid of uniform density biunded by the graphs of the equations

x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across. Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass. The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position. We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y. We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b). To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u). Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a). To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2). (a)+(b)=πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a)-(2/3)(a^2-y^2)^(3/2)=0. Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0. From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant. To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so!
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find the number of red, blue and yellow stars that are in the bag

R + B = Y B^2 = R B^2 + R = Y + 12 Y/B = R - 11 B^3 = 4Y - R Look at these this line: B^2 = R In the other lines replace all Rs with B^2: B^2 + B = Y B^2 + B^2 = Y + 12 Y/B = B^2 - 11 B^3 = 4Y - B^2 Let's re-word the third line: B^2 + B = Y B^2 + B^2 = Y + 12 Y = B^3 - 11B B^3 = 4Y - B^2 See the first line?  Let's replace all of the Ys with B^2 + B: B^2 + B^2 = B^2 + B + 12 B^2 + B = B^3 - 11B B^3 = 4(B^2 + B) - B^2 Simplify: 2B^2 = B^2 + B + 12 B^2 = B^3 - 12B B^3 = 4B^2 + 4B - B^2 More simplifying: B^2 = B + 12 B^2 = B^3 - 12B B^3 = 3B^2 + 4B Consider the first line: B^2 = B + 12 Move everything to one side: B^2 - B - 12 = 0 Same as: (B - 4) (B + 3) = 0 B has to equal 4 or -3. You can't have negative 3 of a thing, so B = 4 Consider the other two lines: B^2 = B^3 - 12B B^3 = 3B^2 + 4B Check those to make sure B = 4 works. 4^2 = 4^3 -12*4 >> 16 = 64 - 48 >> Yes. 4^3 = 3*4^2 + 4*4 >> 64 = 48 + 16 >> Yes. Remember how we said B^2 + B = Y ?  Now that we know B = 4, we can do this: 4^2 + 4 = Y 16 + 4 = Y Y = 20 Remember how we said B^2 = R ?  Now that we know B = 4, we can do this: 4^2 = R R = 16 Answer:  16 red, 4 blue, 20 yellow.
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Find the maximum and minimum values of the function

how do you find the maximum and minimum values of the function on the given interval: y=x-4x/x+1 Assuming your function to be: y = x - 4x/(x + 1), then maxima and minima are given when/where the slope is zero. The slope is dy/dx = 1 – {4/(x + 1) + 4x.(-1).(x + 1)^(-2)} dy/dx = 1 – 4/(x + 1) + 4x/(x + 1)^2 when dy/dx = 0, then (x + 1)^2 -4(x + 1) + 4x = 0 x^2 + 2x + 1 – 4x – 4 + 4x = 0 x^2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x = 1, -3 So, the maxima and minima of the function, y = x - 4x/(x + 1), are found when x = 1 and when x = -3 x = 1 y =  1 – 4/(1 + 1) = 1 – 4/2 = -1 Minimum value is at (1,-1) and equals -1. x = -3 y = -3 + 12/(-3 + 1) = -3 – 12/2 = -9 Maximum value is at (-3,-9) and equals -9.  
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area between curves Xto the fourth+ysquared=12 and x=y

The line x=y bisects the enclosed area of the other curve x^4+y^2=12. The x coords of the points of intersection are given by x^4+x^2-12=0=(x^2+4)(x^2-3), so x=±sqrt(3). The y coords are given by y=x, so the points are (sqrt(3),sqrt(3)) and (-sqrt(3),-sqrt(3)).  The picture shows how the area is bisected, so all we need to do is to find the area of the whole figure and halve it. But x^4+y^2=12 is symmetrical so to find the area of the whole figure, we just look at a quadrant and find its area, then we double the result to find the area enclosed by the line and the curve. When y=0 we have x intercept: x^4=12, so the area of a quadrant is integral of ydx between x=0 and x=12^(1/4), where y=sqrt(12-x^2) and the integrand is sqrt(12-x^2)dx. Let x=sqrt(12)sin(z); dx=sqrt(12)cos(z)dz, so the integrand becomes: sqrt(12-12sin^2(z))sqrt(12)cos(z)dz=12cos^2(z)dz. cos(2z)=2cos^2(z)-1, so 2cos^2(z)=1+cos(2z) and 12cos^2(z)=6+6cos(2z). But x=sqrt(12)sin(z), so if x=0, z=0 and if x=12^(1/4), sin(z)=12^(1/4)/12^(1/2)=12^(-1/4), z=sin^-1(12^(-1/4)). sin(z)=12^(-1/4); cos(z)=sqrt(1-12^(-1/2)). The result of integration is: 6z+3sin(2z) between the limits 0 and sin^-1(12^(-1/4)). Plugging in values we get the area of a quadrant is 3.4033 + 2.7189 = 6.1222 approx., making half the area of the figure12.2443 approx. It doesn't matter that the line x=y splits the area obliquely, it is still half the area.  
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