Guide :

# 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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## Suggested Questions And Answer :

### 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?????

### 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

### 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.

### 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.

### 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.

### 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.

### 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.

### 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.