A set of average city temperatures in May are normally distributed with a mean of 20.66 ^\circ \text{C}20.66
1 answer:
Answer:
0.9962
Step-by-step explanation:
The z score shows the number of standard deviations by which the raw score is above or below the mean. If the raw score is above the mean then the z score is positive but if the raw score is less than the mean then the raw score is negative. The z score is given by:
μ = 20.66°C, σ = 2°C
For x < Singapore temperature, i.e. x < 26
From the normal distribution table P(x < 26) = P(z < 2.67) = 0.9962
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Unfortunately the portion of the graph shown does not show the locations of points A and B, nor does it show sufficient parts of the curve.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.