mathsss (1 Viewer)

[stressedadfff]

hey guys does anyone know the solution to the equations?





i also really need help with the following pls

 

[icycledough]

For the first question, a PDF has 2 criteria: f(x) is greater or equal than 0 for all x and that the area under the curve is 1. The first one is quite evident from the graph. The second criteria just requires the area under the triangle, which is 1/2 x 4 x 1/2 = 1, as needed. For piecewise function, consider 2 parts of the graph, i.e the 2 lines.

You could have something like for x in between and including 0 and 2, the equation is y = x/4, and for x in between 2 and 4 (inclusive of 4), an equation like y = -x/4 + 1

 

[icycledough]

For the second question, part 1, I've attached the answer (because I can't type it out)

 

[stressedadfff]

For the first question, a PDF has 2 criteria: f(x) is greater or equal than 0 for all x and that the area under the curve is 1. The first one is quite evident from the graph. The second criteria just requires the area under the triangle, which is 1/2 x 4 x 1/2 = 1, as needed. For piecewise function, consider 2 parts of the graph, i.e the 2 lines.

You could have something like for x in between and including 0 and 2, the equation is y = x/4, and for x in between 2 and 4 (inclusive of 4), an equation like y = -x/4 + 1

i see, could you explain how you got the constant 1 pls

 

[icycledough]

For part 2, if a = pi / 3, and we know r = 63 / a (from what I circled at the top above), just sub it in and you should get the radius

 

[stressedadfff]

For the second question, part 1, I've attached the answer (because I can't type it out)View attachment 32423

thank youuuuuu!! <3

 

[icycledough]

i see, could you explain how you got the constant 1 pls

As all the possible probabilities summed up add up to 1, that's how you get the constant 1. In terms of the diagram, it's just finding the area under the graph. As it's a triangle, you just go bh/2 (where b = base = 4, h = height = 1/2), then you just sub it in

 

[CM_Tutor]

Note that, in @icycledough's solution, the result

​

is used, and it isn't in the Advanced syllabus, only in MX1.

However, you can add in the bisector of the angle , which will also perpendicularly bisect the chord (104 m) to give a right-angled triangle in which

​

This also saves the need to use the Sine Rule or to recognise that

 

[specificagent1]

is circle geo in adv? the second question there?

 

[icycledough]

Note that, in @icycledough's solution, the result

​

is used, and it isn't in the Advanced syllabus, only in MX1.

However, you can add in the bisector of the angle , which will also perpendicularly bisect the chord (104 m) to give a right-angled triangle in which

​

This also saves the need to use the Sine Rule or to recognise that

Oh, I didn't realise the double sin angle result wasn't in the Advanced syllabus. I just couldn't see another way to go about it without using that.

 

[CM_Tutor]

is circle geo in adv? the second question there?

Circle Geometry, no... but radians and trigonometry, yes... and it's only trig and triangle geometry that you need.