Question

please answer.........

Answer 1

(1)

Given that volume of the hemisphere = 1152pi cm^3.

We know that Volume of the hemisphere = 2/3 pir^3

2/3 pir^3 = 1152pi

2pir^3 = 1152 * 3pi 

2pir^3 = 3456pir^3

r^3 = 3456/2

r^3 = 1728

r = 12cm.

The curved surface area of the hemisphere = 2pir^2

                                                                          = 2 * pi * (12)^2

                                                                          = 2 * pi * 144

                                                                          = 288picm^3.


(2).

The class marks are equally spaced.

class size = 114 - 104

                  = 10.

Therefore, h = 10.

So, We know that Lower limit = a - h/2 = 104 - 10/2 = 104 - 5 = 99.

So, We know that Upper limit = a + h/2 = 104 + 10/2 = 104 + 5 = 109.

Hence, the new class intervals are 99-109,109-119,119-129,129-139,139-149,149-159,159-169.


Hope this helps!