Acompany that manufactures storage bins for grains made a drawing of a silo. the silo has a conical base, as shown below: the figure shows a silo shaped as a closed cylinder with a conical end. the diameter of the silo is 4 ft, the length of the cylindrical part is 6 ft, and the entire length of the silo is 10.5 ft. which of the following could be used to calculate the total volume of grains that can be stored in the silo?

Given:

Cylinder part

D = 4 ft

H = 6 ft

Cone part

H = 10.5 – 6 = 4.5ft

D = 4ft

Required:

Volume of silo

Solution:
V of cylinder = pi*r2*H

V of cylinder = pi*(4/2)^2*(6)

V of cylinder = 75.4 ft3

V of cone = pi*r2*H/3

V of cone = pi*(4/2)^2*(4.5/3)

V of cone = 18.85ft3

Total volume = 94.25ft3

it costs $20 to interview each person for a survey. find the cost to obtain a +- 3% margin of error.

step-by-step explanation:

copy and paste, put it in google

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841