In getting the average of all the measurement you give you need first to sum all the measurements and also make sure that you do it properly. Then divide the sum with the number of measurements. By calculating the answer is 10.288
Answer:
see the explanation
Step-by-step explanation:
The picture of the question in the attached figure
we know that
1) 
----> is given
2) 
----> by central angle
3) ![m\angle RWT=\frac{1}{2}[arc\ ST]](https://tex.z-dn.net/?f=m%5Cangle%20RWT%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20ST%5D)
----> by inscribed angle
![m\angle RWT=\frac{1}{2}[30^o]=15^o](https://tex.z-dn.net/?f=m%5Cangle%20RWT%3D%5Cfrac%7B1%7D%7B2%7D%5B30%5Eo%5D%3D15%5Eo)
4) 
----> Triangle RTW is an isosceles triangle (RT=RW=radius of circle R)
therefore
100(47000-32000)/32000=47% (to nearest percent)
Answer: First Option: Robin's graph is a reflection of Giselle´s graph over the x-axis.
A reflection of the graph f(x)=y is g(x)=-y. In this case y=x^2, then:
g(x)=-x^2 that corresponds with the Robin's function
Answer:
The volume of foam needed to fill the box is approximately 2926.1 cubic inches.
Step-by-step explanation:
To calculate the amount of foaming that is needed to fill the rest of the box we first need to calculate the volume of the box and the volume of the ball. Since the box is cubic it's volume is given by the formula below, while the formula for the basketball, a sphere, is also shown.
Vcube = a³
Vsphere = (4*pi*r³)/3
Where a is the side of the box and r is the radius of the box. The radius is half of the diameter. Applying the data from the problem to the expressions, we have:
Vcube = 15³ = 3375 cubic inches
Vsphere = (4*pi*(9.5/2)³)/3 = 448.921
The volume of foam there is needed to complete the box is the subtraction between the two volumes above:
Vfoam = Vcube - Vsphere = 3375 - 448.921 = 2926.079 cubic inches
The volume of foam needed to fill the box is approximately 2926.1 cubic inches.
