Respuesta :
Answer:
30 inches
Step-by-step explanation:
As, the stability ball of Ana is congruent to the stability ball of Fredrick. This implies that the volume of both the balls will be same.
So, Volume of Ana's ball = Volume of Fredrick's ball = 4500[tex]\pi[/tex] [tex]inch^{3}[/tex].
Now, the volume of a sphere is given by [tex]\frac{4\pi\times r^{3}}{3}[/tex], where 'r' is the radius.
Substituting the value of the volume in this formula, we will obtain the radius.
i.e. [tex]r^{3} = \frac{3V}{4\pi}[/tex]
i.e. [tex]r^{3} = 3375[/tex]
i.e. r = 15 inches
As, diameter is the double of the radius.
So, diameter = 2×radius = 2×15 = 30 inches.
Hence, the diameter of Fredrick's ball the 30 inches.
For Ana's fitness class, the diameter of Fredrick’s stability ball is mathematically given as
d= 30 inches.
What must the diameter of Fredrick’s stability ball be for it to be congruent to Ana’s?
Question Parameter(s):
Ana bought a stability ball that has a volume of 4,500π cubic inches.
Generally, the equation for the volume of a sphere is mathematically given as
[tex]V=\frac{4\pi* r^{3}}{3}[/tex]
Therefore
[tex]r^{3} = \frac{3V}{4\pi}[/tex]
r = 15 inches
In conclusion, diameter is
d= 2×radius
d= 2×15
d= 30 inches.
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