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2 years ago

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Use the following table of Registered Voters at Kensington College, Registered Voters at Kensington College 2009 2010 640 790 De

mocrats Republicans 960 410 Independents 320 300 11) What is the probability that a voter registered in Kensington College in 2009 was not a Republican?

1 answer:

sasho [114]2 years ago

7 0

Step-by-step explanation:

remember, a probability is ultimately always

desired cases / total possible cases

what are the total possible cases for our question here ?

all registered voters in 2009 :

640 + 960 + 320 = 1920

the desired cases (not republican) out of all these 1920 possibilities are

640 + 320 = 960

so, the probability for a registered voter to not be republican was

960 / 1920 = 1/2 = 0.5

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1. 170.083 in³

2. 126π in³

3. 92.106 m³

4. 2412.74 in³

5. 612π m³ and 1922 m³

Step-by-step explanation:

1.

Cylinder:

$V = \pi r^{2}h$ *Plug in numbers*

$(3.14)(2.5)^{2}(7)$ *Square 2.5*

$(3.14)(6.25)(7)$ *Solve*

≈ $137.375in^{3}$

Sphere:

$V = \frac{4}{3}\pi r^{3}$ *Plug in numbers*

$\frac{4}{3} (3.14)(2.5)^{3}$ *Cube 2.5*

$\frac{\frac{4}{3}(3.14)(15.625)}{2}$ *Divide by 2 and Solve*

≈ $32.7083 in^{3}$

Add both volumes

$137.375 + 32.7083$ ≈ $170.083in^{3}$

2.

Cylinder:

$V = \pi r^{2}h$ *Plug in numbers*

$\pi (3)^{2}(10)$ *Square 3*

$\pi (9)(10)$ *Multiply*

$90\pi$

Sphere:

$V = \frac{4}{3}$ π $r^{3}$ *Plug in numbers*

$\frac{4}{3}\pi (3)^{3}$ *Cube 3*

$\frac{4}{3} \pi (27)$ *Multiply*

$36\pi$

Add both Volumes to get total

$90\pi + 36\pi = 126in^{3}$

3.

Sphere:

$V = \frac{4}{3}\pi r^{3}$ *Plug in numbers*

$\frac{4}{3} (3.14)(3)^{3}$ *Cube 3*

$\frac{4}{3} (3.14)(27)$ *Multiply*

$113.04m^{3}$

Cone:

$V = \frac{\pi r^{2}h}{3}$ *Plug in numbers*

$\frac{(3.14)(2)^{2}(5)}{3}$ *Square 2*

$\frac{(3.14)(4)(5)}{3}$ *Solve*

$20.93m^{3}$

Subtract the volumes to get the volume of the blue area

$113.04 - 20.93 = 92.106m^{3}$

4.

Sphere:

$V = \frac{4}{3} \pi r^{3}$ *Plug in numbers*

$\frac{4}{3}\pi (8)^{3}$ *Cube 8*

$\\\frac{4}{3}\pi (512)$ *Multiply*

$\\\\\pi (682.6)$ *Solve*

$2133.66in^{3}$ *Divide by 2 since it's a hemisphere*

Cone:

$V = \frac{\pi r^{2}h}{3}$ *Plug in numbers*

$\frac{\pi (8)^{2}(20)}{3}$ *Square 8*

$\frac{\pi (64)(20)}{3}$ *Multiply and Divide*

$1340.41 in^{3}$

Add both volumes

$1072.33 + 1340.41 = 2412.74in^{3}$

5.

Cylinder:

$V = \pi r^{2}h$ *Plug in numbers*

$\pi (6)^{2}(16)$ *Square 6*

$\pi (36)(16)$ *Multiply*

$576\pi$

Cone:

$V = \frac{\pi r^{2}h}{3}$ *Plug in numbers*

$\frac{\pi (6)^{2}3}{3}$ *Square 6*

$36\pi$

Add both volumes

$576\pi + 36\pi = 612\pi m^{3}$

Alternative: *Multiply π*

$1922m^{3}$

4 0

3 years ago

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