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pychu [463]
3 years ago
14

The letters of the alphabet are written on flashcards and dropped into a bag. A flashcard is randomly selected, kept out of the

bag, and then another flashcard is randomly selected. Which product gives the probability that both flashcards have a vowel (a, e, i, o, or u) written on them?
Mathematics
2 answers:
ladessa [460]3 years ago
6 0
5:26 Hope I helped :)
ArbitrLikvidat [17]3 years ago
5 0
For it being a strict vowel, there is a 5:26 ratio
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<h3>Given:</h3><h3>Large cone:</h3>
  • r= 8 cm
  • h= 20 cm
<h3>Small cone:</h3>
  • r= 4 cm
  • h= 10cm
<h3>Note that:</h3>
  • r: radius
  • h: height
<h3>To find:</h3>
  • The volume of the frustum of the given cone.
<h3>Solution:</h3>
  • Frustum is a part of a cone formed by cutting off the top by a parallel plane.

\large\boxed{Formula: V= \frac{1}{3}\pi{r}^{2}h}

Let's solve!

First, let's find the volume of the smaller cone.

Substitute the values according to the

formula.

V= \frac{1}{3}×\pi×{4}^{2}×10

V= 167.5516082 \: {cm}^{3}

Now, we can round off to the nearest hundredth.

The value in the thousandths place is smaller than 5 so we won't have to round up.

\boxed{V= 167.55 \: {cm}^{3}}

Next, let's find the volume of the bigger cone.

Substitute the values according to the formula.

V= \frac{1}{3}×\pi×{8}^{2}×20

V= 1340.412866 \: {cm}^{3}

Now, we can round off to the nearest hundredth.

The value in thousandths place is smaller than 5 so we won't have to round up.

\boxed{V=1340.41 \: {cm}^{3}}

Now, we can find the volume of the frustum.

We'll have to minus the volume of the smaller cone from the bigger cone.

V= 1340.41-167.55

\large\boxed{V= 1172.86 \: {cm}^{3}}

<u>Hence, the volume of the frustum is 1172.86 cubic centimeters.</u>



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