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2 years ago

Find the volume of this cone. Use 3 for TT. 5in Tr2h 3 8in V V ~ [?]in3

Mathematics

1 answer:

LuckyWell [14K]2 years ago

5 0

Answer:

50 in ^3

Step-by-step explanation:

V = 1/3 pi r^2 h

We know that pi = 3

the diameter is 5 so the radius is 1/2 the diameter of 5/2 = 2.5

The height is 8

V = 1/3 ( 3) ( 2.5)^2 ( 8)

V = 50 in ^3

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Answer:

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3 years ago

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3 years ago

Isabella went into a movie theater and bought 3 drinks and 10 pretzels, costing a total of $54.50. Jason went into the same movi

ruslelena [56]

The price of 1 drink is $ 4 and price of 1 pretzel is $ 4.25

Solution:

Let d" be the price of 1 drink

Let "p" be the price of 1 pretzel

Given that, Isabella went into a movie theater and bought 3 drinks and 10 pretzels, costing a total of $54.50

Therefore, we can frame a equation as,

3 x price of 1 drink + 10 x price of 1 pretzel = 54.50

$3 \times d + 10 \times p = 54.50$

3d + 10p = 54.5 ---------- eqn 1

Jason went into the same movie theater and bought 9 drinks and 5 pretzels, costing a total of $57.25

9 x price of 1 drink + 5 x price of 1 pretzel = 57.25

$9 \times d + 5 \times p = 57.25$

9d + 5p = 57.25 ---------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 3

9d + 30p = 163.5 ----- eqn 3

Subtract eqn 2 from eqn 3

9d + 30p = 163.5

9d + 5p = 57.25

( - ) -----------------------

25p = 106.25

p = 4.25

Substitute p = 4.25 in eqn 1

3d + 10(4.25) = 54.5

3d + 42.5 = 54.5

3d = 54.5 - 42.5

3d = 12

d = 4

Thus price of 1 drink is $ 4 and price of 1 pretzel is $ 4.25

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3 years ago

O DATA ANALYSIS AND STATISTICS Outcomes and event probability A number cube is rolled three times. An outcome is represented by

Simora [160]

ANSWER

EXPLANATION

For each of the three events, there are 8 possible outcomes, so the probability of each event is,

$P(event)=\frac{favorable\text{ }outcomes}{possible\text{ }outcomes}=\frac{favorable\text{ }outcomes}{8}$

• Event A:, there must be only odd numbers in the last two rolls, which are either ,EOO or OOO,. This event has 2 favorable outcomes, ,so its probability is 1/4,.

• Event B,: this is the complementary event to event A. Here the last two rolls must be even numbers - both or only one of the last two: ,OOE, EEO, EOE, OEE, EEE, OEO,. This event has 6 favorable outcomes, so ,its probability is 3/4,.

• Event C,: this event is when the first and last rolls are even numbers in the same event. This is either ,EOE or EEE,. Since this event has 2 favorable outcomes, ,its probability is 1/4,.

5 0

1 year ago

We were never taught this in class, and I understand it, kind of... but i'm just confused with everything.

GaryK [48]

$\bf \cfrac{x^3y}{xy^5}\cdot \cfrac{x^2y^9}{x^8}\implies \cfrac{x^3x^2y^1y^9}{x^1x^8y^5}
\\ \quad \\
\textit{now, using exponent rules, same base, different exponent}
\\ \quad \\
\cfrac{x^{3+2}y^{1+9}}{x^{1+8}y^5}\implies \cfrac{x^5y^{10}}{x^9y^5}\\
----------------------------\\$ $\bf \textit{now, keep in mind that}
\\ \quad \\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad

\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
% negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 

\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}}
\\ \quad \\
thus\\
----------------------------\\
$ $\bf \cfrac{x^5y^{10}}{x^9y^5}\implies \cfrac{x^5y^{10}}{1}\cdot \cfrac{1}{x^9}\cdot \cfrac{1}{y^5}\implies x^5y^{10}\cdot x^{\boxed{-9}}\cdot y^{\boxed{-5}}
\\ \quad \\
x^5x^{-9}y^{10}y^{-5}\implies x^{5-9}y^{10-5}\implies x^{-4}y^5
\\ \quad \\
\cfrac{1}{x^{\boxed{4}}}\cdot \cfrac{y^5}{1}\implies \cfrac{y^5}{x^4}$

8 0

3 years ago