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mariarad [96]
3 years ago
11

Dan is planning a vacation and needs to purchase enough feed to fill the barrel in his pasture so that his cows will not be hung

ry while he is away. If the feed barrel is 7 feet high with a radius of 4 feet, what is the volume of feed the barrel can hold?
Mathematics
1 answer:
Aloiza [94]3 years ago
7 0
The formula for a cylinder is PieR^2H (Pi x Radius squared x Height).
Given the standard form of pi is 3.14, the height is 7, and the radius is 4, we can solve this easily.
The answer is 351.85 ft.

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Answer:

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f(x) = Price

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Use the side spittler theorem to find x given that PQ |BC
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18/12 = x/8

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5 0
3 years ago
Use the properties of exponents to rewrite the expression.<br> (-3yz)(-3yz)(-3yz)(-3yz)
Lera25 [3.4K]

Answer:

Answer is 81y^4z^4

Step-by-step explanation:

The expression is (-3yz) (-3yz) (-3yz) (-3yz).

Since all the four terms have power 1:

(-3yz)^1 (-3yz)^1 (-3yz)^1 (-3yz)^1

We know that we can add the powers if the terms have same base

So,

=(-3yz)^1+1+1+1

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3 0
3 years ago
The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand
Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})

P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})

P(X > 1560) = P(Z > \dfrac{110}{220})

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean \mu_x = np = 52 × 0.3085 = 16.042

The standard deviation =  \sqrt {n \time p (1-p)}

The standard deviation = \sqrt {52 \times 0.3085 (1-0.3085)}

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})

Pr ( Y > 20) = P(Z >1 .338)

From z tables

P(Y > 20) \simeq 0.0903

7 0
3 years ago
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