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3 years ago

Please help!! I need to know this by today

Mathematics

2 answers:

DiKsa [7]3 years ago

6 0

The answer is 94.2 cm. multiply 3.14*10*6 divide 2.

liraira [26]3 years ago

6 0

So to find out the surface area of the bottom of the pencil holder, we can use the formula for the area of a circle.

A(bottom)=πr²
plug in the data we know
A(bottom)=π6²
A(bottom)=π*36
A(bottom)=36π cm²

For the surface area of the side, we need to find out the circumference of the bottom circle and then multiply that by the height.

A(side)=C*h

C=πd
C=π12
C=12π

A(side)=12π*10
A(side)=120π cm²

Total surface area=36π+120π=156π or 490.09cm²

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230.

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The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$