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ElenaW [278]
3 years ago
7

An oil tank has to be drained for maintenance. The tank is shaped like a cylinder that is 5 ft long with a diameter of 1.8 ft .

Suppose oil is drained at a rate of 2.1 ft3 per minute. If the tank starts completely full, how many minutes will it take to empty the tank? Use the value 3.14 for π , and round your answer to the nearest minute. Do not round any intermediate computations.
Mathematics
1 answer:
dangina [55]3 years ago
3 0
\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad 
\begin{cases}
r=radius=\frac{diameter}{2}\\
h=height\\
----------\\
diameter=1.8\\
r=\frac{1.8}{2}=0.9\\
h=5
\end{cases}\implies V=\pi \cdot 0.9^2\cdot 5\\\\
-------------------------------\\\\
\textit{is draining at }\boxed{2.1}\ \frac{ft^3}{min}\textit{ , it has a total of }\boxed{\pi \cdot 0.9^2\cdot 5} \ ft^3
\\\\\\
\textit{it'll take }\cfrac{\pi \cdot 0.9^2\cdot 5}{2.1}\ minutes
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What perfect square that is between 1 and 100 has 27 as one of its factors?
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Answer:

81

Step-by-step explanation:

Its a perfect square and 27*3=81

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Based on the 2009 season, the Texas Rangers have a winning percentage of .533. Use the binomial model to find the probability th
Naddik [55]

Answer:

0.128

Step-by-step explanation:

We know the probability for any event X is given by,

P(X=x)=\binom{n}{x}\times p^{n-x}\times q^{x},

where p is the probability of success and q is the probability of failure.

Here, we are given that p = 0.533.

Since, we have that q = 1 - p

i.e. q = 1 - 0.533

i.e. q = 0.467

It is required to find the probability of 4 wins in the next 5 games i.e. P(X=4) when n = 5.

Substituting the values in the above formula, we get,

P(X=4)=\binom{5}{4}\times 0.533^{5-4}\times 0.467^{4}

i.e. P(X=4)=5 \times 0.533 \times 0.048

i.e. P(X=4)=5 \times 0.533 \times 0.048

i.e. i.e. P(X=4)=0.128

Hence, the probability of 4 wins in the next 5 games is 0.128.

8 0
3 years ago
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2 years ago
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What is the midpoint of EC ?
sladkih [1.3K]

<u>Given</u>:

Given that the graph OACE.

The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)

We need to determine the midpoint of EC.

<u>Midpoint of EC:</u>

The midpoint of EC can be determined using the formula,

Midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Substituting the coordinates E(2t,0) and C(2p, 2r), we get;

Midpoint=(\frac{2t+2p}{2},\frac{0+2r}{2})

Simplifying, we get;

Midpoint=(\frac{2(t+p)}{2},\frac{2r}{2})

Dividing, we get;

Midpoint=(t+p,r)

Thus, the midpoint of EC is (t + p, r)

Hence, Option A is the correct answer.

4 0
3 years ago
(Uniform) Suppose X follows a continuous uniform distribution from 4 to 11. (a) Write down the PDF of X. (b) Find P(X ≤ 7). Roun
love history [14]

Answer:

a) f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11

b) P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286

c) P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857

d) P(X >5 | X \leq 7)

And we can find this probability with this formula from the Bayes theorem:

P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5

Step-by-step explanation:

For this case we assume that the random variable X follows this distribution:

X \sim Unif (a=4, b =11)

Part a

The probability density function is given by the following expression:

f(x) = \frac{1}{b-a} , a \leq x \leq b

f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11

Part b

We want this probability:

P(X \leq 7)

And we can use the cumulative distribution function given by:

F(x) = \frac{x-a}{b-a}= \frac{x-4}{11-4}

And replacing we got:

P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286

Part c

We want this probability:

P(5 < X \leq 7)

And we can use the CDF again and we have:

P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857

Part d

We want this conditional probabilty:

P(X >5 | X \leq 7)

And we can find this probability with this formula from the Bayes theorem:

P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5

7 0
3 years ago
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