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3 years ago

The probability that an event will occur is 2/3 what is the best answer of this event likely occurring

Mathematics

1 answer:

Oduvanchick [21]3 years ago

6 0

It's likely that the event will occur.

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The gas tank on the back of a tanker truck can be equated to a cylinder with a diameter of 8 feet and a length of 19 feet. A gal

Leokris [45]

9514 1404 393

Answer:

driver's tank: 30,427 lb
farmer's tank: 12,811 lb

Step-by-step explanation:

The formula for the volume of a cylinder is ...

V = πr^2·h . . . radius r, height h

The radius of the driver's tank is half its diameter, so is (8 ft)/2 = 4 ft. Then the volume of that tank is ...

V = π(4 ft)^2·(19 ft) = 304π ft^3

Each cubic foot of gasoline has a mass of ...

(1728 in^3/ft^3)(0.0262 lb/in^3) = 45.2736 lb/ft^3

Then the total mass in the driver's full tank is ...

(304π ft^3)(45.2736 lb/ft^3) ≈ 43,238.3 lb

The farmer's tank is a scaled-down version of the driver's tank. It's volume will be scaled by the cube of the linear scale factor, so will be (2/3)^3 = 8/27 of the volume of the driver's tank.

The farmer's tank will hold a mass of (43,238.3 lb)(8/27) ≈ 12,811 lb.

The amount remaining in the driver's tank is 43,238 -12,811 = 30,427 lb.

6 0

3 years ago

An artist designed a metal structure in the shape of a pyramid with dimensions as shown.

amid [387]

Answer:

C. 504

Step-by-step explanation:

Got it correct on the assignment

8 0

3 years ago

Prove 2^n > n for all n equal to or greater than 1. I mostly need help with how to solve the problem when it is greater than

noname [10]

If $n$ is an integer, you can use induction. First show the inequality holds for $n=1$ . You have $2^1=2>1$ , which is true.

Now assume this holds in general for $n=k$ , i.e. that $2^k>k$ . We want to prove the statement then must hold for $n=k+1$ .

Because $2^k>k$ , you have

$2^{k+1}=2\times2^k>2k$

and this must be greater than $k+1$ for the statement to be true, so we require

$2k>k+1$

for $k>1$ . Well this is obviously true, because solving the inequality gives $3k>1\implies k>\dfrac13$ . So you're done.

If you $n$ is any real number, you can use derivatives to show that $2^n$ increases monotonically and faster than $n$ .

7 0

3 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

2 years ago

If you graph the system of inequalities given below, which points would lie in the solution set? Select all the correct answers.

vivado [14]

Its just a matter of subbing...but keep in mind, for it to be a solution, it has to satisfy all 3 inequalities

x - y > = -4
2x - y < = 5
2y + x > 1

solutions are : (-1,3) and (3,5)

5 0

3 years ago

Read 2 more answers