There are 453.592 grams in one pound. A semi-slick, Kevlar tire weighs about 520 grams. What is the tire’s weight in pounds?

Mathematics

2 answers:

marissa [1.9K]3 years ago

7 0

520/453.592 = <span>1.1464047 pounds</span>

olga nikolaevna [1]3 years ago

4 0

<h2>Answer:</h2>

The tire’s weight in pounds is:

1.146404 pounds.

<h2>Step-by-step explanation:</h2>

Conversion factor--

Conversion factor is used to change one unit into another by multiplying or dividing by a specific value.

The unit of conversion of weight from grams into pound is given by:

There are 453.592 grams in one pound.

i.e. 453.592 grams= 1 pounds

i.e. 1 grams= 1/453.592 pounds

The weight of tire is given by: 520 grams

Hence, the weight of tire in pounds is:

520/453.592 grams=1.148937 pounds.

Hence, the weight of tire in pounds is:

1.146404 pounds.

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A system has one each of two different types of components in joint operation. Let X and Y denote the random lengths of the live

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Answer:

The value of E(Y/X) is 2.

Step-by-step explanation:

As the complete question is not given, thus the complete question is found online and is attached herewith.

So the joint density function is given as

$f(x,y)=\left \{ {{\dfrac{x}{8}e^{-\dfrac{x+y}{2}} \,\,\,\,0 \leq x,0\leq y \atop {0}} \right.$

So the marginal function for X is given as

$f_x=\int\limits^{\infty}_0 {f(x,y)} \, dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x+y}{2} }dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x}{2}}e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{8}e^{-\frac{x}{2}}\int\limits^{\infty}_0e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{4}e^{-\frac{x}{2}}$

Now

$f(Y/X)=\dfrac{f(X,Y)}{f(X)}\\f(Y/X)=\dfrac{\dfrac{x}{8}e^{-\frac{x+y}{2} }}{\dfrac{x}{4}e^{-\frac{x}{2}}}\\f(Y/X)=\dfrac{1}{2}e^{-\frac{y}{2} }$

Now the value of E(Y/X) is given as

$E(Y/X)=\int\limits^{\infty}_0 {yf_{Y/X}} \, dy \\E(Y/X)=\int\limits^{\infty}_0 {y\dfrac{1}{2}e^{-\frac{y}{2} } }\, dy\\E(Y/X)=\dfrac{1}{2}\dfrac{\sqrt{2}}{(\dfrac{1}{2})^2}=2$