Determine if the statement is true or false. 66 = 7 (mod 12)

Mathematics

1 answer:

Y_Kistochka [10]1 year ago

6 0

Answer:

false

Step-by-step explanation:

66 mod 12

quotient 5

reminder 6

true statement

66 = 6 (mod 12)

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Given that f(x)=x^2+3x+6 and g(x)= x-3/2, solve for f(g(x)) when x=1

podryga [215]

If you would like to solve for f(g(x)) when x = 1, you can do this using the following steps:

<span>f(x) = x^2 - 3x + 6
g(x) = x - 3/2

f(g(x)) = f(</span>x - 3/2) = (x - 3/2)^2 - 3 * (x - 3/2) + 6
x = 1
f(g(1)) = f(1 - 3/2) = f(-1/2) = (-1/2)^2 - 3 * (-1/2) + 6 = 1/4 + 3/2 + 6 = 1/4 + 6/4 + 24/4 = 31/4 = 7 3/4

The correct result would be 7 3/4. add me as a friend

4 0

2 years ago

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

RSB [31]

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

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