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2 years ago

Help please, try to explain it in steps for me if u can't that's fine, ty!

Mathematics

1 answer:

Iteru [2.4K]2 years ago

3 0

Answer:

Step-by-step explanation:

We are given that Expression One is:

$2\cdot 10^x$

And Expression Two is:

$4\cdot 10^{x+y}$

We are given that the value of Expression Two is 20,000 times greater than the value of Expression One. And we want to find the value of y. In other words:

$4\cdot 10^{x+y}=20000(2\cdot 10^x)$

We can divide both sides by four:

$10^{x+y}=10000\cdot 10^x$

Recall that:

$x^a\cdot x^b=x^{a+b}$

Therefore:

$10^x\cdot 10^y=10000\cdot 10^x$

Divide both sides by 10ˣ:

$10^y=10000$

Solve for y:

$\displaystyle y = \log_{10}10000 = 4$

Our answer is B.

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The table shows the outputs, y, for different inputs, x: Input (x) 3 7 11 15 Output (y) 4 6 8 10 Part A: Do the data in this tab

stiv31 [10]

Answer:

Step-by-step explanation:

x (3,7,11,15)

y (4,6,8,10)

A. This IS a function because u have no repeating x values

B. Data in table :

when x = 11, y = 8

relation : f(x) = 5x - 21....when x = 11

f(11) = 5(11) - 21

f(11) = 55 - 21

f(11) = 34.....so when x = 11, y = 34

therefore, the relation f(x) = 5x - 21 has a greater value then the table

C. f(x) = 5x - 21......f(x) = 99

99 = 5x - 21

99 + 21 = 5x

120 = 5x

120/5 = x

24 = x <====

3 0

3 years ago

Law of Syllogism.

stiks02 [169]

Answer:

true

Step-by-step explanation:

if it is 10,20,30,40etc...

10÷5=2

20÷5=10

30÷5=15

40÷5=20

3 0

3 years ago

Mrs. Cahan asked her class to write fractions on their whiteboards that were equivalent to 2/4. Which fraction is equivalent to

KatRina [158]

Answer:

1/2

Step-by-step explanation:

because 2/4 can be reduced to 1/2

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3 years ago

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :