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3 years ago

Solve log2(x) – log2(3) = 2.

Mathematics

2 answers:

Masteriza [31]3 years ago

5 0

I don’t know if this helps you could try this

Lynna [10]3 years ago

3 0

Answer:

x=9.64385618

Step-by-step explanation:

I used an app to solve it

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Julio believes that the quotient of 6 divided by 1/2 = 3. Explain Julios error and tell how to find the correct answer.

Gnom [1K]

Answer:

it can be written as 6/1 divided by 1/2

then you'll use the keep, switch, flip method. keep 6/1, switch division to multiplication, and flip 1/2 to become 2/1 or 2

now your equation is 6 x 2 = 12

Julio did 6 divided by 2. which could've been fixed if he changed the division sign to multiplication and multiplied

8 0

3 years ago

Please help Which of the following inequalities is graphed on the coordinate plane?

Gekata [30.6K]

Answer:

The inequality which represents the graph is y ≤ -2x + 1 ⇒ A

Step-by-step explanation:

To solve the question you must know some facts about inequalities

If the sign of inequality is ≥ or ≤, then it represents graphically by a solid line

If the sign of inequality is > or <, then it represents graphically by a dashed line

If the sign of inequality is > or ≥, then the area of the solutions should be over the line

If the sign of inequality is < or ≤, then the area of the solutions should be below the line

Let us study the graph and find the correct answer

∵ The line represented the inequality is solid

∴ The sign of inequality is ≥ or ≤

→ That means the answer is A or B

∵ The shaded area is the area of the solutions of the inequality

∵ The shaded area is below the line

∴ The sign of inequality must be ≤

→ That means the correct answer is A

∴ The inequality which represents the graph is y ≤ -2x + 1

3 0

3 years ago

The town council wants to know if the residents support beautification of the downtown area. What is the best design for this st

Natasha_Volkova [10]

Experimental census sample survey

7 0

4 years ago

Help please gread 8 ........................................................

motikmotik

Answer:

see explanation

Step-by-step explanation:

(a)

Since Y is the midpoint of XZ , then

XY = YZ , that is

4a - 4 = 3a + 2 ( subtract 3a from both sides )

a - 4 = 2 ( add 4 to both sides )

a = 6

(b)

XY = 4a - 4 = 4(6) - 4 = 24 - 4 = 20

XZ = 4a - 4 + 3a + 2 = 7a - 2 = 7(6) - 2 = 42 - 2 = 40

6 0

3 years ago

Read 2 more answers

If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

3 years ago