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Oduvanchick [21]

3 years ago

10

You are told that P(A | B) = P(B | A). Which statement below must be true?

2 answers:

Katen [24]3 years ago

6 0

The best answer to the question that is being stated above would be that A and B are both dependent. If I was told that P(A | B) = P (B | A), then the true statement is that both variables inside would be dependent upon each other, no matter if you exchange their places with one another.

Karo-lina-s [1.5K]3 years ago

6 0

Answer:

The answer is A and B are dependent

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Each quilt requires 3 7/8 yards of fabric how many yards will you need for 1, 2, 3,4 quilts.

ICE Princess25 [194]

Step-by-step explanation:

Given,

No. of yards required for each quit = 3 7/8

can also be written as = 59/8

so,

yards,

required for 2 quilts = 2 × 59/8 = 14 3/4

required for 3 quilts = 3 × 59/8 = 23 4/8

required for 4 quilts = 4 × 59/8 = 8 1/2

<em><u>hope </u></em><em><u>this </u></em><em><u>answer </u></em><em><u>helps </u></em><em><u>you </u></em><em><u>dear.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>take </u></em><em><u>care!</u></em>

7 0

3 years ago

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

Olin [163]

Answer:

The correct option is d

$f(1) = -5$

$f(2) = 11$

The correct option is d

The correct option is c

the correct option is b

Step-by-step explanation:

The given equation is

$f(x) = x^4 + x -7 =0$

The give interval is $(1,2)$

Now differentiating the equation

$f'(x) = 4x^3 +7 > 0$

Therefore the equation is positive at the given interval

Now at x= 1

$f(1) = (1)^4 + 1 -7 =-5$

Now at x= 2

$f(2) = (2)^4 + 2 -7 =11$

Now at the interval (1,2)

$f(1) < 0 < f(2)$

i.e

$-5 < 0 < 11$

this tell us that there is a value z within 1,2 and

f(z) = 0

Which implies that there is a root within (1,2) according to the intermediate value theorem

5 0

3 years ago

PLEASE HELP! I Have Attached pictures of the question and poem.

iren2701 [21]

They seem to be very descriptive and use lots of examples of nature.

7 0

4 years ago

What is the volume of this cube? Enter you answer as a decimal in the box.

Katyanochek1 [597]

Solution: V=a^3=3=0.5^3=0.125
so your answer is 0.13, hope this helps!

5 0

4 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago