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3 years ago

Solve for x: 3(x + 1) = −2(x − 1) − 4.

Mathematics

1 answer:

Vlad [161]3 years ago

6 0

Answer:

Multiply the first bracket by 3

Multiply the second bracket by -2

3x+3=-2x+2-4

3x+2x+3=-2x-2 negative (-) number and positive number (+)= negative number

Just rearranging

5x+3=-2

5x+3-3=-2-3

5x=-5

Dividing by 5

5x/5=-5/5

x=-1

Check answer by using substitution method

3(-1+1)=-2(-1-1)-4

0=-2(-2)-4 negative number and negative number = positive number

0=4-4

0=0

Answer is x=-1

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Assume that we have two events, A and B, that are mutually exclusive. Assume further that we know P(A) = 0.3 and P(B) = 0.7. If

Thepotemich [5.8K]

Answer:

See below

Step-by-step explanation:

What is P(A B)?

If A and B are mutually exclusive, then

P(AB) = P(A∩B) = P(∅) = 0

What is P(A | B)?

By definition

P(A | B) = P(A∩B)/P(B).

Since

P(A∩B) = 0, then P(A | B) = 0

Is P(A | B) equal to P(A)?

No, because P(A | B) = 0 and P(A) = 0.3

Are events A and B dependent or independent?

A and B would be independent if

P(A | B) = P(A) and P(B | A) = P(B)

But both P(A |B) and P(B | A) equals 0 and P(A) = 0.3, P(B) = 0.7

Hence, A and B are dependent.

A student in statistics argues that the concepts of mutually exclusive events and independent events are really the same, and that if events are mutually exclusive they must be independent. Is this statement accurate?

No, it is not.

As we already saw, A and B are mutually exclusive but they are not independent.

What general conclusion would you make about mutually exclusive and independent events given the results of this problem?

If A and B are not empty events which are mutually exclusive, they can never be independent.

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3 years ago

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sladkih [1.3K]

Answer:

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Step-by-step explanation:

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3 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

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$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

3 years ago

Pplllleeaaassseeeee hellppppp meeee this is easy but i dont get it

AURORKA [14]

Answer:

choice 1) -6 and choice 2) -5

Step-by-step explanation:

choice 1)

-(1/2)-6)+6 > 8

9 > 8

choice 2)

-(1/2)(-5)+6 > 8

8 1/2 > 8

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3 years ago

Selectie correct answer.

denis-greek [22]

Answer:

OD.0

Step-by-step explanation:

because 2(X+4)-1 = 2x+7

or, 2x+8-1 = 2x+7

or, 2x+7= 2x+7

and it cancels out eachother so it will be 0

4 0

3 years ago