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2 years ago

Please help and explain.

Mathematics

2 answers:

Alexandra [31]2 years ago

6 0

Answer:

-1/3

Step-by-step explanation:

Use the formula, (y2 - y1)/(x2 - x1). In this case, there is a negative slope.

Start from point, (1, 9) and go to point (4, 8).

Use the formula:

(y2 - y1)/(x2 - x1)

Y2 = 8

Y1 = 4

X2 = 9

X1 = 1

(8 - 9)/(4 - 1) = -1/3

Therefore, the slope is -1/3

tino4ka555 [31]2 years ago

4 0

Answer:

-(1/3)

Step-by-step explanation:

The answer is negative 1/3. Slope is the change in y divided by the change in x. Given two points, (4,8) and (7,7), the slope is negative 1/3

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Hi guys, Can anyone help me with this tripple integral? Thank you:)

OleMash [197]

I don't usually do calculus on Brainly and I'm pretty rusty but this looked interesting.

We have to turn K into the limits of integration on our integrals.

Clearly 0 is the lower limit for all three of x, y and z.

Now we have to incorporate

x+y+z ≤ 1

Let's do the outer integral over x. It can go the full range from 0 to 1 without violating the constraint. So the upper limit on the outer integral is 1.

Next integral is over y. y ≤ 1-x-z. We haven't worried about z yet; we have to conservatively consider it zero here for the full range of y. So the upper limit on the middle integration is 1-x, the maximum possible value of y given x.

Similarly the inner integral goes from z=0 to z=1-x-y

We've transformed our integral into the more tractable

$\displaystyle \int_0^1 \int_0^{1-x} \int _0^{1-x-y} (x^2-z^2)dz \; dy \; dx$

For the inner integral we get to treat x like a constant.

$\displaystyle \int _0^{1-x-y} (x^2-z^2)dz = (x^2z - z^3/3)\bigg|_{z=0}^{z= 1-x-y}=x^2(1-x-y) - (1-x-y)^3/3$

Let's expand that as a polynomial in y for the next integration,

$= y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3$

The middle integration is

$\displaystyle \int_0^{1-x} ( y^3/3 +(x-1) y^2 + (2x+1)y -(2x^3+1)/3)dy$

$= y^4/12 + (x-1)y^3/3+ (2x+1)y^2/2- (2x^3+1)y/3 \bigg|_{y=0}^{y=1-x}$

$= (1-x)^4/12 + (x-1)(1-x)^3/3+ (2x+1)(1-x)^2/2- (2x^3+1)(1-x)/3$

Expanding, that's

$=\frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1)$

so our outer integral is

$\displaystyle \int_0^1 \frac{1}{12}(5 x^4 + 16 x^3 - 36 x^2 + 16 x - 1) dx$

That one's easy enough that we can skip some steps; we'll integrate and plug in x=1 at the same time for our answer (the x=0 part doesn't contribute).

$= (5/5 + 16/4 - 36/3 + 16/2 - 1)/12$

$=0$

That's a surprise. You might want to check it.

Answer: 0

6 0

3 years ago

Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know

USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

$P(A/B)= \frac{P(AnB)}{P(B)}$

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

$P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33$

$P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33$

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0

3 years ago

14 + 5x = 3(-x + 3) -11 please show work step by step

mr Goodwill [35]

14+5x=3(-x+3)-11

14+5x=-3x+9-11

14+5x=-3x-2

add 2 to both sides

16+5x=-3x

subtract 5x from both sides

16=-8x

divide both sides by -8

-2=x

Hope this helps! :)

3 0

3 years ago

Mrs Low has some money to buy plates and bowls for her new house.

jekas [21]

Answer:

78=p

$150

$20

6 bowls

Step-by-step explanation:

8 0

3 years ago

−3 1/3 ÷ 9= What I need it now!

Contact [7]

Step 1. Convert 3 1/3 to improper fraction.

- 3 * 3 + 1/3 ÷ 9

Step 2. Simplify 3 * 3 to 9

-9 + 1/3 ÷ 9

Step 3. Simplify 9 + 1 to 10

-10/3 ÷ 9

Step 4. Use this rule: a ÷ b/c = a * c/b

-10/3 * 1/9

Step 5. Use this rule: a/b * c/d = ac/bd

-10 * 1/3 * 9

Step 6. Simplify 10 * 1 to 10

-10/3 * 9

Step 7. Simplify 3 * 9 to 27

-10/27

8 0

3 years ago