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Luden [163]
3 years ago
8

Simplify the equation -x^3 + 5x + 7x^2 +8x^3 - (-x^3 +x)​

Mathematics
2 answers:
sammy [17]3 years ago
4 0

Let's simplify step-by-step.

−x^3 + 5x + 7x^2 + 8x^3 − ( −x^3 + x )

Distribute the Negative Sign:

= −x^3 + 5x + 7x^2 + 8x^3 + −1 ( −x^3 + x )

= −x^3 + 5x + 7x^2 + 8x^3 + −1 ( −x^3 ) + −1x

= −x^3 + 5x + 7x^2 + 8x^3 + x^3 + −x

Combine Like Terms:

= −x^3 + 5x + 7x^2 + 8x^3 + x^3 + −x

= ( −x^3 + 8x^3 + x^3 ) + ( 7x^2 ) + ( 5x + −x )

= 8x^3 + 7x^2 + 4x

Answer:

= 8x^3 + 7x^2 + 4x

monitta3 years ago
3 0

Answer:

8x³ + 6x² + 5x

Step-by-step explanation:

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Suppose a game show called Bayesian Boxes has the contestant sample one marble from a box and guess what type of box it is. When
8_murik_8 [283]

Answer:

P(C/R) = 0.4298

Step-by-step explanation:

Missing word <em>"arble, what is the chance it came from box Type C? (Give the decimal answer rounded to at least four places )"</em>

<em />

Solution:

Given that P(A) = 75% = 0.75

P(B) = 20% = 0.20

P(C) = 0.05

The chance of drawing a Red marble fro,

P(R/A) = 2% = 0.02

P(R/B) = 25% = 0.25

P(R/C) = 98% = 0.98

If he select Box A than the probability of Red marble drawing

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If he select Box  B than the probability of Red marble drawing

= P(B) * P(R/B)

= 0.20 * 0.25

= 0.5

If he select Box C than the probability of Red marble drawing

= P(C) * P(R/C)

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= 0.049

So total probability of drawing Red marble is

P(R) = 0.015 + 0.050 + 0.049

P(R) = 0.114

If the Contestant draw a Red marble. Probability that it comes from box type (C) using Boyes theorem is

P(C/R) = {P(C) * P(R/C)} / P(R)

P(C/R) = 0.049 / 0.114

P(C/R) = 0.429824

P(C/R) = 0.4298

3 0
3 years ago
Solve For Q<br><br> q/10=4<br><br> Please Answer This For Me. Thanks.
Maurinko [17]

\frac{q}{10} = 4\\q = 4 * 10\\q = 40

Answer: 40.

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4 years ago
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Answer:

Ok, so I think I may have got it right.

Step-by-step explanation:

So in each row going left to right is the following numbers

1st row: 2, 3, 4, 5

2nd row: 3, 4, 5, 6

3rd row: 4, 5, 6, 7

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Exponential is the answer
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