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iogann1982 [59]
2 years ago
15

Vito has a fair 20-sided die with the faces

Mathematics
1 answer:
Hitman42 [59]2 years ago
5 0

Answer:

15, C

Step-by-step explanation:

Any number is equally likely to appear. Therefore, we have 3/20 * 100 = 15. This is C.

Hope this helped!

~clouddragon

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Which expression is equivalent to (x2 +9x – 1)(-4x + 3)? brainly
Andrews [41]

Answer:

- 4x³ - 33x² + 31x - 3

Step-by-step explanation:

Each term in the second factor is multiplied by each term in the first factor, that is

x²(- 4x + 3) + 9x(- 4x + 3) - 1 (- 4x + 3) ← distribute all 3 parenthesis

= - 4x³ + 3x² - 36x² + 27x + 4x - 3 ← collect like terms

= - 4x³ - 33x² + 31x - 3

4 0
3 years ago
Six times one hundred thirty one .four
Nonamiya [84]

Answer:

786

Step-by-step explanation:

6•1=6

6•30=180

6•100=600

600+180+6=786

6 0
3 years ago
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I’ll give you points for help
Ivahew [28]
The answer is option B. Hope this helps:)
3 0
3 years ago
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Can someone help please I need this done in 20 minutes and no links I have been getting them all day so please help me with this
Natasha2012 [34]
2.8 mi
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4.8+7(3)=8+21=29mi
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3 0
3 years ago
The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0
Zinaida [17]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i)

And if we replace we got:

E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8

So we expect about 0.8 defective computes in a batch of 4 selected.

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

Solution to the problem

For this case we have the following distribution given:

X           0             1               2               3             4

P(X)  0.4096    0.4096    0.1536    0.0256    0.0016

And we satisfy that P(X_i) \geq 0 and \sum P(X_i) =1 so we have a probability distribution. And we can find the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And if we replace we got:

E(X) = 0*0.4096 +1*0.4096+ 2*0.1536+ 3*0.0256 +4*0.0016 = 0.8

So we expect about 0.8 defective computes in a batch of 4 selected.

5 0
4 years ago
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