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3 years ago

Round 490,618 to the nearest hundred

Mathematics

2 answers:

schepotkina [342]3 years ago

7 0

Answer: 490,600

Step-by-step explanation:

Tomtit [17]3 years ago

4 0

Answer:

490600

Step-by-step explanation:

The answer would be 490600 because when you are rounding you start at the ones. So 8, you can round 8 up to ten which adds the the ten that was there which results in 20. 20 isn't large enough to be rounded up so it has to be rounded down 20, the 6 is affected by the 2 (because it couldn't be rounded up) so it stays the same therefore your answer is 490600. :D

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Which ordered pair is a solution of the equation y=-9+4 (10,-86)

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a boy is 26 years younger than his father in 3 years time his age will be one third his father's age. Find the boy's present age

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Answer:

61 years old

Step-by-step explanation:

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2 years ago

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A certain kind of sheet metal has, on average, 3 defects per 18 square feet. Assuming a Poisson distribution, find the probabili

Fofino [41]

Answer:

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have that:

3 defects per 18 square feet.

So for 31 square feet, we have to solve a rule of three

3 defects - 18 square feet

x defects - 31 square feet

$18x = 3*31$

$x = \frac{31*3}{18}$

$x = 5.17$

So $\mu = 5.17$

Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.

Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.

$P(X \leq 3) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

$P(X \geq 4) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057$

$P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294$

$P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760$

$P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242$

Finally

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758$

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

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3 years ago

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Answer:

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Step-by-step explanation:

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