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3 years ago

SOMEONE PLEASE HELP! I don’t know how to do this !

Mathematics

1 answer:

BARSIC [14]3 years ago

3 0

9514 1404 393

Answer:

136

Step-by-step explanation:

The x-value at one end of the beam will be where f(x) = 0.

0 = -x²/272 +17

x² = 17(272) = 4624

x = ±√4624 = ±68

The beam extends from x=-68 to x=68, so has a width of ...

68 -(-68) = 136

Note that the x in the equation for f(x) is not the same as the x in the picture corresponding to width.

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The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,

professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

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 interval.

Mean of 0.6 times a day

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What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103$

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

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Find ????⋅ ???? if |????|=3 , |????|=9 , and the angle between ???? and ???? is π2 radians.

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Answer:

$a{\cdot}b=0$

Step-by-step explanation:

It is given that,

Magnitude of a A, $|A|=3$

magnitude of B, $|B|=9$

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Dot product,

$a{\cdot} b=|a||b|\ \cos\theta\\\\a{\cdot} b=3\times 9\times \ \cos(90)\\\\a{\cdot} b=0$

So, a.b is equal to 0.

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Step-by-step explanation:

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