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jok3333 [9.3K]
4 years ago
5

Two number cubes are rolled what is P(odd and even)

Mathematics
2 answers:
kupik [55]4 years ago
6 0
P(odd and even):

P(odd): HINT: Since it's 2 cubes there are double the odd numbers(1,1,3,3,5,5)
6/12=1/2
P(even):6/12=1/2
After you have both of the probabilities, you multiply them.
1/2*1/2=1/4 or 0.25
ddd [48]4 years ago
5 0
Since there are exactly three odd numbers and three even numbers on a standard number cube. The probability of rolling two odds is 50%. Since there 6 numbers on a cube, three are odd. With two number cubes there are 12 numbers and six of them are odd, which makes is a 6/12 probability which alo equals 50%. The same goes for even which means there is a 50/50 chance of getting two evens or two odds.
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Answer:

x =  \frac{46}{3}

Step-by-step explanation:

A straight angle is equal to 180° degrees. Since both 9x +4 and 3x -18 is supplementary then I can make this claim below

(9x + 4 ) + (3x - 8) = 180

All we have to do now is to solve for x

(9x + 4) + (3x - 8) = 180 \\ 9x + 4+ 3x - 8 = 180 \\ 12x - 4 = 180 \\ 12x = 184 \\ x = \frac{184}{12}  \\ x =  \frac{46}{3}

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3 years ago
The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine
salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

                                              f(x)=\frac{e^{\frac{-x}{1000} }}{1000}

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

              b. P(1000<x<2000) = 0.2325

              c. P(x<1000) = 0.6321

              d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = \int\limits^b_a {P(x)} \, dx

Then, for the electronic component, probability will be:

P(a<x<b) = \int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx

P(a<x<b) = \frac{1000}{1000}.e^{\frac{-x}{1000} }

P(a<x<b) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

a. For a component to last more than 3000 hours:

P(3000<x<∞) = e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = e^{-3}

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}

P(1000<x<2000) = e^{-2}-e^{-1}

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}

P(0<x<1000) = e^{-1}-1

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

0.1 = 1-e^\frac{-x}{1000}

-e^{\frac{-x}{1000} }=-0.9

{\frac{-x}{1000} }=ln0.9

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

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167/32 - 10/32 = 157/32

= 4 29/32

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