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3 years ago

Two dice are rolled. What is the probability of getting a sum equals 5?

Mathematics

1 answer:

Ugo [173]3 years ago

3 0

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What is the decimal equivalent of 23/9

Gekata [30.6K]

Answer:

the ans in decimal will be 2.55 bcz of 2 s.f ( significant figures)

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4 years ago

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What two integers have a sum of -31 and a difference of -55?

nekit [7.7K]

Answer:

Integer 1 is x

Integer 2 is y

x + y = -31

x - y = -55

Add:

2x = -86

x = -43

Substitute:

-43 + y = -31

y = 12

Therefore, the two integers are: 12 and -43

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3 years ago

Find a number close to 132 that divides easily by 3

Sergio [31]

132 divides easily by 3

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3 years ago

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The average lifetime of a certain new cell phone is 3.4 years. The manufacturer will replace any cell phone failing within three

melisa1 [442]

Answer:

68% of these phones last 3.87 years.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The average lifetime of a certain new cell phone is 3.4 years.

This means that $m = 3.4, \mu = \frac{1}{3.4} = 0.2941$

$P(X \leq x) = 1 - e^{-0.2941x}$

68% of these phones last how long (in years)?

This is x for which:

$P(X \leq x) = 0.68$

$P(X \leq x) = 1 - e^{-0.2941x}$

Then

$0.68 = 1 - e^{-0.2941x}$

$e{-0.2941x} = 0.32$

$\ln{e{-0.2941x}} = \ln{0.32}$

$-0.2941x = \ln{0.32}$

$x = -\frac{\ln{0.32}}{0.2941}$

$x = 3.87$

68% of these phones last 3.87 years.

4 0

3 years ago

Classify a triangle with each set of side lengths as acute, right, or obtuse

Serga [27]

Answer: Acute- (2,3,3.5) Right-(2.5,6,6.5) obtuse- (7,8,12), (7,9,14)

Step-by-step explanation:

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3 years ago