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madreJ [45]
3 years ago
12

Divide

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
8 0

Hello There!

<u>The answer is....</u>

<u />

A. -1 \frac{1}{4} .

Hopefully, this helps you!!

AnimeVines

vladimir1956 [14]3 years ago
5 0

Hello!!

Your answer will be is :

<h2><em>A. </em>-1~1/4.<em></em></h2><h2><em></em></h2><h2><u>EXPLANATION :</u></h2><h2><u /></h2>

Convert the mixed numbers to improper fractions: 3\frac{3}{4} =\frac{15}{4}

= -\frac{15}{4} ÷ 3

Convert element to fraction: 3 = \frac{3}{1}

= -\frac{15}{4} ÷ \frac{3}{1}

Apply the fraction rule: \frac{a}{b} ÷ \frac{c}{d} = \frac{a}{b}~x~\frac{d}{c}

Cross - cancel common factor:  3

= -\frac{5}{4}

Convert improper fractions to mixed numbers: \frac{5}{4} = 1\frac{1}{4}

= -1\frac{1}{4}

Hope It Helps. . .

#LearnWithBrainly

A~n~s~w~e~r~:

Jace ^-^

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2nd blank: 2

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Read 2 more answers
Under average driving conditions, the life lengths of automobile tires of a certain brand are found to follow an exponential dis
wariber [46]

Answer:

a) P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

b) P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}, x>0

And 0 for other case. Let X the random variable that represent "life lengths of automobile tires of a certain brand" and we know that the distribution is given by:

X \sim Exp(\lambda=\frac{1}{30000})

The cumulative distribution function is given by:

F(X) = 1- e^{-\frac{x}{\mu}}

Part a

We want to find this probability:

P(X>30000) and for this case we can use the cumulative distribution function to find it like this:

P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

Part b

For this case w want to find this probability

P(X>30000|X>15000)

We have an important property on the exponential distribution called "Memoryless" property and says this:

P(X>a+t| X>t)=P(X>a)  

On this case if we use this property we have this:P(X>30000|X>15000)=P(X>15000+15000|X>15000)=P(X>15000)

We can use the definition of the density function and find this probability:

P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

7 0
3 years ago
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