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Zigmanuir [339]
3 years ago
5

Evaluate the expressiin when a=6, b=3, and c=2: c(b+a)-a= please help

Mathematics
1 answer:
Setler [38]3 years ago
8 0

Answer:

\huge{ \fbox{ \tt{12}}}

Step-by-step explanation:

\underline{ \sf{Given}} : \sf a=6, \:  b=3 \:  and  \: c=2

\underline{ \sf{To \: find }}  :   \sf{Value \: of \: the \: given \: expression}

\underline{ \underline{ \mathfrak{Let's \: solve}}} :

\bigstar{ \tt{ \: c \: ( \: b \:  +  \: a \: ) -  \: a}}

\text{Step \: 1 \: } : Plug the values of a , b and c

\mapsto{ \tt{ \: 2 \: ( \: 3 \:  +  \: 6 \: ) \:  -  \: 6}}

\text{Step \: 2} : \sf{ \: Add \: the \: numbers : 6 \: and \: 3}

\mapsto{ \sf{ \: 2 \: * \: 9 \:  -  \: 6 \: }}

\text{Step \: 3 \: } : \sf{ \: Multiply \: the \: numbers : 9 \: and \: 2}

\mapsto{ \sf{18 - 6}}

\text{Step \: 4 \:} :  \sf{Subtract \: 6 \: from \: 18}

\mapsto{ \sf{ \: 12}}

Note : Just follow the PEMDAS rule ( See attached picture )

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

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A publication released the results of a study of the evolution of a certain mineral in the​ Earth's crust. Researchers estimate
san4es73 [151]

Answer:

a) E(A)=\frac{1+6}{2}=3.5 ppm

b) P(2.875

c) P(X

Step-by-step explanation:

If we work with the limits defined from 5 to 10 then part b and c from this question not makes sense. If we work with the limits 1 and 6 all the parts for the question makes sense because if we work with 5 and 10 the only thing that we can find is the expected value E(A) = \frac{5+10}{2}= 7.5

Assuming the following correct question : "A publication released the results of a study of the evolution of a certain mineral in the​ Earth's crust. Researchers estimate that the trace amount of this mineral x in reservoirs follows a uniform distribution ranging between 1 and 6 parts per million"

Solution to the problem

Let A the random variable that represent " amount of the mineral x ". And we know that the distribution of A is given by:

A\sim Uniform(1 ,6)

Part a

For this uniform distribution the expected value is given by E(X) =\frac{a+b}{2} where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

E(A)=\frac{1+6}{2}=3.5 ppm

Part b

For this case we can use the cumulative distribution function for the uniform distribution given by:

F(X=x)= \frac{x-a}{b-a} = \frac{x}{6-1} =\frac{x-1}{5} , 1 \leq X \leq 6

And we want this probability:P(2.875Part c

For this case we want this probability:

P(X

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Step-by-step explanation:

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the answer to this is B.


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