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nikdorinn [45]
1 year ago
8

Factor: x(3y-5)+3(3y-15)

Mathematics
1 answer:
alina1380 [7]1 year ago
8 0

3xy-5x+9y-45

Step-by-step explanation:

Step by Step Solution

STEP1:STEP2:Pulling out like terms

 2.1     Pull out like factors :

   3y - 15  =   3 • (y - 5) 

Equation at the end of step2: (x • (3y - 5)) + 9 • (y - 5) STEP3:Equation at the end of step 3 x • (3y - 5) + 9 • (y - 5) STEP4:Trying to factor a multi variable polynomial

 4.1       Split       3xy-5x+9y-45 

 4.1       Split       3xy-5x+9y-45 

             into two 2-term polynomials

             -5x+3xy   and    +9y-45

             This partition did not result in a factorization. We'll try another one:

             3xy-5x   and    +9y-45

             This partition did not result in a factorization. We'll try another one:

             3xy+9y   and    -5x-45

             This partition did not result in a factorization. We'll try another one:

             3xy-45   and    +9y-5x

             This partition did not result in a factorization. We'll try another one:

             -45+3xy   and    +9y-5x

             This partition did not result in a factorization. We'll try 

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3 years ago
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Answer:

a.) f(x) = \frac{1}{30} where 90 < x < 120

b.) \frac{2}{3}

c.)  \frac{2}{3}

d.)  \frac{1}{2}

Step-by-step explanation:

Let

X be a uniform random variable that denotes the actual charging time of battery.

Given that, the actual recharging time required is uniformly distributed between 90 and 120 minutes.

⇒X ≈ ∪ ( 90, 120 )

a.)

Probability density function , f (x) = \frac{1}{120 - 90} = \frac{1}{30} where 90 < x < 120

b.)

P(x < 110) = \int\limits^{110}_{90} {\frac{1}{30} } \, dx

               = \frac{1}{30}[x]\limits^{110}_{90}  = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}

c.)

P(x > 100 ) = \int\limits^{120}_{100} {\frac{1}{30} } \, dx

                 = \frac{1}{30}[x]\limits^{120}_{100}  = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}

d.)

P(95 < x< 110)  = \int\limits^{110}_{95} {\frac{1}{30} } \, dx

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Gemiola [76]

The earning of the salesperson is an illustration of a linear function.

The possible functions in the two scenarios are: \mathbf{I(s) = 0.1s + 2500} and \mathbf{I(s) = 0.05s + 2000}\\

The function is given as:

\mathbf{I(s) = 0.1s + 2000}

When the base salary is increased, a possible function is:

\mathbf{I(s) = 0.1s + 2500}

This is so, because 2500 is greater than 2000

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\mathbf{I(s) = 0.05s + 2000}\\

This is so, because 0.05 is less than 0.1

So, the possible functions in the two scenarios are:

\mathbf{I(s) = 0.1s + 2500} and \mathbf{I(s) = 0.05s + 2000}\\

See attachment for the graphs of both functions

Read more about linear equations at:

brainly.com/question/21981879

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