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chubhunter [2.5K]
2 years ago
7

Factor out the greatest common factor. If the greatest common factor is 1, just retype the

Mathematics
1 answer:
harina [27]2 years ago
4 0

Answer:

Below.

Step-by-step explanation:

The GCF of 9 , 27 and 45 = 9.

GCF of the d terms is d^2.

So the GCF of the whole expression is 9d^2.

Factors are 9d^2( 1 - 3d^3 + 5d)

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Simplify the expression
Stolb23 [73]

Answer:

A. {x}^{3}  - x -  \frac{6}{2x + 3}

Step-by-step explanation:

\frac{2 {x}^{4}  + 3 {x}^{3}  - 2 {x}^{2}  - 3x - 6}{2x + 3}  \\  \\  = \frac{(2 {x}^{4}  + 3 {x}^{3} ) - (2 {x}^{2}   + 3x) - 6}{2x + 3}  \\  \\  = \frac{ {x}^{3} (2 {x}  + 3  ) - x(2 {x}  + 3) - 6}{2x + 3}  \\  \\  =  \frac{ {x}^{3} (2x + 3)}{2x + 3}  -  \frac{x(2x + 3)}{2x + 3}  -  \frac{6}{2x + 3}  \\  \\   \purple { \bold{=  {x}^{3}  - x -  \frac{6}{2x + 3}  }}

8 0
2 years ago
Just.. please help this
Helga [31]

Answer:

D)

Step-by-step explanation:

3 0
3 years ago
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Is the following relation a function? (-2,3) (0,1) (5,3) (5,8) (7,5)
Rashid [163]
No it is not because for the x value of 5 there are two different y values.
5 0
3 years ago
there is 1,000 cm3 of aluminum available to cast a trophy that will be in the shape of a right square pyramid. Is this enough al
enyata [817]

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Step-by-step explanation: I hope this help you

3 0
3 years ago
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The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and me
Dovator [93]

Answer:

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

The problem states that:

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.

To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.

There are 5 weekdays, with a mean of 0.1 calls per day.

The weekend is 2 days long, with a mean of 0.2 calls per day.

So:

\mu = \frac{5(0.1) + 2(0.2)}{7} = 0.1286

If today is Monday, what is the probability that Ben receives a total of 2 phone calls in a week?

This is P(X = 2). So:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 2) = \frac{e^{-0.1286}*0.1286^{2}}{(2)!} = 0.0073

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

3 0
2 years ago
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