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3 years ago

4x^2−4x+10 and 2x^2+5x-82x^2+5x−8.

Mathematics

2 answers:

Hitman42 [59]3 years ago

3 0

2(2x^2-2x+5) and 2(40x^2-5x+4)

Step-by-step explanation:

Phantasy [73]3 years ago

3 0

Answer:

4x^2−4x+10 = 2(2x^2−2x+5)

2x^2+5x-82x^2+5x−8 = 2(−40x^2+5x−4)

Step-by-step explanation:

4x^2−4x+10

Factor 4x^2−4x+10

4x^2−4x+10

=2(2x^2−2x+5)

2x^2+5x-82x^2+5x−8

Factor 2x^2+5x−82x^2+5x−8

−80x^2+10x−8

=2(−40x^2+5x−4)

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kolezko [41]

Answer:

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3 years ago

Mr. Mole left his burrow that lies below the ground and started digging his way at a constant rate deeper into the ground.

Marizza181 [45]

Answer:

$A(t)=-1.8t-4.5$

Step-by-step explanation:

Mr.Mole's digging can be described by a linear model of general equation:

$A(t)=m*t+c$

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The constant "c" is his starting position. To find this value, assign the given data at t = 5 min to the equation above

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3 years ago

The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays. Assuming the

AysviL [449]

Answer:

1.83% probability there are no car accidents on that stretch on Monday

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 number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays.

This means that $\mu = 4$

What is the probability there are no car accidents on that stretch on Monday?

This is P(X = 0).

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

$P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.0183$

1.83% probability there are no car accidents on that stretch on Monday

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3 years ago

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sashaice [31]

Answer:

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Step-by-step explanation:

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