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3 years ago

Answer for this question

Mathematics

2 answers:

Firlakuza [10]3 years ago

8 0

-31x is the answer to the problem

natulia [17]3 years ago

7 0

Grouping,
looking at the coefficients (2,-7,-4, and 14)
you notice that 2,-4, and 14 are all factors of 2,
however you also notice that -7 and 14 arefactors of 7, so you want to group the terms containing -7 and 14 up (and 2 and -4 together)

so rearranging stuff
2x^3 -4x -7x^2+14

grouping
(2x^3-4x) + (-7x^2+14)

factoring stuff, (reverse distribution process)
notice you can factor out a 2x from the first group, and a -7 from the 2nd group, hence you will get

2x(x^2-2) -7(x^2-2)

you can factor out the (x^2-2) from both terms, hence you are left with

(2x-7)(x^2-2)

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PLEASE HELP ME! How do you factor out the coefficient of a variable?? Example: 5c - 15

SpyIntel [72]

Divide both by a number that can go into both terms so it will be 5 .
$\frac{5c-15}{5}$ which is 5(c-3)

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3 years ago

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A wildlife biologist determines that there are approximately 200 deer in a region of a national park. The population grows at a

Amanda [17]

Answer:

FV= 200*(1.07^n)

Step-by-step explanation:

Current population= 200 deer

Growth rate= 7% per year

To calculate the future value of deer in any given years, we need to use the following formula:

FV= PV*(1+g)^n

FV= future value

PV= present value

g= growth rate

n= number of years

FV= 200*(1.07^n)

For example, the population in 5 years:

FV= 200*(1.07^5)

FV= 281

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2 years ago

CHOOSE THE CORRECT ANSWER FROM THE BRACKET 1. The sum of 2 integers is zero. If one of them is -5, then the other is------------

Mandarinka [93]

Answer:

c) +5

Step-by-step explanation:

x + y = 0

-5 + (5) = 0

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3 years ago

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The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

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.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

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

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

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3 years ago

I need help with this question please

Lana71 [14]

Answer:

I) 23

K) 96

J) 77

Step-by-step explanation:

All angles of a triangle equal up to 180

6x+4x-3+x+7=180

11x+4=180

11x=176

x=16

6 0

2 years ago