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3 years ago

Vertex form of y=-4x^2+4x-6

Mathematics

1 answer:

AlexFokin [52]3 years ago

6 0

Answer:

Step-by-step explanation:

y = -4x^2 + 4x + 6

Factor out the leading coefficient:

y = -4(x^2 - x) + 6

Complete the square:

y = -4(x^2 - x + (½)^2) + 4(½)^2 + 6

= -4(x-½)^2 + 7

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Factor gcf 5ab^2 + 10ab

Mrrafil [7]

<span>5ab^2 + 10ab

5ab^2 = 5ab *(b)
10ab = 5ab *(2)

so GCF of </span>5ab^2 + 10ab = 5ab

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3 years ago

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6 2/3 divided by 2 6/7

Arturiano [62]

Answer:

7/3 in decimal form it's 2.3 repeated

Step-by-step explanation:

5 0

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.

8 0

3 years ago

Needing help to find the inequality

tiny-mole [99]

http://goblues.org/faculty/grantm/files/2015/12/Practice-Test-3.7-Solutions.pdf

I think it's question number three. Sorry if the link dosen't work.

4 0

3 years ago

Given P(x) = 2x3+ 3x2- x + 4, find P(0) and P(-2).

pshichka [43]

9514 1404 393

Answer:

P(0) = 4

P(-2) = 2

Step-by-step explanation:

Put the value where the variable is and do the arithmetic.

P(0) = 2·0³ +3·0² -0 +4

P(0) = 4

P(-2) = 2·(-2)³ +3·(-2)² -(-2) +4 = 2(-8) +3(4) +2 +4

P(-2) = -16 +12 +2 +4

P(-2) = 2

6 0

3 years ago