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Kamila [148]
3 years ago
9

Evaluate f(x)=6x² – 12 when x=-3. A -30 B 42 C -48 D 312

Mathematics
2 answers:
pentagon [3]3 years ago
6 0

Answer:

B

Step-by-step explanation:

f(x) = 6x² - 12

f(-3) = 6(-3)² - 12

f(-3) = 6(9) - 12

f(-3) = 54 - 12

f(-3) = 42

Best of Luck!

Colt1911 [192]3 years ago
5 0

Answer:

-48

Step-by-step explanation:

its -66 but ok

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11,20,29, Find the 30th term.
Mekhanik [1.2K]

Answer:

372

There are already 3 terms. So, now you have to find the 30th term.

The difference of 30 - 3 is 27. Now, since you add 9 for every term, multiply it by 27.

27 x 9 = 243.

The third term is 29. So you add 29.

Your answer will be 372. Hope it helps.

4 0
3 years ago
A researcher conducts an experiment comparing two treatment conditions with 20 scores in each treatment condition.
Savatey [412]
To determine the number of subjects that are needed for the experiments, we multiply the number of independent variables with the number of scores. For example, there are n independent variables then, there are approximately,

                                number of subjects = 20 x n = 20n
6 0
3 years ago
Three siblings are comparing their Ages Jenna is the youngest. Her brother Carl is 6 years younger than her . Her sister eve is
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Answer:

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Step-by-step explanation:

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8 0
3 years ago
Which of the following is the standard form for y+4= 1.5(x-4)
MatroZZZ [7]
Answer: y= 1.5x -2 we find this by solving
6 0
3 years ago
Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t
andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ.e^{-\lambda.x}

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - e^{-\lambda.x}

(a) <u>Probability</u> of distance at most <u>100m</u>, with λ = 0.0143:

F(100) = 1 - e^{-0.0143.100}

F(100) = 0.76

<u>Probability</u> of distance at most <u>200</u>:

F(200) = 1 - e^{-0.0143.200}

F(200) = 0.94

<u>Probability</u> of distance between <u>100 and 200</u>:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = \frac{1}{\lambda}

E(X) = \frac{1}{0.0143}

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = \sqrt{\frac{1}{\lambda^{2}} }

σ = \sqrt{\frac{1}{0.0143^{2}} }

σ = 69.93

<u>Distance exceeds the mean distance by more than 2σ</u>:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - e^{-0.0143*209.79})

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

\int\limits^m_0 f({x}) \, dx = 0.5

\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx = 0.5

\lambda.\frac{e^{-\lambda.x}}{-\lambda} = -e^{-\lambda.x} + e^{0}

1 - e^{-\lambda.m} = 0.5

-e^{-\lambda.m} = - 0.5

ln(e^{-0.0143.m}) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0
3 years ago
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