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Arlecino [84]
3 years ago
5

G(x) = 3x -1. Find x such that g(x)= 2​​0​.

Mathematics
1 answer:
sergiy2304 [10]3 years ago
3 0

Step-by-step explanation:

i. 3x-1=20

or, 3x= 21

x= 7

ii. 3x+1= 22

3x= 21

x= 7

iii. 2x-5=99

2x= 104

x=52

iv. g(3)= 3+5/2= 11/2

g(0)= 0+ 5/2=5/2

g(-3)= -3+5/2 = -1/2

x+5/2= 0

so, x= -5/2

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Like terms" are terms whose variables (and their exponents such as the 2 in x2) are the same. In other words, terms that are "like" each other. Note: the coefficients (the numbers you multiply by, such as "5" in 5x) can be different.

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Please help asap. Thank you! :)
Kruka [31]

When he cubed ,

Result is x³

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{x}^{6}

<h3>Therefore, k = 6</h3>

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2 years ago
For each table of values, write an expression that relates the input to the output.
Arlecino [84]

Answer:

x3, +3

Step-by-step explanation:

Each number gets multiplied by 3 and then you add 3.

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2 years ago
P - 3 1/6 = -2 1/2<br> what is p?
goldfiish [28.3K]
<span>P - 3 1/6 = -2 1/2

3 1/6- 21/2 

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3 years ago
Read 2 more answers
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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