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Lostsunrise [7]
3 years ago
5

Four expressions are shown below:

Mathematics
2 answers:
Marina86 [1]3 years ago
8 0

Answer:

32x+8 and 4(8x+2)

Step-by-step explanation:

1. simplify the expression 4(8x+2) =32x+8

Serggg [28]3 years ago
3 0

Answer:

Step-by-step explanation:

4(7x + 2 + x) = 4(8x + 2) = 4*8x + 4*2 = 32x + 8

4(7x + 2 + x) are equivalent to 4(8x + 2) and 32x + 8

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Step-by-step explanation:

count how many zeros

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Which is the graph of x^2/16 + y^2/ 9 =1
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I can’t see the graph for b and d…
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For the following​ distribution, decide whether you expect the​ mean, median, or mode to give the best representation of the cen
valentinak56 [21]

Answer:

The median because it is unaffected by outliers.

Step-by-step explanation:

While mean is one of the most well known and used central tendency measurements, it has some weaknesses. Mean will show inaccurate data if there are outliers that have a very huge/small value. Income is one of the things that can hugely vary among people. A billionaire can pull the data up and make the mean show result that higher than it should be. In this case, the median is better to show the data. The mode is used for nominal data, while income is ordinal.

5 0
3 years ago
Solve the equation for a. 55a – 44a = 11
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7 0
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
4 years ago
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