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AleksandrR [38]
4 years ago
13

Which elephant weight is 2.85 tons when you rounded to the nearest hundredth

Mathematics
2 answers:
aalyn [17]4 years ago
7 0
I think it’d be 2.9 , because you’re rounding the hundredth which is .05 , so you’d move up the .8 to a .9
Lera25 [3.4K]4 years ago
4 0

Answer:

Step-by-step explanation

Hundredth is expressed as 1/100 0r 0.01.

Rounding off of 2.85 tons = 3.00 tons

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1/33

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I don't know if i'm correct but thats what I got. Due to 20 minus 19 is 1. and 17 minus -16 equals 33.

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3 years ago
Write down<br> 1<br> 16<br> as<br><br> (a) a decimal, [1]<br> (b) a percentage.
Ede4ka [16]

Answer:

(a) 1.0625

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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
4 years ago
Get brainiest !! Easyyy question
Reika [66]

Answer:

Step-by-step explanation:

hmmm

7 0
3 years ago
Read 2 more answers
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