Mathematics, 19.11.2019 05:31 ashleywoo21361

Let x denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. suppose that for banner-tailed kangaroo rats, x has an exponential distribution with parameter λ = 0.01352. (a) what is the probability that the distance is at most 100 m? at most 200 m? between 100 and 200 m? (round your answers to four decimal places.)

Answers: 3

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Answer from: econsta3

Answer with Step-by-step explanation:

We are given that

X=Distance (m) that animal moves from its birth site to the first territorial vacancy it encounters.

Parameter= $\lambda=0.01352$

Exponential distribution of probability is given by

$f(x)=\lambda e^{-\lambda x}, x\geq 0$

f(x)=0, x < 0

Cumulative distribution is given by

$F(X)=P(X\leq x)=1-e^{-\lambda x }, x\geq 0$

Where $\lambda=Parameter=0.01352$

a.We have to find the probability that the distance is at most 100m.

$P(X\leq 100)=1-e^{-0.01352\times 100}=0.7413$

Hence, the probability that the distance is at most 100 m=0.7413

b. $P(X\leq 200)=1-e^{-0.01352\times 200}=0.9331$

Hence, the probability that the distance is at most 200m =0.9331

c. $P(100\leq X\leq 200)=P(X\leq 200)-P(X\leq 100)$

$P(100\leq X\leq 200)=0.9931-0.7413=0.1918$

Hence, the probability that the distance between 100 m and 200 m=0.1918

Answer from: Quest

Ithink it’s b, hope this .

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

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Let x denote the distance (m) that an animal moves from its birth site to the first territorial vaca...