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maxonik [38]
4 years ago
6

Help with number 1 and 2 thanks

Mathematics
1 answer:
lidiya [134]4 years ago
7 0
So for #1 follow,

i. \:  \frac{4}{3}  >  \frac{5}{4}
Cross multiply and get,

4 \times 4 = 16 > 3 \times 5 = 15

16 > 15

Or,

\frac{4}{3}  = 1 \frac{1}{3}  = 1.33

\frac{5}{4}  =  1\frac{1}{4}  = 1.25

ii. \:  -  \frac{2}{3}  >  -  \frac{3}{4}

-  \frac{2}{3}  = .666 >   - \frac{3}{4}  = 0.75

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3 years ago
Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and relea
Talja [164]

Answer:

a) For this case the random variable X follows a hypergometric distribution.

b) E(X)= n\frac{M}{N}=10 \frac{5}{25}=2

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1

c) P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565

d) P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474

Step-by-step explanation:

The hypergometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

E(X)= n\frac{M}{N}

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}

a. What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

b. Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

E(X)= n\frac{M}{N}=10 \frac{5}{25}=2

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1

c. What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565

d. What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474

4 0
3 years ago
PLEASE HELP i have to turn this in in an hour PLEASE DONT GUESS
mina [271]

Answer:

x = 29°

Step-by-step explanation:

the Δ on the right has 3 angles whose measures are 30°, 74°, and 76° (you get 76° because it forms a linear pair with an angle that is 104°; you get 74° by adding 30 and 76 and subtracting the sum from 180°.

the Δ on the left has 3 angles whose measures are 45°, 29°, and 106° (you get 106° because it forms a linear pair with an angle that is 74°; you get 29° by adding 45 and 106 and subtracting the sum from 180°.

∡x is a vertical angle with the angle that measures 29° and vertical angles are congruent

4 0
3 years ago
Read 2 more answers
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