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3 years ago

Find the area of the isosceles triangle

Mathematics

2 answers:

AveGali [126]3 years ago

4 0

175m^2
I got the by multiplying 1/2, by the base (14), by the height (25).

1/2 * 14 * 25

icang [17]3 years ago

4 0

First we are gonna find the height using trigonometry.

h^2 = 25^2 - 7^2
h = 24

Height is 24m

Now find area using 1/2 x b x h

14/2 x 24 = 168 m^2

Therefore area is 168m^2

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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:

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

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3 years ago

IS THE ANSWER B JUST TRYING TO CHECK

Yuki888 [10]

Answer:

No my dude its c

Step-by-step explanation:

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Answer:

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