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3 years ago

HELLPPP!!!!!!!!!!!!!!!!

Mathematics

2 answers:

mars1129 [50]3 years ago

6 0

The answer is (4,2).......

Alex Ar [27]3 years ago

6 0

(4,2) is your answer

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A diver went 25.35 feet below the surface of the ocean and then 18.5 feet further down he then rose 12.45 feet. Write and solve

Paladinen [302]

Answer:

X= 25.65 + 16.5 - 12.45

X=29.70

Step-by-step explanation:

4 0

3 years ago

The marching band organises 80 students in the six lines how many students are in each line and how many students are left over

Bogdan [553]

80 ÷6=13
13×6=78
80-78=2

4 0

3 years ago

Read 2 more answers

What is the vertical asymptote of the function 3/x+5

kobusy [5.1K]

Since we cannot divide by zero, x=0 is forbidden, and x=0 is the vertical asymptote of this function.

7 0

3 years ago

Find the sum of the series and show your process for finding each term leading to the final sum.

Harrizon [31]

Answer:

$\sum_{k=3}^5(-2k+5)=-9$

Step-by-step explanation:

The given series is:

$\sum_{k=3}^5(-2k+5)$

The series is not very lengthy so we can write the expanded form;

$\sum_{k=3}^5(-2k+5)=(-2(3)+5)+(-2(4)+5)+(-2(5)+5)$

This implies that;

$\sum_{k=3}^5(-2k+5)=-1+-3+-5$

$\sum_{k=3}^5(-2k+5)=-9$

5 0

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