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3 years ago

Solve the problem: -4/9 -8/9

Mathematics

1 answer:

Trava [24]3 years ago

7 0

Answer:

$\frac{-4}{3}$

Step-by-step explanation:

When subtracting fractions with the same denominator (bottom number[s]) you can treat the numerators (top number[s]) like subtraction with whole numbers.

$-4-8$ is equal to -12. Therefore, $\frac{-4}{9} - \frac{8}{9}$ will equal $\frac{-12}{9}$ . But we're not there yet - this fraction can be simplified. We need to find the GCF, or greatest common factor, of -12 and 9. This would be another number that can be divided evenly into these two numbers. The first possible GCF that comes to mind is 3. Let's test that out:

$-12/3 = -4$

$9/3 = 3$

It works! Now we can use this information to simplify the fraction. We replace -12 with -4 and replace 9 with 3. This gives us our final answer, $\frac{-4}{3}$ .

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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:

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

4 0

3 years ago

A proton initially has and then 8.00 s later has (in meters per second). (a) For that 8.00 s, what is the proton's average accel

svet-max [94.6K]

Answer:

Some details are missing in the question, here are the details ; A proton initially has v = -6.5i + 17j + 13k and then 8.00s later has v = -2.8i + 17j - 9.3k (in meters per seconds).

a) proton's average acceleration in unit vector notation = 0.46i - 2.78k

b) Magnitude = 2.85m/s2

c) angle between and the positive direction of the x axis = 279.39 degree (counter clockwise)

Step-by-step explanation:

The detailed and step by step explanation is as shown in the attachment