I need help please answer asap

If jill climed 100 feet more how many feet above sea level would she be

Anit [1.1K]

You have to have a better question to answer that, 100 feet more what?

5 0

3 years ago

Based on the family the graph below belongs to, which equation could represent the graph?

tatuchka [14]

Answer: y=-0.6x^3- 1

Step-by-step explanation:

In the graph, first we must see the points where the line intersects the axis.

We can see that it goes throug the point (0, -1)

Now, of the given options the only two that are 0 when x is -1, are:

f1(x) =y = 0.6*x^2 - 1

f2(x) =y = -0.6x^3 - 1

We also can see that for all the positive values of x, y has a negative value, so let's value bot functions in x = 2, and see what happens.

f1(2) = 0.6*4 - 1 = 1.4

f2(2) = -0.6*8 - 1 = -5.8

so the correct option is

y=-0.6x^3- 1

3 0

3 years ago

Read 2 more answers

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

Can you help me with this please

djyliett [7]

The answer to this problem is simple all you have to do is pemdas and you get the answer 0.6

5 0

3 years ago

***Please Help***

grin007 [14]

A 7 to the right 2 up

8 0

3 years ago

I need help please answer asap​

I need help please answer asap