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3 years ago

Please help 10 points i only have 21

Mathematics

2 answers:

vaieri [72.5K]3 years ago

7 0

The answer is C. (3 1/4 cups) :)

strojnjashka [21]3 years ago

3 0

It’s c I had this question

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Solve the system by graphing:

Georgia [21]

Answer:

The graphs intersect at (6,-1)

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3 years ago

2^4x3^2 as a single exponent

dolphi86 [110]

Step-by-step explanation:

2x2=4x2=8x2=16

3x3=9

16x9=144

hope it helps!

if not, sry.

7 0

3 years ago

What is the solution to this inequality?<br> –16x > –80

Kisachek [45]

Answer:

x < 5

Step-by-step explanation:

–16x > –80

Divide each side by -16, remembering to flip the inequality

-16x/-16< -80/-16

x < 5

7 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

4 years ago

Find an explicit rule for the nth term of the sequence.

REY [17]

9514 1404 393

Answer:

(a) an = -4·2^(n-1)

Step-by-step explanation:

The form for the general term is ...

an = a1·r^(n-1) . . . . . a1 is the first term; r is the common ratio

For a1=-4 and r=-8/-4 = 2, the rule will be ...

an = -4·2^(n-1)

8 0

3 years ago

Read 2 more answers