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3 years ago

Reflected over the x-axis, then translated 6 units left

1 answer:

7 0

Take the original coordinate (x,y)
to reflect over x-axis change it to (x,-y)
*basically change the sign of y*
to translate 6 units left (x-6,y)
* basically subtract 6 from x
So your new point should be (x-6,-y)

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Which of the following are true statements using the Associative Property of Multiplication?

dangina [55]

Answer:

All of the following statements are true because the products, three or four numbers, remain the same regardless of how the numbers are grouped.

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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:

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

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4 years ago

17. Mary Palm's checking account had a starting balance of $785.63. She wrote a check for $57.00 for groceries and a check for $

Scrat [10]

Answer:B 660.88

Step-by-step explanation:

First, 785.63-57=728.63

Then 728.63 - 125=603.63

Last add 603.63 to 125.00

To give 660.88

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3 years ago

Which fraction is between -5/8 and -1/2

Rzqust [24]

Answer:

The answer is −1.125...

Step-by-step explanation:

Hope this helps! :)

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4 years ago

When you are determining The probability of an event why must the probability be between 0 and 1

slega [8]

The probability of an event cannot be less than 0 because 0 means it's impossible. The probability of an event cannot be more than 1 because 1 means that it's certain that it will happen. That's why the probability must be between 0 and 1.

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4 years ago