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Nat2105 [25]
3 years ago
11

parallelogram ABCD is reflected over the y axis, followed by a reflection over the x axis, and then rotated 180 degrees about th

e origin,. What is the location of point A after the transformations are complete? a (-5,-1) b. (5,1) c.(-5,1) d.(5,-1)

Mathematics
1 answer:
mixer [17]3 years ago
7 0

Answer: Option c.

Step-by-step explanation:

The missing figure is attached.

Given a point (x,y):

1. If it is reflected over the y-axis:

(x,y) → (-x,y)

2. If it is reflected over the x-axis:

(x,y) → (x,-y)

3. If it is rotated 180 degrees about the origin:

(x,y) → (-x, -y)

In this case, you can identify that the point A is:

A(-5,1)

Then:

If it is reflected over the y-axis:

(-5,1) → (5,1)

If then it is reflected over the x-axis:

(5,1) → (5,-1)

Finally, if this is followed by a rotationf of 180 degrees about the origin, this is:

(5,-1) → (-5,1)

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A new bank customer with ​$5,000 wants to open a money market account. The bank is offering a simple interest rate of ​1.8%.
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Answer:

a) 900 dollars

b) 5900 dollars

Step-by-step explanation:

The complete question is

A new bank customer with $5,000 wants to open a money market account. The bank is offering a simple interest rate of 1.8%. a. How much interest will the customer earn in 10 years? b. What will the account balance be after 10 years?

solution

a) The simple interest SI = P*R*T

Where P is the principal amount

R is the rate of interest and

T is the time period

Substituting the given values, we get -

SI

= 5000 *\frac{1.8}{100} *10\\= 900 dollars

b) Amount is the sum of Principal and simple interest

A = 5000 + 900 = 5900 dollars

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2 years ago
Which best describes bias, if any, from the sample surveyed?
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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
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lutik1710 [3]

Answer:

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4. n(4 - n)

Step-by-step explanation:

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For 2, you can also only factor out the constant, but remember that you include the 1 when dividing 5 by 5.

For 3, you can factor out the constant and the variables.

For 4, you can only factor out the variable.

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