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stellarik [79]
3 years ago
14

Parallelogram WXYZ and parallelogram W" X" Y" Z" are congruent.

Mathematics
1 answer:
Nadya [2.5K]3 years ago
3 0
Reflections, rotations, and translations all change the location, but not the size or shape of the parallelogram.  Congruence describes the line lengths and angle sizes - all of which will be maintained.  Thus, reflections, rotations, and translations will maintain congruence.

All of the options use only reflection, rotation, and translation, so all of these could be the transformation in question.
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For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distr
pshichka [43]

Answer:

See the proof below.

Step-by-step explanation:

Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."

Solution to the problem

For this case we can assume that we have N independent variables X_i with the following distribution:

X_i Bin (1,p) = Be(p) bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:

Z = \sum_{i=1}^N X_i

From the info given we know that N \sim Bin (M,q)

We need to proof that Z \sim Bin (M, pq) by the definition of binomial random variable then we need to show that:

E(Z) = Mpq

Var (Z) = Mpq(1-pq)

The deduction is based on the definition of independent random variables, we can do this:

E(Z) = E(N) E(X) = Mq (p)= Mpq

And for the variance of Z we can do this:

Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2

Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2

And if we take common factor Mpq we got:

Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]

And as we can see then we can conclude that   Z \sim Bin (M, pq)

8 0
3 years ago
a supermarket sells 2 kg and 4 kg of sugar a shipment of 1100 bags of sugar has a total mass of 2900 kg. How many 2 kg bags and
Vaselesa [24]

There are 750 (2 kg) bags and 350 (4 kg) bags in the shipment​

Step-by-step explanation:

The given is:

  • A supermarket sells 2 kg and 4 kg of sugar a shipment of 1100 bags of sugar
  • The total mass of the sugar is 2900 kg

We need to find how many 2 kg bags and 4 kg bags are in the shipment

Assume that x represents the number of 2 kg bags and y represents the number of 4 kg bags

∵ x is the number of 2 kg bags

∵ y is the number of 4 kg bags

∵ The total number of bags are 1100

- Add x and y, then equate the sum by 1100

∴ x + y = 1100 ⇒ (1)

∵ The total mass of sugar is 2900 kg

- Multiply x by 2 and y by 4 to find the mass of the bags, then

  add the two product and equate them by 2900

∴ 2x + 4y = 2900 ⇒ (2)

Now we have a system of equations to solve it

Multiply equation (1) by -2 to eliminate x

∴ -2x - 2y = -2200 ⇒ (3)

- Add equations (2) and (3)

∴ 2y = 700

- Divide both sides by 2

∴ y = 350

- Substitute the value of y in equation (1) to find x

∵ x + 350 = 1100

- Subtract 350 from both sides

∴ x = 750

There are 750 (2 kg) bags and 350 (4 kg) bags in the shipment​

Learn more:

You can learn more about the system of equations in brainly.com/question/6075514

#LearnwithBrainly

4 0
3 years ago
Estimate: 43 divide by 6
Nadya [2.5K]

Answer:

7

Step-by-step explanation:

42/ 6 = 7

8 0
2 years ago
Read 2 more answers
Help ASAP!!!!!!!!!!!!!!!!!​
Karolina [17]

Answer:158 centimeters

Step-by-step explanation:7/10 in centimeters is 70 cm and 1/2 meter in centimeters is 50 so, 38+70+50=158cm

7 0
3 years ago
Read 2 more answers
Anybody help me pls :,)
Korvikt [17]

Answer:

2/3 = 3/2

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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