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Aneli [31]
3 years ago
8

classify the polygon by its number of sides. state whether the polygon appears to be regular or not regular

Mathematics
2 answers:
UkoKoshka [18]3 years ago
8 0

Answer: the correct answer is the third option




Kazeer [188]3 years ago
3 0

Answer:

<h2>Pentagon, regular.</h2>

Step-by-step explanation:

The given figure is a polygon which is a figure with at least three sides which enclose certain area.

If a polygon has three sides, it's called a triangle.

If a polygon has four sides, it's called a quadrilateral.

If a polygon has five sides, it's called a pentagon.

Now, polygons can be regular or irregular, this depends on the equivalence of its sides, if the polygon has all equal sides, then it's called regular polygon.

In this case, we have a polygon with 5 equal sides, therefore it's a pentagon, which is a regular polygon with 5 equal sides.

Therefore, the right answer is the last choice.

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tickets at a particular movie theatre have different rates for adults and children. on a friday the theatre sold 4 adult tickets
maria [59]
Using elimination method. Using substitution will cause some fraction problems.

c- child tickets
a- adult tickets

4a + 7c = $83
5a + 6c = $90

multiply by -5

-20a - 35c = -$415
20a + 24c = $360

-11c= -$55
c = $5 child tickets

4a + 7(5)= $83

4a + 35 = $83

4a = $48

a= $12 adult tickets

4 0
3 years ago
HHEELLLPPP MEEhbbnbjhjjh jhjhjhjh
Ber [7]

Answer:

-3x + 36

it is correct answer

4 0
3 years ago
A company that offers tubing trips down a river rents tubes for a person to use for $20 and “cooler” tubes to carry food and wat
aivan3 [116]

The group rent 11 river rent tubes and 4 cooler tubes.

Step-by-step explanation:

Rent for river rent tube = $20

Rent for cooler tubes = $12.50

Total spent = $270

Total tubes rented = 15

Let,

x be the number of river rent tube.

y be the number of cooler tubes.

According to given statement;

x+y=15    Eqn 1

20x+12.50y=270    Eqn 2

Multiplying Eqn 1 by 20

20(x+y=15)\\20x+20y=300\ \ \ Eqn\ 3

Subtracting Eqn 2 from Eqn 3

(20x+20y)-(20x+12.50y)=300-270\\20x+20y-20x-12.50y=30\\7.50y=30

Dividing both sides by 7.5

\frac{7.5y}{7.5}=\frac{30}{7.5}\\y=4

Putting y=4 in Eqn 1

x+4=15\\x=15-4\\x=11

The group rent 11 river rent tubes and 4 cooler tubes.

Keywords: linear equations, subtraction

Learn more about linear equations at:

  • brainly.com/question/8520610
  • brainly.com/question/846474

#LearnwithBrainly

6 0
3 years ago
Traveling with the current, a boat covers a distance 1.5 times greater than traveling against the current in the same amount of
jeka94
Let the speed of the current equal c
and the speed of the boat in still water equal b.

b + c = 1.5 (b - c)
b + c = 1.5b - 1.5c
0.5b = 2.5c
b = 5c

The speed of the current is 1.5 mph so
b = 5 * 1.5 = 7.5 mph
4 0
3 years ago
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
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