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Phantasy [73]
3 years ago
8

In a cinema theatre, why do we normally not hear echoes even though the sounds are loud?​

Mathematics
1 answer:
ikadub [295]3 years ago
4 0

Answer:

An echo is formed as a result of reflection of the original sound which returns to the listener. You might have experienced Echo in an EMPTY ROOM but no echo is produced after the room is occupied. To avoid echo from being  produced the echo has to be absorbed. Echo in a cinema theatre is absorbed by the WALLS of the theatre.

They are not plain. The walls are corrugated or shaped to form tiny holes or empty spaces to ABSORB the ECHO. Modern Theatre's have ABSORBENT materials sticked to the walls of the theatre.

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10x 6x+8 Solve For X
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Answer:

I had x to be 2

Step-by-step explanation:

a square has all it side to be equal

so 10x=6x+8

you will group like terms 10x-6x=8

4x=8 you divide both side by the coefficient of x which is four so X=2

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A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th
umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

C=3X^{2}+X+2

Compute the expected cost of repair as follows:

E(C)=E(3X^{2}+X+2)

        =3E(X^{2})+E(X)+2

Compute the expected value of <em>X</em> as follows:

E(X)=np

         =4\times 0.09\\=0.36

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

V(X)=np(1-p)

         =4\times 0.09\times 0.91\\=0.3276\\

The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

V(X)=E(Y^{2})-(E(Y))^{2}

Then the formula of E(Y^{2}) is:

E(Y^{2})=V(X)+(E(Y))^{2}

Compute the value of E(Y^{2}) as follows:

E(Y^{2})=V(X)+(E(Y))^{2}

          =0.3276+(0.36)^{2}\\=0.4572

The expected repair cost is:

E(C)=3E(X^{2})+E(X)+2

         =(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73

Thus, the expected repair cost is $3.73.

4 0
3 years ago
Aidan has $65 in his bank account. His brother Michael also has a bank account. They want to calculate the amount of money in bo
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Answer:

m, the amount of money in Michael's bank account

t, the total between the two bank accounts

the equation could be 65+m= t

hope this helps :)

4 0
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