All categories

1 year ago

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

Mathematics

1 answer:

ikadub [295]1 year 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.

You might be interested in

Hailey paid $13 for 1 3/7 kg of sliced salami <br><br> What was the cost per kilogram of salami

JulijaS [17]

Translate words into numbers.

13/(1 3/7)

First convert to improper fraction.

13/(10/7)

Then, note that division is the same as multiplying by the divisor's reciprocal.

13x7/10=9.1

answer: $9.10 per kilogram

4 0

2 years ago

Read 2 more answers

Here are Babe Ruth's home run totals for the New York Yankees during the years 1920 - 1934:54, 59, 35, 41, 46, 25, 47, 60, 54, 4

Irina-Kira [14]

Given data is 54, 59, 35, 41, 46, 25, 47, 60, 54, 46, 49, 46, 41, 34, 22

Arranging it in increasing order:

22 25 34 35 41 41 46 46 46 47 49 54 54 59 60

Number of data items, n=15.

Finding Median, First Quartile, and Third Quartile

$Finding\;\;median\\M=\frac{n+1}{2}th\;\;term =\frac{15+1}{2}=8th\;\;term\;\; i.e. \;\;46$

22 25 34 35 41 41 46 46 46 47 49 54 54 59 60

Now finding First Quartile from set: 22 25 34 35 41 41 46

Q₁ = middle term i.e. 35

Now finding Third Quartile from set: 46 47 49 54 54 59 60

Q₃ = middle term i.e. 54

So we have Q₁ = 35, M = 46, and Q₃ = 54. Lower bound is 22 and Upper bound is 60.

While drawing the given data on number line and marking First quartile, Median, Third quartile on number line. We can clearly observe that option A matches with given data points.

Hence, option A i.e. Box-plot A is the correct choice.

5 0

2 years ago

Read 2 more answers

Divide 3p2 + 6p + 2 by 3p2.

kondor19780726 [428]

We need to divide

$\frac{3p^2+6p+2}{3p}$

So, we can divide each terms of the numerator by 3p. So,

$\frac{3p^2}{3p} +\frac{6p}{3p} +\frac{2}{3p}$

= $p+2+\frac{2}{3p}$

\frac{2}{3p} [/tex] cannot be further simplify. So, the final answer is:

$p+2+\frac{2}{3p}$

4 0

2 years ago

Read 2 more answers

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined: