In Exercise 6, suppose that there is a probability of .01 that a digit is incorrectly sent over a communication channel (i. e., that a digit sent as a 1 is received as a 0, or a digit sent as a 0 is received as a 1). Consider a message that consists of exactly 60% 1s. a. What is the proportion of 1s received at the end of the channel

The proportion of 1s received at the end of the channel is of 0.598.

Step-by-step explanation:

In this question:

60% of the bits sent are 1, 100 - 60 = 40% of the bits sent are 0.

For each bit receive, 0.01 probability of an error, so 0.99 probability it is correct.

a. What is the proportion of 1s received at the end of the channel?

99% of 60%(sent 1s, received 1).

1% of 40%(sent 0s, received 1). So

The proportion of 1s received at the end of the channel is of 0.598.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

b. 1.25 hr

step-by-step explanation:

if you have a constant speed of 60 mph, then you have a ratio between 60 mi. to 1 hr. dividing both sides by 60 we get the ratio

1 mi. to

then mult. both sides by 75 we get the ratio

75 mi. to

simplifying we get

75 mi. to

or 1.25 hr.