I hope this helps you
33.3+30.6
63.9
Answer:
25
Step-by-step explanation:
To calculate rate, divide the number of cans by the time to get "Cans per minute".
300/15min = 20/min
Let m represent minutes and c for cans.
We write an equation for the problem:
c = 20m
We want to know the time for 500 cans, so substitute 500 for c.
500 = 20m
Isolate m and solve:
m = 500/20
m = 25
It will take 25 minutes to put 500 labels.
Answer:
40 mL of vinegar,
280 mL of dressing
Step-by-step explanation:
Let v = the milliliters ( mL ) of 100% vinegar,
Then it should be that 320 - v = mL of dressing.
v + .12( 320 - v ) = .23( 320 ) - So in this case the mL of vinegar is associated with the percent of vinegar composed. 100 percent is, in other words, 1, and is multiplied by " v " the mL of 100% vinegar. 1 
v is also v, and so is written as such in our equation. 0.12 is the decimal form of the 12% vinegar, associated with the mL of dressing - as the italian dressing is composed of 12% vinegar. 23% is 0.23 in decimal form, multiplied by the mL of of vinegar in the mixture, 320 mL.
Let's solve for the mL of 100% vinegar, subtracting from 320 to receive the mL of dressing,
v + .12( 320 - v ) = .23( 320 ) - Distribute " .12 "
v + 38.4 - 0.12v = .23( 320 ) - Multiply " 0.32 " by " 320 "
v + 38.4 - 0.12v = 73.6 - Combine like terms and add / subtract
0.88v = 35.2 - Divide 0.88 on either side
v = 40 mL of vinegar,
320 - v = 320 - 40 = 280 mL of dressing
Answer:
(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010
(b)
(1 – p) = 0.010
(c)
E(x) = 25000 x 0.010
= 259.2
Explanation has given below.
Step-by-step explanation:
Solution:
(a) Probability that a triplet is decoded.
2 out of three
P = 0.94, n = 3
m= no of correct bits
m bit (3, 0.94)
At p(m≤1) = B (1; 3, 0.94)
= 0.010
(b) Using your answer to part (a),
(1 – p) = 0.010
Error for 1 bit transmission error.
(c) How does your answer to part (a) change if each bit is repeated five times (instead of three?
P( m ≤ 2 )
L = Bit (5, 0.94)
= B (2; 5, 0.94)
= 0.002
(d) Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?
Given:
h = 25000
Bits are switched during transmission between two computers = 6% = 0.06
m = Bit (25000, 0.06)
E(m) = np
= 25000 x 0.06
= 1500
m = Bit (25000, 0.01)
E(m) = 25000 x 0.010
= 259.2
