Let x and y be the length and width of the rectangle respectively
2(x+y)=32...(1)
x=2y-5...(2)
From (1),we have
2x+2y=32...(3)
By sub.(2) into (3),we have
2(2y-5)+2y=32
4y-10+2y=32
6y=42
y=7
By sub.y=7 into (2),we have
x=2*7-5
x=14-5
x=9
Length =9m
Width =7m
Answer:
64 slices
Step-by-step explanation:
The first step to solving this is to find out how many slices of bread there were in total. We know that each loaf of bread had 16 slices, and we know that there were 8 loaves. To get the amount of slices, we multiply the amount of slices in each loaf by the number of loaves there are.
16⋅8=128 slices of bread in total
Now that we know the total number of slices, we need to find how many sandwiches they made. If they used two slices per sandwich, we would divided the total number of bread slices by two.
128÷2=64.
I suppose the / are absolute value lines.... so ill answer it that way, -(-2) simplifies to 2. -/-2/ simplifies to -2
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
