Answer:
Short piece: 9 ft
Long piece: 41 ft
Step-by-step explanation:
The first step to solving this problem is to translate the given information into some equations. Since we know the total length of the rope is 50 feet, the solution when adding the equation for the short piece and the equation for the long piece must be 50. The information from the second sentence can be translated into a mathematical expression. The phrase "5 more than 4 times" has two key words. They are "times" and "more". "Times" implies multiplication while "more" implies addition. Therefore this sentence becomes the expressions 5+4x, where x is the length of the short piece.
The equations we have from the problem statement are:
L = 5+4x where L represents the length of the long piece and x represents the length of the short piece
x + L = 50
Substituting the equation for the long piece into the equation for total length:
x + (5+4x) = 50
5 + 5x = 50
5x = 45
x = 9 ft
Substituting x = 9 into the equation for the long piece:
L = 5 + 4(9)
L = 5 + 36
L = 41 ft
Checking the answers by substituting x = 9 and L = 41 into the equation for total length:
x + L = 50
9 + 41 = 50
50 = 50
Answer:
0.614125
Step-by-step explanation:
Given that a manufacturing process with a quality inspection station has on an average 15% of parts are defective.
As soon as one defective part is found, the process is stopped.
We find that number of defectives would be binomial because each part randomly selected has a constant probability of 0.15 being defective
Probability that at least 11 total parts will be inspected before the process is stopped/8 parts have been inspected without finding a defective part
=
= Probability of 9th, 10th, 11th should not be defective
= 
Radioactive half-life is the time it takes for half an amount of radioactive material to decay into something else. In this case, it is assumed that the decay product is not radioactive or otherwise hazardous.
We must use the radioactive decay formula to determine at what time the radiation reaches a safe level.
A = Ao[e^(-0.693)(t)(t 1/2) where t 1/2 is the half-life, t is elapsed time, Ao is the original quantity, A is the future quantity.
We are given a half-life of 2.4 days , an Ao of 1.25 and an A of 1.00:
1.00 = (1.25)e^(-0.693)(2.4)t
1.00/1.25 = e^(-1.6632)t
0.8 = e^(-1.6632)t
t = 0.135 days = 3 hrs 15 min
This is the amount of time to a "safe level" using only radioactive decay, not venting or other means.
Answer:
b= 10
Step-by-step explanation:
