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Sagot :
Given:
Customers can expect to lose an average of 2.8 psi (pounds per square inch) per day when the bike is not in use.
1.One bike owner fills his tires to 98 psi on the first day of August
a) Make a table showing the expected daily psi for the tires if no air is added for a week:
Each day the lose is 2.8psi
Day 1: is August 1st
b) What type of sequence can be used to represent the expected psi of the tires?
As between each term of the sequence there is a common difference of -2.8, the sequence is an arithmetic sequence. (It can also be written as a linear relationship)
c) Write the formula for the nth term in the sequence:
Use the next formula to write an arithmetic sequence:
[tex]a_n=a_1+d(n-1)[/tex]a1 is the first term
d is the common difference
[tex]\begin{gathered} a_n=98+(-2.8)(n-1) \\ a_n=98-2.8n+2.8 \\ a_n=100.8-2.8n \end{gathered}[/tex]The formula for the nth term is:
[tex]a_n=100.8-2.8n[/tex]d) Based on the above sequence, on what day would the tire go flat (0 psi)?
Find n when the nth term is 0:
[tex]0=100.8-2.8n[/tex]Solve n:
[tex]\begin{gathered} \text{Subtract 100.8 in both sides of the equation:} \\ 0-100.8=100.8-100.8-2.8n \\ -100.8=-2.8n \\ \\ \text{Divide both sides of the equation into -2.8:} \\ \frac{-100.8}{-2.8}=\frac{-2.8}{-2.8}n \\ \\ 36=n \\ \\ n=36 \end{gathered}[/tex]Then, the tire go flat after 36days, in September 5
e) If the bike owner does not check the tires until day 15, what is the expected psi?
Find 15th term:
[tex]\begin{gathered} a_{15}=100.8-2.8(15) \\ a_{15}=100.8-42 \\ a_{15}=58.8 \end{gathered}[/tex]Then, in day 15 the expected psi is 58.8
f) Find the 25 of the maximum (100)
[tex]100\cdot0.25=25[/tex]Find the psi for which the pressure is in the range of increase accidents:
[tex]100psi-25psi=75psi[/tex]Find the n (day) when nth term is 75:
[tex]\begin{gathered} 75=100.8-2.8n \\ \\ 75-100.8=-2.8n \\ \\ -25.8=-2.8n \\ \\ \frac{-25.8}{-2.8}=n \\ \\ n=9.21 \end{gathered}[/tex]Then, the pressure of the tires will be in the range for increased accidents after approximately 9 daysg) How does the length of time before there is concern for increased accidents change if the bike owner chooses to routinely fill the bike tires to 92 psi?
First term: 92
common difference -2.8
[tex]\begin{gathered} a_n=92-2.8(n-1) \\ a_n=92-2.8n+2.8 \\ a_n=94.8-2.8n \end{gathered}[/tex]Using the formul above solve n when the nth term is 75:
[tex]\begin{gathered} 75=94.8-2.8n \\ 75-94.8=-2.8n \\ -19.9=-2.8n \\ \frac{-19.8}{-2.8}=n \\ \\ n=7.07 \end{gathered}[/tex]Then, the length of the time before there is concern for increased accidents decreases (approximately 2 days)
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