How many times is 1 trillion greater than 1 billion
1 answer:
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Answer:
B. 55.10
Step-by-step explanation:
Given:
1 pair of Shoes for 1st year = $50
2 pair of Sock for 1st year = $2 each = 22= $4
CPI for year 1 = Price of shoes for 1st year + Price of Sock for 1st year= $50 + $4 = $54
Now
1 pair of Shoes for 2nd year = $51
2 pair of Sock for 2nd year = $2.05 each = 2 2.05 =$4.10
CPI for Year 2 = Price of shoes for 2nd year + Price of Sock for 2nd year= $51 + $4.1 = $55.10
Hence CPI for Year 2 is $55.10
Let the cost of gasoline in the year 2000 be represented b the equation
y = a + b*x
where
x = months, counted from January
y = cost, dollars
The given data in the table is
Month: Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
x, months: 1 2 3 4 5 6 7 8 9 10 11 12
y, dollars: --- --- --- --- 1.76 2.13 --- --- --- --- --- ---
When x = 5, y = 1.76.
Therefore
a + 5b = 1.76 (1)
When x = 6, y = 2.13
Therefore
a + 6b = 2.13 (2)
Subtract equation (1) from (2).
a + 6b - (a + 5b) = 2.13 - 1.76
b = 0.37
From (1), obtain
a = 1.76 - 5b
= 1.76 - 5*0.37
= -0.09
The required equation is
y = 0.37x - 0.09
The graph shows the line, with the given data for May and June.
Answer: D. y = 0.37x - 0.09
y = a + b*x
where
x = months, counted from January
y = cost, dollars
The given data in the table is
Month: Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
x, months: 1 2 3 4 5 6 7 8 9 10 11 12
y, dollars: --- --- --- --- 1.76 2.13 --- --- --- --- --- ---
When x = 5, y = 1.76.
Therefore
a + 5b = 1.76 (1)
When x = 6, y = 2.13
Therefore
a + 6b = 2.13 (2)
Subtract equation (1) from (2).
a + 6b - (a + 5b) = 2.13 - 1.76
b = 0.37
From (1), obtain
a = 1.76 - 5b
= 1.76 - 5*0.37
= -0.09
The required equation is
y = 0.37x - 0.09
The graph shows the line, with the given data for May and June.
Answer: D. y = 0.37x - 0.09