Answer :
Certainly! Let's work through the problem step-by-step.
### a) Finding the formula which connects [tex]\( C \)[/tex] and [tex]\( U \)[/tex]
We know that the charge [tex]\( C \)[/tex] of the telephone company is partly constant and partly varies with the number of units of call [tex]\( U \)[/tex]. We can express this relationship as:
[tex]\[ C = a + bU \][/tex]
where:
- [tex]\( a \)[/tex] is the constant part of the cost.
- [tex]\( bU \)[/tex] is the part of the cost that varies linearly with the units [tex]\( U \)[/tex] of call.
Given:
1. The cost for 90 units is [tex]$1120$[/tex]. So,
[tex]\[ 1120 = a + 90b \][/tex]
2. The cost for 120 units is [tex]$1216$[/tex]. So,
[tex]\[ 1216 = a + 120b \][/tex]
We now have a system of linear equations:
[tex]\[ \begin{cases} 1120 = a + 90b \\ 1216 = a + 120b \end{cases} \][/tex]
To find [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we can solve this system of equations.
Subtract the first equation from the second to eliminate [tex]\( a \)[/tex]:
[tex]\[ (1216 - 1120) = (a + 120b) - (a + 90b) \][/tex]
[tex]\[ 96 = 30b \][/tex]
[tex]\[ b = \frac{96}{30} \][/tex]
[tex]\[ b = \frac{16}{5} \][/tex]
Now, substitute [tex]\( b \)[/tex] back into the first equation to find [tex]\( a \)[/tex]:
[tex]\[ 1120 = a + 90 \cdot \frac{16}{5} \][/tex]
[tex]\[ 1120 = a + 90 \cdot 3.2 \][/tex]
[tex]\[ 1120 = a + 288 \][/tex]
[tex]\[ a = 1120 - 288 \][/tex]
[tex]\[ a = 832 \][/tex]
Thus, the formula connecting [tex]\( C \)[/tex] and [tex]\( U \)[/tex] is:
[tex]\[ C = 832 + \frac{16}{5}U \][/tex]
### b) Finding [tex]\( C \)[/tex] when [tex]\( U \)[/tex] is 150 units
Now we need to find the cost [tex]\( C \)[/tex] when the number of units [tex]\( U \)[/tex] is 150. Using the formula we derived:
[tex]\[ C = 832 + \frac{16}{5} \cdot 150 \][/tex]
First, calculate the varying part [tex]\( \frac{16}{5} \cdot 150 \)[/tex]:
[tex]\[ \frac{16}{5} \cdot 150 = 16 \cdot 30 = 480 \][/tex]
Now, add this to the constant part:
[tex]\[ C = 832 + 480 \][/tex]
[tex]\[ C = 1312 \][/tex]
Therefore, the cost [tex]\( C \)[/tex] for 150 units is [tex]\( 1312 \)[/tex].
### Summary:
1. The formula connecting [tex]\( C \)[/tex] and [tex]\( U \)[/tex] is:
[tex]\[ C = 832 + \frac{16}{5}U \][/tex]
2. The cost [tex]\( C \)[/tex] when [tex]\( U \)[/tex] is 150 units is:
[tex]\[ C = 1312 \][/tex]
### a) Finding the formula which connects [tex]\( C \)[/tex] and [tex]\( U \)[/tex]
We know that the charge [tex]\( C \)[/tex] of the telephone company is partly constant and partly varies with the number of units of call [tex]\( U \)[/tex]. We can express this relationship as:
[tex]\[ C = a + bU \][/tex]
where:
- [tex]\( a \)[/tex] is the constant part of the cost.
- [tex]\( bU \)[/tex] is the part of the cost that varies linearly with the units [tex]\( U \)[/tex] of call.
Given:
1. The cost for 90 units is [tex]$1120$[/tex]. So,
[tex]\[ 1120 = a + 90b \][/tex]
2. The cost for 120 units is [tex]$1216$[/tex]. So,
[tex]\[ 1216 = a + 120b \][/tex]
We now have a system of linear equations:
[tex]\[ \begin{cases} 1120 = a + 90b \\ 1216 = a + 120b \end{cases} \][/tex]
To find [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we can solve this system of equations.
Subtract the first equation from the second to eliminate [tex]\( a \)[/tex]:
[tex]\[ (1216 - 1120) = (a + 120b) - (a + 90b) \][/tex]
[tex]\[ 96 = 30b \][/tex]
[tex]\[ b = \frac{96}{30} \][/tex]
[tex]\[ b = \frac{16}{5} \][/tex]
Now, substitute [tex]\( b \)[/tex] back into the first equation to find [tex]\( a \)[/tex]:
[tex]\[ 1120 = a + 90 \cdot \frac{16}{5} \][/tex]
[tex]\[ 1120 = a + 90 \cdot 3.2 \][/tex]
[tex]\[ 1120 = a + 288 \][/tex]
[tex]\[ a = 1120 - 288 \][/tex]
[tex]\[ a = 832 \][/tex]
Thus, the formula connecting [tex]\( C \)[/tex] and [tex]\( U \)[/tex] is:
[tex]\[ C = 832 + \frac{16}{5}U \][/tex]
### b) Finding [tex]\( C \)[/tex] when [tex]\( U \)[/tex] is 150 units
Now we need to find the cost [tex]\( C \)[/tex] when the number of units [tex]\( U \)[/tex] is 150. Using the formula we derived:
[tex]\[ C = 832 + \frac{16}{5} \cdot 150 \][/tex]
First, calculate the varying part [tex]\( \frac{16}{5} \cdot 150 \)[/tex]:
[tex]\[ \frac{16}{5} \cdot 150 = 16 \cdot 30 = 480 \][/tex]
Now, add this to the constant part:
[tex]\[ C = 832 + 480 \][/tex]
[tex]\[ C = 1312 \][/tex]
Therefore, the cost [tex]\( C \)[/tex] for 150 units is [tex]\( 1312 \)[/tex].
### Summary:
1. The formula connecting [tex]\( C \)[/tex] and [tex]\( U \)[/tex] is:
[tex]\[ C = 832 + \frac{16}{5}U \][/tex]
2. The cost [tex]\( C \)[/tex] when [tex]\( U \)[/tex] is 150 units is:
[tex]\[ C = 1312 \][/tex]
