Respuesta :
Answer:
(a). 33.05 feet.
(b). In year 4150.
Step-by-step explanation:
Let [tex]x=0[/tex] represent year 1900.
We have been given the average sea-level in 1900 at London-bridge was 33 feet. In 1990 it was 33.08 feet.
(a). We have two points (0,33) and (90,33.08).
Let us find slope using both points as:
[tex]m=\frac{33.08-33}{90-0}=\frac{0.08}{90}=0.000889[/tex]
Let us find equation of the line representing the given information as:
[tex]y-33=0.000889(x-0)[/tex]
[tex]y-33=0.000889x[/tex]
[tex]y=0.000889x+33[/tex]
To find the average sea-level was in 1961, we will substitute [tex]x=61[/tex] in our equation as:
[tex]y=0.000889(61)+33[/tex]
[tex]y=0.054229+33[/tex]
[tex]y=33.054229\approx 33.05[/tex]
Therefore, the average sea level in 1961 was 33.05 feet.
(b). To solve part (b), we will substitute [tex]y=35[/tex] in our equation as:
[tex]35=0.000889x+33[/tex]
[tex]35-33=0.000889x+33-33[/tex]
[tex]2=0.000889x[/tex]
[tex]x=\frac{2}{0.000889}[/tex]
[tex]x=2249.718785\approx 2250[/tex]
Since [tex]x=0[/tex] represent year 1900, so we will add 1900 to 2250 as:
[tex]1900+2250=4150[/tex]
Therefore, in 4150, the average sea-level will be 35 feet.
The true statements are:
a. The average sea-level in 1961 is 33.05 feet
b. The average sea-level will be 35 feet in 4150
The given parameters are:
- The average sea-level in 1900 is 33 feet
- The average sea-level in 1990 is 33.08 feet
(a) The average sea level in 1961
Using linear interpolation, we have:
[tex]\frac{y_2 - y_1}{x_2 -x_1} = \frac{y -y_1}{x -x_1}[/tex]
So, the equation becomes
[tex]\frac{33.08 - 33}{1990 -1900} = \frac{y -33}{1961 -1900}[/tex]
Evaluate the differences
[tex]\frac{0.08}{90} = \frac{y -33}{61}[/tex]
Multiply both sides by 61
[tex]y - 33 = 61 \times \frac{0.08}{90}[/tex]
[tex]y - 33 = 0.05[/tex]
Add 33 to both sides
[tex]y = 33.05[/tex]
Hence, the average sea-level in 1961 is 33.05 feet
(b) The year when the average sea level is 35 feet
Using linear interpolation, we have:
[tex]\frac{y_2 - y_1}{x_2 -x_1} = \frac{y -y_1}{x -x_1}[/tex]
So, the equation becomes
[tex]\frac{33.08 - 33}{1990 -1900} = \frac{35 -33}{x-1900}[/tex]
Evaluate the differences
[tex]\frac{0.08}{90} = \frac{2}{x-1900}[/tex]
Take the inverse of both sides
[tex]\frac{90}{0.08} = \frac{x-1900}{2}[/tex]
[tex]1125 = \frac{x-1900}{2}[/tex]
Multiply both sides by 2
[tex]2250 = x-1900[/tex]
Add 1900 to both sides
[tex]x = 4150[/tex]
Hence, the average sea-level will be 35 feet in 4150
Read more about interpolation at:
https://brainly.com/question/12862813
