Mathematics, 26.08.2019 09:30 Kimmi4807

This is a homeork problem from my class
"the life expectancy for females born during 1980-1985 was aproximately 77.6 years. this grew to 78 years during 1985-1990 and to 78.6 years during 1990-1995. construct a model for this data by finding a quadratic function whose graph passes through the points (0,77.6), (5,78), and (10,78.6). use this model to estimate the life expectancy for females born between 1995 and 2000 and for those born between 2000 and 2005."
must show all work and use elimination method

Answers: 2

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Answers

This is a quadratic function:

f(x)=ax²+bx+c

We have three points:
(0 , 77.6)
(5 , 78)
(10 , 78.6)

Then, we make the function wiht these points:
(0 , 77.6)

x=0
f(x)=77.6

a(0)²+b(0)+c=77.6
c=77.6

(5 , 78)
a(5)²+b(5)+77.6=78
25a+5b=78-77.6
25a+5b=0.4   (1)

(10 , 78.6)
a(10)²+b(10)+77.6=78.6
100a+10b=78.6-77.6
100a+10b=1 (2)

With the equations (1) and (2) we have a system of equations:
100a+10b=1
25a + 5b=0.4

We solve this system of equations by method of elimination.

   100a+10b=1
-4(25a+5b=0.4)

   -10b=-0.6 ⇒    b=-0.6/-10=0.06

   100a+10b=1
-2(25a+5b=0.4)

   50a = 0.2 ⇒ a=0.2/50=0.004

We have a, b and c:
a=0.004
b=0.06
c=77.6

Therefore, the quadratic funtion is:
f(x)=0.004x²+0.06x+77.6
x=year of the beginning of the interval - 1980

The life expectancy for females born between 1995 and 2000 is when x=1995-1980=15
Therefore:
f(15)=0.004(15)²+0.06(15)+77.6
f(15)=0.004(225)+0.06(15)+77.6
f(15)=79.4

The life expectancy for females born between 2000 and 2005 is when:
x=2000-1980=20
therefore:
f(20)=0.004(20)²+0.06(20)+77.6
f(20)=1.6+1.2+77.6
f(20)=80.4

The funtion is:
f(x)=0.004x²+0.06x+77.6
x=year of the beginning of the interval - 1980

The life expectancy for females born between 1995 and 2000 is 79.4 years.
The life expectancy for females born between 2000 and 2005 is 80.4 years.

Answer from: tyme66

Solutions

To solve this problem we have to use the quadratic function. An equation where the highest exponent is (usually "x") is a square (2).

Quadratic function = f(x)=ax²+bx+c

Three points have been given to us

⇒ (0 , 77.6)
⇒ (5 , 78)
⇒ (10 , 78.6)

Calculations

(0 , 77.6)

x=0
f(x)=77.6

a(0)²+b(0)+c=77.6
c=77.6

(5 , 78)
a(5)²+b(5)+77.6=78
25a+5b=78-77.6
25a+5b=0.4 (1)

(10 , 78.6)
a(10)²+b(10)+77.6=78.6
100a+10b=78.6-77.6
100a+10b=1 (2)

As you see equations number (1) and (2) have a system

100a+10b=1
  25a + 5b=0.4

We solve this system we have to use elimination.  In the elimination method you can add or subtract the equations to get an equation into one variable.

   100a+10b=1
-4(25a+5b=0.4)
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
   -10b=-0.6

⇒    b=-0.6/-10=0.06

   100a+10b=1
-2(25a+5b=0.4)
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
   50a = 0.2

⇒ a=0.2/50=0.004

Now we have a, b and c:

α = 0.004
β = 0.06
c = 77.6

Therefore, the quadratic function is:

f(x)=0.004x²+0.06x+77.6

x=year of the beginning of the interval - 1980

The life expectancy for females born between 1995 and 2000 is when

x=1995-1980=15

Therefore we do

f(15)=0.004(15)²+0.06(15)+77.6
f(15)=0.004(225)+0.06(15)+77.6
f(15)=79.4

Females between 2000 and 2005

x=2000-1980=20

Now we do

f(20)=0.004(20)²+0.06(20)+77.6

f(20)=1.6+1.2+77.6

f(20)=80.4

The life expectancy for females born between 1995 and 2000 = 79.4 years.

The life expectancy for females born between 2000 and 2005 = 80.4 years.

Answer from: Quest

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.

Me if you can and if you finished this lesson on edg- comment you.

Answer from: Quest

answer: b.48 for

step-by-step explanation:

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This is a homeork problem from my class
"the life expectancy for females born during 1980-1985...

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