In 1992, the life expectancy of males in a certain country was 61.6 years. In 1995, it was 65.2 years. Let E represent the life expectancy in year t and let t represent the number of years since 1990.

The linear function E(t) that fits the data is... E(t)=_____t + _______ (round to nearest 10th)
Use the function to predict the life expectancy of males in 2006.
E(14) = ___________ (round to nearest 10th)

I come up with:
E(t)=6t + 68.3--- for the 1st answer and

E(14) = 74.9
i.e. 74.88 rounded to nearest 10th

Is it right? if not please, please show and correct me!

To find the linear function E(t) that fits the given data, we need to determine the slope and the y-intercept.

The slope, denoted as m, is calculated by finding the change in the life expectancy divided by the change in years:
m = (Change in life expectancy) / (Change in years)

From 1992 to 1995, the life expectancy increased by 65.2 - 61.6 = 3.6 years.
The corresponding change in years is 1995 - 1992 = 3 years.

So, the slope is: m = 3.6 / 3 = 1.2.

Next, we need to determine the y-intercept, which represents the life expectancy at t = 0, i.e. in the year 1990.

To do this, we can substitute one of the given data points into the equation E(t) = mt + b.

Using the data from 1992, we have:
61.6 = (1.2)(2) + b

Simplifying the equation:
61.6 = 2.4 + b

Rearranging to solve for b:
b = 61.6 - 2.4 = 59.2.

Therefore, the linear function that fits the data is: E(t) = 1.2t + 59.2.

Now, let's predict the life expectancy of males in 2006, which is t = 16. We can substitute this value into the function E(t) = 1.2t + 59.2.

E(16) = 1.2(16) + 59.2 = 19.2 + 59.2 = 78.4.

Rounding to the nearest tenth, the predicted life expectancy of males in 2006 is approximately 78.4 years.

So, your first answer, E(t) = 1.2t + 59.2, is correct.
However, your second answer, E(14) = 74.9, is incorrect. The correct answer is E(14) = 78.4 (rounded to the nearest tenth).