The average daily minimum temperature for Pablo’s hometown can be modeled by the function f(x)=15.3sin(πx6)+44.1 , where f(x) is the temperature in °F and x is the month.

x = 0 corresponds to January.

What is the average daily minimum temperature in June?

Round to the nearest tenth of a degree if needed.

Use 3.14 for π .

really stuck could someone help out??

since the period of your sine curve should be 12

2π/k = 12
k = 6/π
your equation probably should have been:
f(x)=15.3sin(πx/6)+44.1 instead of what you typed.

since Jan ----> x = 0
June ----> x = 5

so just sub in x = 5 and evaluate

Let me know what you got.

i did but im still stuck

stuck how? Just replace x with 5 and evaluate!

too bad you could not be bothered to show us what you did.

f(x) = 15.3sin(πx/6)+44.1
f(5) = 15.3sin(5π/6)+44.1
= 15.3(1/2)+44.1
...

Okay so I worked on the same problem and I got this answer, rounded to the nearest tenth like the problem says; 10.9

Here's my work:

15.3 sin (3.14(5)/6)+44.1
15.3 sin (15.7/6)+44.1
15.3 sin(2.616666667)+44.1
I5.3 sin(46.71666667)
I got=10.91958335

Unfortunately, I can't put in screenshots here :|
Hopefully my answer is correct!

To find the average daily minimum temperature in June, we need to substitute the month of June into the given function.

In this case, x = 6 corresponds to June since x = 0 corresponds to January.

The given function is f(x) = 15.3sin(πx/6) + 44.1, where f(x) represents the temperature in °F for a given month x.

So, to find the average daily minimum temperature in June, we plug in x = 6 into the function:

f(6) = 15.3sin(π(6)/6) + 44.1

Simplifying this expression, we get:

f(6) = 15.3sin(π) + 44.1

Since sin(π) equals 0, we have:

f(6) = 15.3(0) + 44.1

Which simplifies to:

f(6) = 0 + 44.1

Therefore, the average daily minimum temperature in June, according to the given function, is 44.1°F.