For competition,the cheerleading team,is preparing a dance routine that must last 4 minutes,with a variation of +- 5 seconds. find the least and greatest possible times for the routine in minutes and seconds. find the least and greatest possible times in seconds

Ideal timeline of the dance routine = 4 minutes = 4 × 60 seconds = 240 seconds

Variation allowed in the dance routine timeline = +- 5 seconds

Let the timeline of the dance routine be T

⇒ 240 seconds - 5 seconds < T < 240 seconds + 5 seconds

⇒ 235 seconds < T < 245 seconds

⇒ minutes < T < minutes

⇒ 3.92 minutes < T < 4.08 minutes

So the least possible time of the dance routine can be 3.92 minutes (or 235 seconds) and the greatest possible time of the dance routine can be 4.08 minutes (or 245 seconds)

It is a linear function.

Step-by-step explanation:

 It is a linear equation because it is in a straight line. The equation can be written in standard form as in y=mx+b, so it is a linear function. You can find an equation for this function by first picking a point on the line. I'd perfer to choose the line that goes through 3. As you know in the equation y=mx+b, the b in this equation is the y-intercept. Your line goes through 3. So the equation would be y=mx+3. Then you just need to find the slope by using the rise/run equation. Go down 3 times and 2 to the left. Your destination will then be the point on (0,-2). That is your rise over run. The equation then would be y= 3/2+3.