A turboprop plane flying with the wind flew 400 mi in 4 h. Flying against the wind, the plane required 5 h to travel the same distance. Find the rate of the plane in calm air and the rate of the wind.

Rate of the plane in calm air is  90 miles/hr and rate of the wind of wind is 10 miles/hr

Step-by-step explanation:

Average speed of plane in with wind = 400 /4 = 100 miles/hr

Average speed of plane against wind = 400/5 = 80 miles/hr

Consider the speed of plane in wind be  x miles/hr and speed of plane against wind  be  y  miles/hr

As such speed of plane in wind would be  x + y miles/hr and speed of plane  against wind would be  x - y miles/hr . i.e

x+y = 100

x-y = 80

by solving these two equation, we get

2x=180

x= 90 miles/hr

y=100-90

y= 10 miles/hr

hence, Rate of the plane in calm air is  90 miles/hr and rate of the wind of wind is 10 miles/hr

answer: the answer is 127 degrees

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )