3 months, I remember a question like this awhile back.
35,23,32,25,27,31,29,30
1st: Order data from smaller to greater: 23,25,27,29,30,31,32,35
2nd Find the median of the set (central value). Since the number of data is even, then the median is the average between 29 & 30 = (29+30)/2 =29.5,(we will not use the median value, just for indication)
3rd, Find the central value of the subset on the left of 29 (that is 23,25,27)
This lower median =25 . This is called Q1
4th , Do the same with the subset on the right of 30 (that is 31,32,35).
This higher median = 32 is called Q3 (for your info the 1st median=29.5 is Q2)
The IQR or inter quartile range is Q3 - Q1 (or upper - lower Q)
IQR = 32-25) =7
Answer:
The answer to your question is 13.2 ft
Step-by-step explanation:
Data
length of the ladder = hypotenuse = 16 ft
adjacent side = 9 ft
opposite side = ?
Process
1.- To solve this problem use the Pythagorean theorem
c² = a² + b²
Solve for b²
b² = c² - a²
2.- Substitution
b² = 16² - 9²
3.- Simplification
b² = 256 - 81
b² = 175
b = 13.2 ft
4.- Conclusion
The ladder is 13.2 ft resting against the building.
Answer:
1.7 seconds
Step-by-step explanation:
we have
![H=-16t^2+vt+s](https://tex.z-dn.net/?f=H%3D-16t%5E2%2Bvt%2Bs)
where
v is the initial velocity (in feet per second)
s represents the initial height (in feet)
In this problem we have
---> because the object is dropped
![s=45\ ft](https://tex.z-dn.net/?f=s%3D45%5C%20ft)
substitute
![H=-16t^2+45](https://tex.z-dn.net/?f=H%3D-16t%5E2%2B45)
Remember that
When the chestnut hit the ground the value of H is equal to zero
so
For H=0
![-16t^2+45=0](https://tex.z-dn.net/?f=-16t%5E2%2B45%3D0)
solve for t
![16t^2=45\\t^2=2.8125\\t=\pm1.7\ sec](https://tex.z-dn.net/?f=16t%5E2%3D45%5C%5Ct%5E2%3D2.8125%5C%5Ct%3D%5Cpm1.7%5C%20sec)
therefore
The solution is t=1.7 sec