Answer:
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs is the answer.
Step-by-step explanation:
given that Ty and Gajge are football players.
Carries is 15 for both and average is the same 4 for both.
But on scrutiny we find that maximum and minimum and 6 and 2 for Ty.
Hence range for Ty = 6-2 =4 (2 runs on eithre side of mean)
But for Gajge, highest is 19 and lowest is 2.
i.e. range = 19-2 =17 very much higher than that of Ty
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs.
Answer:
x<-2.5 =(2x+5)-(2x-5) 2. 5-5
3.0
Step-by-step explanation:
PLz make brainlestiset answer
Answer:

Step-by-step explanation:
We are given that

Side of base=4 cm
l=w=4 cm
Height,h=12 cm
We have to find the rate at which the water level rising when the water level is 4 cm.
Volume of pyramid=


Substitute the value

Differentiate w.r.t t

Substitute the values


Answer:
The standard parabola
y² = -18 x +27
Length of Latus rectum = 4 a = 18
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given focus : (-3 ,0) ,directrix : x=6
Let P(x₁ , y₁) be the point on parabola
PM perpendicular to the the directrix L
SP² = PM²
(x₁ +3)²+(y₁-0)² = 
x₁²+6 x₁ +9 + y₁² = x₁²-12 x₁ +36
y₁² = -18 x₁ +36 -9
y₁² = -18 x₁ +27
The standard parabola
y² = -18 x +27
Length of Latus rectum = 4 a = 4 (18/4) = 18