Answer: 250
10 times itself or 10^2 = 100
(100)(5)=500
500/2= 250
7 is bases number two , I'm not sure it sounds like similar makes more sense , but not sure for 2
Answer:
A) (1 s, 2.3 s)
B) (-4 m/s², 3.8 m/s²)
Step-by-step explanation:
The car's position which is the distance is given by the equation;
s(t) = t³ - 5t² + 7t
A) Velocity is the first derivative of the distance. Thus;
v(t) = ds/dt = 3t² - 10t + 7
At v = 0, we have;
3t² - 10t + 7 = 0
Using quadratic formula, we have;
t = 1 and t = 2.3
Thus, time at velocity of 0 is t = (1 s, 2.3 s)
B) acceleration is the derivative of the velocity. Thus;
a(t) = dV/dt = 6t - 10
At velocity of 0, we got t = 1 and t = 2.3
Thus;
a(1) = 6(1) - 10 = -4 m/s²
a(2.3) = 6(2.3) - 10 = 3.8 m/s
Thus, a(t) at v = 0 gives; (-4 m/s², 3.8 m/s²)
Answer:
−35.713332 ; 0.313332
Step-by-step explanation:
Given that:
Sample size, n1 = 11
Sample mean, x1 = 79
Standard deviation, s1 = 18.25
Sample size, n2 = 18
Sample mean, x2 = 96.70
Standard deviation, s2 = 20.25
df = n1 + n2 - 2 ; 11 + 18 - 2 = 27
Tcritical = T0.01, 27 = 2.473
S = sqrt[(s1²/n1) + (s2²/n2)]
S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]
S = 7.284
(μ1 - μ2) = (x1 - x2) ± Tcritical * S
(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284
(μ1 - μ2) = - 17.7 ± 18.013332
-17.7 - 18.013332 ; - 17.7 + 18.013332
−35.713332 ; 0.313332