<- Distributive Property
<- Combine Like Terms
If you're trying to solve for 0:
<- Subtracted 18 from both sides
<- Divided both sides by 60 and then simplified.
<- Fraction Form
<- Decimal Form
Give Brainliest for simple answer plz :P
IF MARTHA EARNS 3.25/HOUR FRO BABYSITTING (SHE IS UNDERPAID!)
IN 1.5 HOURS=3.25*1.5=4.88 DOLLARS
IN 2.5 HOURS=3.25*2.5=8.13 DOLLARS
IN 3.5 HOURS=3.25*3.5=11.38 DOLLARS
The average rate of change over some interval [a, b] is equal to the slope of the secant line from (a, h(a)) to (b, h(b)).
h(t) is a quadratic function, so its graph is a parabola, and in particular it's one that has a minimum of -4 when t = 2.
The secant line over an interval [a, b] will have a negative slope if the distance from a to 2 is larger than the distance from b to 2.
(A) If a = 4 and b = 5, then |a - 2| = 2 and |b - 2| = 3, so the slope and hence average rate of change is positive.
(B) If a = -1 and b = 5, then |a - 2| = 3 and |b - 2| = 3, so this ARoC is zero.
(C) If a = 0 and b = 4, then |a - 2| = 2 and |b - 2| = 2, so this ARoC is also zero.
(D) If a = -1 and b = 4, then |a - 2| = 3 and |b - 2| = 2, so this ARoC is negative.
Answer:
13.05MbpS < x < 22.15 Mbps
Step-by-step explanation:
Confidence interval CI can be expressed in the form;
Lower bound < x < upper bound
Given;
CI = (13.046, 22.15)
Lower bound = 13.046 = 13.05 Mbps (to two s.f.)
Upper bound = 22.15 Mbps
So, the Confidence interval is given as;
13.05 MbpS < x < 22.15 Mbps
