Answer:
only a stop since no buildings would exist for ten years. ... the present day Anna, Highland and Mantua, were bypassed by the railroad ... Today, Anna finds itself as a small, stable, attractive community offering a small ...
Answer:
x y^2 (11 x y - 4)
Step-by-step explanation:
Simplify the following:
4 x^2 y^3 + 2 x y^2 - 2 y - (-7 x^2 y^3 + 6 x y^2 - 2 y)
Factor y out of -7 x^2 y^3 + 6 x y^2 - 2 y:
4 x^2 y^3 + 2 x y^2 - 2 y - y (-7 x^2 y^2 + 6 x y - 2)
-y (-2 + 6 x y - 7 x^2 y^2) = 2 y - 6 x y^2 + 7 x^2 y^3:
4 x^2 y^3 + 2 x y^2 - 2 y + 2 y - 6 x y^2 + 7 x^2 y^3
Grouping like terms, 4 x^2 y^3 + 2 x y^2 - 2 y + 2 y - 6 x y^2 + 7 x^2 y^3 = (4 x^2 y^3 + 7 x^2 y^3) + (2 x y^2 - 6 x y^2) + (2 y - 2 y):
(4 x^2 y^3 + 7 x^2 y^3) + (2 x y^2 - 6 x y^2) + (2 y - 2 y)
x^2 y^3 4 + x^2 y^3 7 = 11 x^2 y^3:
11 x^2 y^3 + (2 x y^2 - 6 x y^2) + (2 y - 2 y)
x y^2 2 + x y^2 (-6) = -4 x y^2:
11 x^2 y^3 + -4 x y^2 + (2 y - 2 y)
2 y - 2 y = 0:
11 x^2 y^3 - 4 x y^2
Factor x y^2 out of 11 x^2 y^3 - 4 x y^2:
Answer: x y^2 (11 x y - 4)
Answer:
a. i. x ≤ 25 ii. 16 < x < 35 iii. 25 < x ≤ 95
b. The second and third car seats are appropriate for a 35 lb child.
Step-by-step explanation:
a. Model those ranges with compound inequalities
Let x represent the car seat.
i. A car seat designed for a child weighing up to and including 25 lb is described by the inequality.
x ≤ 25
ii. A car seat designed for a child weighing between 16 lb and 35 lb is described by the inequality.
16 < x < 35
iii. A car seat designed for a child weighing between 25 lb and 95 lb inclusive is described by the inequality.
25 < x ≤ 95
b. Which car seats are appropriate for a 33-lb child?
Since 35 lb is included in the range of the inequalities for the second and third card seats, the second and third car seats are appropriate for a 35 lb child.
1.) 17 dollars 2.) 3 packs
Answer:
P(system works) = 0.6144
Step-by-step explanation:
Each compoent has an 80% = 0.8 probability of working correctly.
1 and 2
We need at least one of them working.
Either neither works, or at least one works.
Each one has a 1 - 0.8 = 0.2 probability of not working.
0.2*0.2 = 0.04
0.04 of neither working.
1 - 0.04 = 0.96 probability of at least one working
3 and 4
We need both to work, each with a 0.8 probability.
Calculate P(system works).
At least one of the first two, with 0.96 probability
3 and 4, each with 0.8 probability.
0.96*0.8*0.8 = 0.6144
So
P(system works) = 0.6144
