Answer:
it would be 12 to the second power.. i think or 24 to the second power
Step-by-step explanation:
<h2>
<em>step </em><em>-</em><em>1</em><em> </em></h2>
<em>Changes made to your input should not affect the solution:</em>
<em>Changes made to your input should not affect the solution: (1): "x2" was replaced by "x^2". 1 more similar replacement(s).</em>
<em>x3+x2-8x-12 is not a perfect </em><em>cube</em>
<em>Factoring: x3+x2-8x-12 </em>
<em>Thoughtfully split the expression at hand into groups, each group having two terms :</em>
<em>Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: x3+x2 </em>
<em>Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: x3+x2 Group 2: -8x-12 </em>
<em>Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: x3+x2 Group 2: -8x-12 Pull out from each group separately :</em>
<em>Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: x3+x2 Group 2: -8x-12 Pull out from each group separately :Group 1: (x+1) • (x2)</em>
<em>Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: x3+x2 Group 2: -8x-12 Pull out from each group separately :Group 1: (x+1) • (x2)Group 2: (2x+3) • (-4)</em>
Answer:
the probability is P=0.012 (1.2%)
Step-by-step explanation:
for the random variable X= weight of checked-in luggage, then if X is approximately normal . then the random variable X₂ = weight of N checked-in luggage = ∑ Xi , distributes normally according to the central limit theorem.
Its expected value will be:
μ₂ = ∑ E(Xi) = N*E(Xi) = 121 seats * 68 lbs/seat = 8228 lbs
for N= 121 seats and E(Xi) = 68 lbs/person* 1 person/seat = 68 lbs/seat
the variance will be
σ₂² = ∑ σ² (Xi)= N*σ²(Xi) → σ₂ = σ *√N = 11 lbs/seat *√121 seats = 121 Lbs
then the standard random variable Z
Z= (X₂- μ₂)/σ₂ =
Zlimit= (8500 Lbs - 8228 lbs)/121 Lbs = 2.248
P(Z > 2.248) = 1- P(Z ≤ 2.248) = 1 - 0.988 = 0.012
P(Z > 2.248)= 0.012
then the probability that on a randomly selected full flight, the checked-in luggage capacity will be exceeded is P(Z > 2.248)= 0.012 (1.2%)
Answer:
ok
Step-by-step explanation:
So lets just assume that y = 1 since 2x is most likely an even number.
Then we can say that 2x = 8
8 divided by 2 is 4
so a point on this line could be (4,1)
Hope this helped
