Answer:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
x-(3*x^3+8*x^2+5*x-7)=0
Step by step solution :Step 1 :Equation at the end of step 1 : x-((((3•(x3))+23x2)+5x)-7) = 0 Step 2 :Equation at the end of step 2 : x - (((3x3 + 23x2) + 5x) - 7) = 0 Step 3 :Step 4 :Pulling out like terms :
4.1 Pull out like factors :
-3x3 - 8x2 - 4x + 7 =
-1 • (3x3 + 8x2 + 4x - 7)
Checking for a perfect cube :
4.2 3x3 + 8x2 + 4x - 7 is not a perfect cube
Trying to factor by pulling out :
4.3 Factoring: 3x3 + 8x2 + 4x - 7
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 3x3 - 7
Group 2: 8x2 + 4x
Pull out from each group separately :
Group 1: (3x3 - 7) • (1)
Group 2: (2x + 1) • (4x)
Please 5 star and like. Appreciated
11 kl = 11 000 l.
Hope this helps !
Photon
Answer:
P=0.147
Step-by-step explanation:
As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2
We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.
We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.
So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7
P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02
P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007
P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12
P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147
66.6 is the answer for this onel
Okay so i cant just post what the answer is i have to type something.
but anyways this is the answer: t= 4.56
