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sveticcg [70]
3 years ago
9

(7n-8n^2+3n)-(2n^2+4n-3n)

Mathematics
2 answers:
artcher [175]3 years ago
8 0

Answer:

-10n^2+9n

Step-by-step explanation:

distribute the - to everything in the parenthesis | (-2n^2-4n+3n) then put them all in order from greatest variable or coefficient to lowest and add the two equations together |

  (-8n^2+10n)

+ (-2n^2-1n)

____________

-10n^2+9n

Usimov [2.4K]3 years ago
4 0

Answer:

(6n

3

−8n

2

−4n−3)

Step-by-step explanation:

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NeX [460]

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623168

Step-by-step explanation:

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4 years ago
A fire company keeps two rescue vehicles. Because of the demand on the vehicles and the chance of mechanical failure, the probab
VashaNatasha [74]

Answer:

(a) P (Both vehicles are available at a given time) = 0.81

(b) P (Neither vehicles are available at a given time) = 0.01

(c) P (At least one vehicle is available at a given time) = 0.99

Step-by-step explanation:

Let A = Vehicle 1 is available when needed and B = Vehicle 2 is available when needed.

<u>Given</u>:

The availability of one vehicle is independent of the availability of the other, i.e. P (A ∩ B) = P (A) × P (B)

P (A) = P (B) = 0.90

(a)

Compute the probability that both vehicles are available at a given time as follows:

P (Both vehicles are available) = P (Vehicle 1 is available) ×

                                                              P (Vehicle 2 is available)

                                  P(A\cap B)=P(A)\times P(B)

                                                  =0.90\times0.90\\=0.81

Thus, the probability that both vehicles are available at a given time is 0.81.

(b)

Compute the probability that neither vehicles are available at a given time as follows:

P (Neither vehicles are available) = [1 - P (Vehicle 1 is available)] ×

                                                                   [1 - P (Vehicle 2 is available)]

                                    P(A^{c}\cap B^{c})=[1-P(A)]\times [1-P(B)]\\

                                                       =(1-0.90)\times (1-0.90)\\=0.10\times0.10\\=0.01

Thus, the probability that neither vehicles are available at a given time is 0.01.

(c)

Compute the probability that at least one vehicle is available at a given time as follows:

P (At least one vehicle is available) = 1 - P (None of the vehicles are available)

                                                          =1-[P(A^{c})\times P(B^{c})]\\=1-0.01.....(from\ part\ (b))\\  =0.99

Thus, the probability that at least one vehicle is available at a given time is 0.99.

6 0
3 years ago
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nika2105 [10]

Answer:

22

Step-by-step explanation:

The y values increase by 2 every time so 18+2=22

8 0
3 years ago
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