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3 years ago

Please help with some RSM homework:(

Mathematics

1 answer:

bija089 [108]3 years ago

4 0

Answer:

k<3.5 or k>3.5

Step-by-step explanation:

k²-7k+12.25>0

(k-3.5)²

k²-7k+12.25

k²-2k·3.5+(√12.25)²

(k-3.5)²

k-3.5<0 or k-3.5>0

k-3.5 or k-3.5

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Which inequality below compares 1 and -7 correctly? A. 1>-7 or B. 1<-7

AfilCa [17]

1>-7 is the answer hope it helps you

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3 years ago

Read 2 more answers

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loris [4]

Answer:

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2 years ago

I need an explanation and answer

Minchanka [31]

Answer:

a ║ c

b ║ f

Step-by-step explanation:

when parallel lines are cut by a transversal line, the supplementary angles are angles whose angle measure adds up to 180°

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a ║ c

b ║ f

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3 years ago

Meghan used her cell phone for 45 minutes each day for n days. Her total minutes of cell phone usage was 540 minutes. Write an e

makvit [3.9K]

Answer:

Is there answer options because if not then it would possibly be y = 45x + 540 or it could be 45x = 540 or it could be 45x/540 so it can be any of those :)

Step-by-step explanation:

3 0

3 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

Step-by-step explanation:

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

Given:

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

Please help with some RSM homework:(​

Please help with some RSM homework:(