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cestrela7 [59]
3 years ago
11

In 56 years ,Kevin will be 9 times as old as he is right now. How old is he right now?

Mathematics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

7

Step-by-step explanation:

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Complete all 3 questions.
nalin [4]

Answer:

1.x^2-5x

2.2x^2+15x+18

3. x^2+20x+18

Step-by-step explanation:

1.  x(x-5) x*x-5*x = x^2-5x

2. (x+6)(2x+3) x*2x+x*3+6*2x+6*3= 2x^2+15x+18

3.(2x^2+15x+18)-(x^2-5x)=x^2+20x+18

3 0
3 years ago
Find each function value.
Naddik [55]
1. 35
2. 21
3. 11

Just substitute. 
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3 years ago
Please help answer question 1
lara [203]

Answer:

One Solution ; x=4.5

y=-1.5

Step-by-step explanation:

Line A: 2x+2y=8

÷<u>2</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

x+y=4

Line B: x+y=3 -----> y=x-3

x+(x-3)=4

2x-3=4

2x=7

x=4.5

y=-1.5

6 0
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
How are adding, subtracting, dividing, and multiplying rational expressions are similar and different?
Arisa [49]

Rational expressions are multiplied and divided the same way numeric fractions are.

6 0
3 years ago
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