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insens350 [35]
3 years ago
15

How do. You solve {4+3[2^3÷2

Mathematics
1 answer:
IgorC [24]3 years ago
7 0

Answer:

16

Step-by-step explanation:

Simplify the following:

4 + 3×2^3/2

3×2^3/2 = (3×2^3)/2:

4 + (3×2^3)/2

Combine powers. (3×2^3)/2 = 3×2^(3 - 1):

4 + 3×2^(3 - 1)

3 - 1 = 2:

4 + 3×2^2

2^2 = 4:

4 + 3×4

3×4 = 12:

4 + 12

4 + 12 = 16:

Answer: 16

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70 and 30

Step-by-step explanation:

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Solve by completing the square. Round<br> x^2-4x=5
Novosadov [1.4K]

Answer: -1 and 5.

Step-by-step explanation:

x - 4x = 5 : Move the constant to the left.

x^2 - 4x - 5 = 0 : Rewrite the expression.

x^2 + x - 5x - 5 = 0 : Factor the expressions.

x(x+1) - 5(x+1) = 0 : Factor the expression.

(x+1)(x-5) = 0 : Seperate into possible cases.

x + 1 = 0, x - 5 = 0 : Solve the equations.

x = -1, x = 5 : The equation has 2 solutions.

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The U.S. Department of Transportation estimates that citizens own more than 12,720,000 registered boats. Which shows this number
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Which function has the range (-infinity,-2]U[0,infinity)?
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Step-by-step explanation:

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A consumer organization estimates that over a​ 1-year period 20​% of cars will need to be repaired​ once, 5​% will need repairs​
gtnhenbr [62]

Answer:

a) 0.5476

b) 0.0676

c) 0.4524

Step-by-step explanation:

Given this information, we can conclude that 74% of the cars won't need any repairs over a 1-year period (100 - 20 - 5 - 1 = 74%). And 26% will need at least 1 repair over a 1-year period.

P(car doesn't need repair) = 0.74

P (car needs repair) = 0.26

If you own two cars, the probability that:

<u>a) Neither will need repair:</u>

We need that car 1 won't need repair AND car 2 won't need repair.

=P(Car 1 doesn't need repair) x P(Car 2 doesn't need repair)

= 0.74 x 0.74 = 0.5476

The probability that neither will need repair is 0.5476.

<u>b) Both will need repair:</u>

We need that car 1 needs repair AND car 2 needs repair.

P(Car 1 needs repair) x P(Car 2 needs repair)

= 0.26 x 0.26 = 0.0676

The probability that both will need repair is 0.0676

<u>c) At least one car will need repair</u>

Car 1 needs repair or Car 2 needs repair or both need repair.

To solve this one, it's easier to use the complement of P(neither needs repair)

1 - P(neither needs repair)

1 - (0.74)(0.74)  = 1 - 0.5476 = 0.4524

The probability that at least one car will need repair is 0.4524

4 0
4 years ago
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