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zysi [14]
3 years ago
8

-5k+4(k+6)=-4(2k+1)...

Mathematics
2 answers:
VikaD [51]3 years ago
4 0
Hey there! 

First: Distribute the problem 

Like
:  4(k) and 4(6) (other half) -4(2k) and 4!) 

Problem become: -5k + 4k + 24 = -8k + 1

Second: Combine your like terms 

Like: -5k + 4k + 24 = -8k - 4

Problem becomes: -k + 24 = -8k - 4

Third: add the value 8k to your sides

Like: -k + 24 + 8k \\ \\ =  -8k - 4 + 8k

Problem becomes: 7k + 24 = -4

Fourth: subtract by 24  on each of your sides

Like: 
7k + 24 - 24 \\ \\  = -4 - 24

Problem becomes: 7k = -28

Fifth: divide both sides by 7

Like: \frac{7k}{7} =  \frac{-28}{7}

Answer: k =-4

Good luck on your assignment and enjoy your day! 

~LoveYourselfFirst:)
Alex787 [66]3 years ago
3 0
-5k + 4(k + 6) = -4(2k + 1)        Given
-5k + 4k + 24 = -8k - 4            Distributive Property
          -k + 24 = -8k - 4            Combine like terms
         7k + 24 = -4                   Add 8k to both sides
                 7k = -28                 Subtract 24 from both sides
                   k = -4                   Divide both sides by 7

Answer:
k = -4

Hope this helps!
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Suppose X has an exponential distribution with mean equal to 23. Determine the following:
e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

Mean equal to 23.

This means that m = 23, \mu = \frac{1}{23} = 0.0435

(a) P(X >10)

P(X > 10) = e^{-0.0435*10} = 0.6473

So

P(X > 10) = 0.6473

(b) P(X >20)

P(X > 20) = e^{-0.0435*20} = 0.4190

So

P(X > 20) = 0.4190

(c) P(X <30)

P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

P(X > x) = e^{-\mu x}

0.05 = e^{-0.0435x}

\ln{e^{-0.0435x}} = \ln{0.05}

-0.0435x = \ln{0.05}

x = -\frac{\ln{0.05}}{0.0435}

x = 68.87

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