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3 years ago

10

Solve the system by substitution. 2x - y = -3, 2x-2y = -2

1 answer:

Sonja [21]3 years ago

5 0

Answer:// Solve equation [1] for the variable y

[1] y = 2x - 3

// Plug this in for variable y in equation [2]

[2] -2•(2x-3) + 2x = 2

[2] - 2x = -4

// Solve equation [2] for the variable x

[2] 2x = 4

[2] x = 2

// By now we know this much :// Solve equation [1] for the variable y

[1] y = 2x - 3

// Plug this in for variable y in equation [2]

[2] -2•(2x-3) + 2x = 2

[2] - 2x = -4

// Solve equation [2] for the variable x

[2] 2x = 4

[2] x = 2

// By now we know this much :

y = 2x-3

x = 2

// Use the x value to solve for y

y = 2(2)-3 = 1

y = 2x-3

x = 2

// Use the x value to solve for y

y = 2(2)-3 = 1

Step-by-step explanation:

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The caller times at a customer service center has an exponential distribution with an average of 22 seconds. Find the probabilit

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The probability that a randomly selected call time will be less than 30 seconds is 0.7443.

Step-by-step explanation:

We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.

Let X = caller times at a customer service center

The probability distribution (pdf) of the exponential distribution is given by;

$f(x) = \lambda e^{-\lambda x} ; x > 0$

Here, $\lambda$ = exponential parameter

Now, the mean of the exponential distribution is given by;

Mean = $\frac{1}{\lambda}$

So, $22=\frac{1}{\lambda}$ ⇒ $\lambda=\frac{1}{22}$

SO, X ~ Exp( $\lambda=\frac{1}{22}$ )

To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;

$P(X\leq x) = 1 - e^{-\lambda x}$ ; x > 0

Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)

P(X < 30) = $1 - e^{-\frac{1}{22} \times 30}$

= 1 - 0.2557

= 0.7443

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