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4 years ago

8/3(z+11)=y solve for z.

Mathematics

1 answer:

bagirrra123 [75]4 years ago

3 0

Answer:

z = (3/8)y - 11

Step-by-step explanation:

Undo what has been done to z, in reverse order. Here, 11 is added and the sum is multiplied by 8/3.

To undo the multiplication, multiply the equation by the inverse, 3/8:

(3/8)(8/3)(z+11) = (3/8)y

z +11 = (3/8)y . . . . . . . . . simplify

To undo the addition of 11, add the opposite of 11 to the equation.

z +11 -11 = (3/8)y -11

z = (3/8)y -11 . . . . . . . . . simplify

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Step-by-step explanation:

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3 years ago

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3 years ago

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The length of life (in days) of an alkaline battery has an exponential distribution with an average life of 340 days. (Round you

iren2701 [21]

Answer:

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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:

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

In this problem, we have that:

$m = 340$ . So

$\mu = \frac{1}{340} = 0.0029$

$P(X \leq x) = 1 - e^{-0.0029x}$

(a) What is the probability that an alkaline battery will fail before 185 days?

$P(X \leq x) = 1 - e^{-0.0029x}$

$P(X \leq 185) = 1 - e^{-0.0029*185} = 0.4196 = 0.42$

0.42 = 42% probability that an alkaline battery will fail before 185 days

(b) What is the probability that an alkaline battery will last beyond 2 years? (Use the fact that there are 365 days in one year.)

2 years = 2*365 = 730 days.

Either it lasts 730 or less days, or it last more. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 730) + P(X > 730) = 1$

We want P(X > 730). So

$P(X > 370) = 1 - P(X \leq 730) = 1 - (1 - e^{-0.0029*730}) = 0.12$

0.12 = 12% probability that an alkaline battery will last beyond 2 years

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