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3 years ago

8.35x - 1.5 = 71.98 (a) 8.6 (b) 8.8 (c) 9 (d) 10.3

1 answer:

3 0

<h3> Hola! :D ¡te invito a recibir ayuda de un latinoamericano puto!</h3><h2><u> _____________________________________ </u></h2><h2> 8.35x - 1.5 = 71.98</h2><h2> - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -</h2>

<u>El -1,5 qlero pasaría al otro lado positivo</u>

<u>Ahora, se suma 71.98 + 1.5 = 73,48</u>

<u>El 8.35 qlero que está multiplicando, pasa al otro lado pero dividiendo</u>

<u>Dividimos</u>

<h3> x = 8,8</h3><h2> - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -</h2><h2> <u>(b) 8,8</u> es la opción correcta</h2>

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M Explain why this is true<br> PLEASE HELP!!!

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3 years ago

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3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

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The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

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Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

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This means that,

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$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

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lbvjy [14]

Answer:

Step-by-step explanation:

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3 years ago