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4 x 5 - (2² - 2) =? (18 ÷ 3) + (3² - 7) =?

OverLord2011 [107]

4x5-(2 (2) - 2 ) = 14
(18/ 3) + (3 (2) - 7) 8

7 0

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

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

• What is probability that the red bulb will need to be replaced at the first inspection?

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

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

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

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

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

5 0

3 years ago

Assume you have such a watch. if a minimum of 18.0% of the original tritium is needed to read the dial in dark places, for how m

laila [671]

Given that the half life of 3H is 12.3 years.

The amount of substance left of a radioactive substance with half life of $t_{ \frac{1}{2} }$ after t years is given by

$N(t)=N_0\left(\frac{1}{2} \right)^{ \frac{t}{t_{\frac{1}{2}} }$

Therefore, the number of years it will take for 18% of the original tritinum to remain is given by

$18 \% = 100 \% \left(\frac{1}{2} \right)^{ \frac{t}{12.3}} \\ \\ \Rightarrow\left(\frac{1}{2} \right)^{ \frac{t}{12.3}} =0.18 \\ \\ \Rightarrow\frac{t}{12.3}\ln\left(\frac{1}{2} \right)=\ln0.18 \\ \\ \Rightarrow\frac{t}{12.3}= \frac{\ln0.18}{\ln\left(\frac{1}{2} \right)} = \frac{-1.715}{-0.6931} =2.474\\ \\ \Rightarrow t=2.474(12.3)=30.4$

Therefore, the number of years that the time could be read at night is 30.4 years.

5 0

3 years ago

You earn $15 every 3 weeks. at this rate how much will you earn in 5 weeks

zhannawk [14.2K]

Answer:

$25

Step-by-step explanation:

We know that in 3 weeks you get $15 now you can divide 3 by 15 to get $5 per week so since we know how much you earn per week you multiply $5 by 5 weeks to get $25

4 0

3 years ago

Read 2 more answers

(05.01)

baherus [9]

Answer:

Always

Step-by-step explanation:

6 0

3 years ago