How to write a solution to the equation x^2=5 using exponents and radicals

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

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

<em>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. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

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

5 0

3 years ago

Someone help me please. I don't just want the answer, I need an explanation on how it's done too.

NARA [144]

Answer:

Below.

Step-by-step explanation:

f) (a + b)^3 - 4(a + b)^2

The (a+ b)^2 can be taken out to give:

= (a + b)^2(a + b - 4)

= (a + b)(a + b)(a + b - 4).

g) 3x(x - y) - 6(-x + y)

= 3x( x - y) + 6(x - y)

= (3x + 6)(x - y)

= 3(x + 2)(x - y).

h) (6a - 5b)(c - d) + (3a + 4b)(d - c)

= (6a - 5b)(c - d) + (-3a - 4b)(c - d)

= -(c - d)(6a - 5b)(3a + 4b).

i) -3d(-9a - 2b) + 2c (9a + 2b)

= 3d(9a + 2b) + 2c (9a + 2b)

= 3d(9a + 2b) + 2c (9a + 2b).

= (3d + 2c)(9a + 2b).

j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)

= a^2b^3(2a + 1) + 6ab^2(2a + 1)

= (2a + 1)( a^2b^3 + 6ab^2)

The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:

(2a + 1)ab^2(ab + 6)

= ab^2(ab + 6)(2a + 1).

4 0

3 years ago

Create an expression, to equal 13, using any grouping of addition, subtraction, multiplication, and division. You must use all t

aleksandrvk [35]

(29 - 2 + (6 * 2)) / 3 = 13

4 0

3 years ago

Read 2 more answers

Tuition is $29,637. increase is 5.1%. what is new cost?

Veseljchak [2.6K]

Answer:

31,148.48

Step-by-step explanation:

29,637*.051= 1,511.48

29,637+1,511.48=31,148.48

6 0

3 years ago

A line passes through the points A(-1,7) and B(1,3). Write an equation fo this line

Sunny_sXe [5.5K]

Answer:

y = -2 + 5. Step-by-step explanation: To find the equation of a line when you only have two points you need to use the slope formula

Step-by-step explanation:

6 0

3 years ago