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My name is Ann [436]
3 years ago
13

Given h(x) = 3x - 3, find h(3).

Mathematics
1 answer:
Daniel [21]3 years ago
4 0

Answer:

<h2>h(3) = 6</h2>

Step-by-step explanation:

h(x) = 3x - 3

To find h(3) , substitute the value of x that's 3 into h(x). That is for every x in f (x) replace it with 3

We have

h(3) = 3(3) - 3

= 9 - 3

We have the final answer as

<h3>h(3) = 6</h3>

Hope this helps you

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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
Find a fraction equivalent to<br> that has a denominator of 10.
quester [9]

Answer:

1/10

Step-by-step explanation:

any number (1-9) as the number above the fraction line (numerator) with the number 10 below the fraction line is a fraction with a denominator of 10.

if it was 10/10, it will = 1

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3 years ago
Is 5/10 equivalent to 1/6​
Citrus2011 [14]

Answer:

No, its not equivalent

1/2 is equivalent though

Please mark as brainliest

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(8-7)/(-3-0)= -1/3
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Simple math. <br> Question 11.
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Answer: By 9.1%

Step-by-step explanation:

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2 years ago
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