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4 years ago

Given: f(x) = x - 7 and h(x) = 2x + 3 Write the rule for h(f(x)).

Mathematics

1 answer:

Andrei [34K]4 years ago

3 0

Answer:

h(f(x)) = 2x - 11

Step-by-step explanation:

h(f(x)) is a "composite function." We use the function f(x) as the input to the function h(x):

h(x) = 2x + 3, or 2( x ) + 3; replacing x with f(x), or x - 7, we get:

h(f(x)) = 2(x - 7) + 3, or

h(f(x)) = 2x - 14 + 3, or

h(f(x)) = 2x - 11

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Round 17.834 to the nearest tenth. Your answer

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Answer:

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Step-by-step explanation:

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4 years ago

In 2012, a health insurance plan for a 35-year-old and his or her spouse costs $301 per month. That rate increased to $430 p

katrin2010 [14]

Answer: See explanation

Step-by-step explanation:

Let the cost for insuring the applicant = a.

Let the cost for insuring the spouse = b

Let the cost for insuring the first child= c

Let the cost for insuring the second child = d

A 35-year-old health insurance plan and that of his or her spouse costs $301 per month. This means that:

a + b = $301.

That rate increased to $430 per month if a child were included. This means the cost of a child will be:

= $430 - $301

= $129

The rate increased to $538 per month if two children were included. This means the cost for the second child will be:

= $538 - $430

= $108

The rate dropped to $269 per month for just the applicant and one child. His will be the cost of the applicant and a single child. This can be written as:

a + $129 = $269

a = $269 - $129

a = $140

Since a + b = $301

$140 + b = $301

b = $301 - $140

b = $161

Applicant = $140

The spouse = $161

The first child = $129

The second child = $108

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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

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

• 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 =

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

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3 years ago