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Mice21 [21]
3 years ago
9

If f(x) = 5x + 40, what is f(x) when x = –5? –9 –8 7 15

Mathematics
2 answers:
ki77a [65]3 years ago
8 0

Answer:

f(x) is equal to 15 when  x=-5

Step-by-step explanation:

Given a function f(x) this means that the function ''f'' depends on the variable ''x'', If you want to obtain the value of f(x) when x=a,being ''a'' a possible value of x (this means ''a'' is in the domain of f(x)) you only need to replace x=a in the f(x) equation

If x=-5 ⇒

f(x)=5x+40\\f(-5)=5.(-5)+40=-25+40=15

f(-5)=15 this means that when x=-5 the function value is 15

Butoxors [25]3 years ago
7 0

Answer:

its 15

Step-by-step explanation:

i did this problem before

Hope i helped : )

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3 years ago
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Identify the type of observational study​ (cross-sectional, retrospective, or​ prospective) described below. A research company
Sindrei [870]

Answer:

Prospective study

Step-by-step explanation:

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Considering the information given and the observational study exposed to the question, we can conclude that we are talking about a prospective study because data is collected over the next 10 years.

I hope you find this information useful and interesting! Good luck!

7 0
3 years ago
Under average driving conditions, the life lengths of automobile tires of a certain brand are found to follow an exponential dis
wariber [46]

Answer:

a) P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

b) P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

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And 0 for other case. Let X the random variable that represent "life lengths of automobile tires of a certain brand" and we know that the distribution is given by:

X \sim Exp(\lambda=\frac{1}{30000})

The cumulative distribution function is given by:

F(X) = 1- e^{-\frac{x}{\mu}}

Part a

We want to find this probability:

P(X>30000) and for this case we can use the cumulative distribution function to find it like this:

P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

Part b

For this case w want to find this probability

P(X>30000|X>15000)

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P(X>a+t| X>t)=P(X>a)  

On this case if we use this property we have this:P(X>30000|X>15000)=P(X>15000+15000|X>15000)=P(X>15000)

We can use the definition of the density function and find this probability:

P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

7 0
3 years ago
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Answer:

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Answer:

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