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3 years ago

7. Find the range of the function f(x) = 4x – 1 for the domain {-1, 0.1.2.3). (I point)

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

3 0

Answer: { ${ -5 , -1 , 3 , 7 , 11 }$ }

Step-by-step explanation:

Given :

$f(x) = 4x - 1$

domain = { $-1 , 0 , 1 , 2 , 3$ }

To find the range , we will substitute the domain as the value of x into the function given , that is

when x = - 1

$f(x) = 4 (-1) - 1$

$f(x) = -5$

when x = 0

$f(x) = -1$

when x = 1

$f(x) = 4(1) - 1$

$f(x) = 3$

When x = 2

$f(x) = 7$

when x = 3

$f(x) = 11$

Therefore: the range = ${ -5 , -1 , 3 , 7 , 11 }$

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Determine if x+3 is a factor of -3x^3+6x^2+6x+9. How do u know

krek1111 [17]

By the factor theorem, if x + 3 is a factor of f(x) = -3x^3 + 6x^2 + 6x + 9, then f(-3) = 0
f(-3) = -3(-3)^3 + 6(-3)^2 + 6(-3) + 9 = -3(-27) + 6(9) - 18 + 9 = 81 + 54 - 9 = 126.

Therefore, x + 3 is not a factor of the given function

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2 years ago

After a large scale earthquake, it is predicted that 15% of all buildings have been structurally compromised.a) What is the prob

Westkost [7]

Answer:

a) 13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) 17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) 17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) 75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

Step-by-step explanation:

For each building, there are only two possible outcomes after a earthquake. Either they have been damaged, or they have not. The probability of a building being damaged is independent from other buildings. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

15% of all buildings have been structurally compromised.

This means that $p = 0.15$

20 buildings

This means that $n = 20$

a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756$

17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?

Either they find 4 or less, or they find more than 4. The sum of the probabilities of these events is 1. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0388 + 0.1368 + 02293 + 0.2428 + 0.1821 = 0.8298$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8298 = 0.1702$

17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) What is the probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised?

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X = 5) = C_{20,5}.(0.15)^{5}.(0.85)^{15} = 0.1028$

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2293 + 0.2428 + 0.1821 + 0.1028 = 0.7570$

75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?

The expected value of the binomial distribution is:

$E(X) = np$

$E(X) = 20*0.15 = 3$

The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

3 0

3 years ago

// ANSWER PLS //

Bogdan [553]

Answer:

Step-by-step explanation:

3 0

3 years ago

HELP PLZ IM GIVING 25 BRAIN:EIST PLZZZ

Virty [35]

Answer:

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Distance = Speed ÷ Time

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8 0

3 years ago

Read 2 more answers

a) Suppose you are able to download 1 free song to your tablet daily and each additional download costs $0.59. If your monthly b

icang [17]

If 1 additional song costs $0.59, then there would be 26 additional songs downloaded. Because $15.34 divided by $0.59 is 26.

Let x be the number of songs downloaded.

x = 28 + (15.34/0.59)
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x = 54

You have 54 songs downloaded on your tablet.

Hope this helps :)

7 0

3 years ago