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stellarik [79]
2 years ago
15

Twice the difference of a number and 16 is the same as three times a number less than 18

Mathematics
1 answer:
Maksim231197 [3]2 years ago
4 0

Answer:

Step-by-step explanation:

Let x be the "number."

"twice the difference of a number and 16" can be written as:  

2(x-16)

"is the same as three times a number less than 18" can be written as:

= 3(x<18)

I wonder if this was meant to say "is the same as three times a number less  18," instead of "less than 18."  If so, we would have:

=3(x-18)

1.  Using the sentence as written:  2(x-16) = 3(x<18)

2.  Using the sentence as amended:  2(x-16) = =3(x-18)

 Going with this interpretation:  

 2(x-16) = 3(x-18)

2x - 32 = 3x - 54

x = 22

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If u(x) = x5 – x4 + x2 and v(x) = –x2, which expression is equivalent to (u/v)(x)
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Answer:

c) -x^3 + x^2 - 1

Step-by-step explanation:

Given: u (x) = x^5 - x^4 +x^2 and v(x) = -x^2

(u/v)(x) = u(x)/v(x)

Now plug in the given functions in the above formula, we get

= (x^5 - x^4 + x^2) / -x^2

We can factorize the numerator.

In x^5 - x^4 + x^2. the common factor is x^2, so we can take it out and write the remaining terms in the parenthesis.

= x^2 (x^3 - x^2 + 1) / - x^2

Now we gave x^2 both in the numerator and in the denominator, we can cancel it out.

(u/v)(x) = (x^3 - x^2 + 1) / -1

When we dividing the numerator by -1, we get

(u/v)(x) = -x^3 + x^2 - 1

Answer: c) -x^3 + x^2 - 1

Hope you will understand the concept.

Thank you.

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Every day your friend commutes to school on the subway at 9 AM. If the subway is on time, she will stop for a $3 coffee on the w
Shtirlitz [24]

Answer:

1.02% probability of spending 0 dollars on coffee over the course of a five day week

7.68% probability of spending 3 dollars on coffee over the course of a five day week

23.04% probability of spending 6 dollars on coffee over the course of a five day week

34.56% probability of spending 9 dollars on coffee over the course of a five day week

25.92% probability of spending 12 dollars on coffee over the course of a five day week

7.78% probability of spending 12 dollars on coffee over the course of a five day week

Step-by-step explanation:

For each day, there are only two possible outcomes. Either the subway is on time, or it is not. Each day, the probability of the train being on time is independent from other days. So we use the binomial probability distribution to solve this problem.

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.

In this problem we have that:

The probability that the subway is delayed is 40%. 100-40 = 60% of the train being on time, so p = 0.6

The week has 5 days, so n = 5

She spends 3 dollars on coffee each day the train is on time.

Probabability that she spends 0 dollars on coffee:

This is the probability of the train being late all 5 days, so it is P(X = 0).

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

P(X = 0) = C_{5,0}.(0.6)^{0}.(0.4)^{5} = 0.0102

1.02% probability of spending 0 dollars on coffee over the course of a five day week

Probabability that she spends 3 dollars on coffee:

This is the probability of the train being late for 4 days and on time for 1, so it is P(X = 1).

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

P(X = 1) = C_{5,1}.(0.6)^{1}.(0.4)^{4} = 0.0768

7.68% probability of spending 3 dollars on coffee over the course of a five day week

Probabability that she spends 6 dollars on coffee:

This is the probability of the train being late for 3 days and on time for 2, so it is P(X = 2).

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

P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304

23.04% probability of spending 6 dollars on coffee over the course of a five day week

Probabability that she spends 9 dollars on coffee:

This is the probability of the train being late for 2 days and on time for 3, so it is P(X = 3).

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

P(X = 3) = C_{5,3}.(0.6)^{3}.(0.4)^{2} = 0.3456

34.56% probability of spending 9 dollars on coffee over the course of a five day week

Probabability that she spends 12 dollars on coffee:

This is the probability of the train being late for 1 day and on time for 4, so it is P(X = 4).

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

P(X = 4) = C_{5,4}.(0.6)^{4}.(0.4)^{1} = 0.2592

25.92% probability of spending 12 dollars on coffee over the course of a five day week

Probabability that she spends 15 dollars on coffee:

Probability that the subway is on time all days of the week, so P(X = 5).

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

P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778

7.78% probability of spending 12 dollars on coffee over the course of a five day week

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ludmilkaskok [199]

Answer:

B.

Step-by-step explanation:

Domain is the x-coordinate and the range is the y-coordinate.

B. Domain: {7}

Range: {-4, -2, 1, 4, 6}

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