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3 years ago

3x-91>-87 AND 17x-16>18

Mathematics

2 answers:

astra-53 [7]3 years ago

6 0

Answer:

( 2, ∞ )

Step-by-step explanation:

Given compound inequality,

3x-91 > -87 and 17x-16 > 18,

3x > -87 + 91 and 17x > 18 + 16

3x > 4 and 17x > 34

x > $\frac{4}{3}$ and x > 2

Since, if x > $\frac{4}{3}$

x ∈ ( $\frac{4}{3}$ , ∞ )

If x > 2

x ∈ ( 2, ∞ )

Now, x > $\frac{4}{3}$ and x > 2

$\implies (\frac{4}{3}, \infty)\cap (2, \infty)$

= ( 2, ∞ )

Hence, the possible solution of the given compound inequality is (2, ∞)

kirill [66]3 years ago

3 0

$3x-91 > -81\ \ \ \ |+91\\\\3x > 10\ \ \ \ |:3\\\\x > \dfrac{10}{3}\\\\17x-16 > 18\ \ \ \ |+16\\\\17x > 34\ \ \ \ |:17\\\\x > 2\\\\Answer:\ 2 < x < \dfrac{10}{3}\to x\in\left(2,\ \dfrac{10}{3}\right)$

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Answer:

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Let, $h=0$ , $k=35$ , $x=8$ , $y=21$

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3 years ago

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

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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

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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.

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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).

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

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

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