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3 years ago

15

Please help me with 12!

2 answers:

ICE Princess25 [194]3 years ago

5 0

6 tickets because 1 for an adult one day and 1 for a child for one day then one for adults 2 day one for kids 2 days one for kids weekly and one for adults weekly so 6 different types of tickets

Liono4ka [1.6K]3 years ago

3 0

Four different colours I believe. Because it says each ticket type will be a different colour. Therefor, adult (1), child (2), 1-day (3), 2-day (4).

Hope this was helpful and correct!

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Answer:

the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

Step-by-step explanation:

Given that:

Mean = 30000

Standard deviation = 9000

sample size = 100

The probability that the mean student loan debt for these people is between $31000 and $33000 can be computed as:

$P(31000 < X < 33000) = P( X \leq 33000) - P (X \leq 31000)$

$P(31000 < X < 33000) = P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}} )- P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}} )$

$P(31000 < X < 33000) = P( Z \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}} )- P(Z \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}} )$

$P(31000 < X < 33000) = P( Z \leq \dfrac{3000}{\dfrac{9000}{10}}}) -P(Z \leq \dfrac{1000}{\dfrac{9000}{10}}})$

$P(31000 < X < 33000) = P( Z \leq 3.33)-P(Z \leq 1.11})$

From Z tables:

$P(31000 < X$

$P(31000 < X$

Therefore; the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

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Step-by-step explanation:

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Answer:

please give me a brainless 0

Inequality :

α1+(n−1)a−(⌊n÷m⌋×(a−b))≥x

The following is as far as I get:

α1+(n−1)a−(⌊n÷m⌋×(a−b))≥x

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