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2 years ago

12

A hotel advertises vacation packages on its website, and it charges the same amount per night. The table shows the total cost of

a stay for different numbers of nights.
A table with the cost for a hotel stay is shown.
What is the unit rate per night?

2 answers:

kupik [55]2 years ago

7 0

It cost 105 dollar per rate night.

True [87]2 years ago

6 0

Answer:

105$ per night.......

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A data set has a normal distribution with a mean of 47 and a standard deviation of 4.6. Use this information to scale the horizo

gulaghasi [49]

Horizontal axis with 1, 2, and 3 standard deviations above the mean are 51.6, 56.2, 60.8

Horizontal axis with 1, 2, and 3 standard deviations below the mean are 42.4, 37.8, 33.2

<u>Step-by-step explanation:</u>

A data set has a normal distribution with a mean of 47 and a standard deviation of 4.6

Mean, M ⇒ 47
Standard deviation ⇒ 4.6

From this information, you have to scale the horizontal axis with the mean of this distribution and values at 1, 2, and 3 standard deviations above and below the mean.

Horizontal axis with 1, 2, and 3 standard deviations above the mean :

1 standard deviation above the mean ⇒ M + SD

⇒ 47 + 4.6 = 51.6

2 standard deviations above the mean ⇒ M + 2SD

⇒ 47 + (2 × 4.6) = 56.2

3 standard deviations above the mean ⇒ M + 3SD

⇒ 47 + (3 × 4.6) = 60.8

Horizontal axis with 1, 2, and 3 standard deviations below the mean :

1 standard deviation below the mean ⇒ M - SD

⇒ 47 - 4.6 = 42.4

2 standard deviations below the mean ⇒ M - 2SD

⇒ 47 - (2 × 4.6) = 37.8

3 standard deviations below the mean ⇒ M - 3SD

⇒ 47 - (3 × 4.6) = 33.2

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Carl drove from his house to work at an average speed of 35 miles per hour. The drive took him 25 minutes. If the

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Carls speed going home is 1.2km an hour

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Answer:

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The auto parts department of an automotive dealership sends out a mean of 4.34.3 special orders daily. What is the probability t

Svetlanka [38]

Answer:

$P(X = 5) = 0.166$

Step-by-step explanation:

Given

$Mean = 4.3$

$x = 5$

Required

Determine the probability that the order is 5

This question can be answered using Poisson distribution.

$P(X = x) = \frac{\alpha ^x e^{-\alpha}}{x!}$

Where

$\alpha$ is used to represent the mean

and

$\alpha = 4.3$

$x = 5$

<em /> $e = 2.71828$ <em> ---- Euler's constant</em>

So, we have:

$P(X = x) = \frac{\alpha ^x e^{-\alpha}}{x!}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5!}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5* 4 * 3 * 2 *1}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{120}$

$P(X = 5) = \frac{1470.08443 * 0.01356859825}{120}$

$P(X = 5) = \frac{19.9469850243}{120}$

$P(X = 5) = 0.1662248752$

$P(X = 5) = 0.166$ <em>Approximated</em>

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3 years ago