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For nitrogen to be a liquid, its temperature must be within 12.78 °F of –333.22 °F. Which equation can be used to find the maxim

ziro4ka [17]

To determine the maximum and minimum temperatures at which nitrogen remains a liquid, use the equation |x - (-333.32)| = 12.78, where x is equal to the maximum and minimum temperatures. Using this equation, the maximum temperature is -320.54 degrees Fahrenheit while the minimum is -346.10 degrees Fahrenheit.

3 0

3 years ago

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Pls help .,)

Nat2105 [25]

C.

because a, b, and d aren't porportional

but c x is divided by 2

6 0

3 years ago

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One of the events in the Winter Olympics is the Men's 500-meter Speed Skating.

poizon [28]

Answer:

Mean: $\bar x = 40.61$

$Median =40.2$

$Mode = 43.4$

Step-by-step explanation:

Given

See attachment for data

Solving (a): The mean

Mean is calculated as:

$\bar x = \frac{\sum x}{n}$

From the attached:

$n = 14$

So, the mean is:

$\bar x = \frac{43.4+43.4+43.4+43.1+43.2+40.2+40.2+40.1+40.3+39.44+39.17+38.03+38.19+36.45}{14}$

$\bar x = \frac{568.58}{14}$

$\bar x = 40.61$

Solving (b): The median

$n = 14$ ; this is an even number. So, the median is:

$Median = \frac{1}{2}(n+1)$

$Median = \frac{1}{2}(14+1)$

$Median = \frac{1}{2}(15)$

$Median = 7.5th$

This implies that the median is the average of the 7th and 8th item.

Next, is to order the data (in ascending order): <em>36.45, 38.03, 38.19, 39.17, 39.44, 40.1, 40.2, 40.2, 40.3, 43.1, 43.2, 43.4, 43.4, 43.4.</em>

The 7th and 8th items are: 40.2 and 40.2

The median is:

$Median = \frac{1}{2}(40.2 + 40.2)$