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3 years ago

If I know the area of a triangle, how can I find the area of the smallest rectangle that would fit around it?

Mathematics

1 answer:

sasho [114]3 years ago

3 0

Explanation :

What you would do is you take the area of the rectangle and subtract the area of the triangle

The area of a rectangle is Length * Width

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3 years ago

A 100 mL vial of 25% solution is available in stock. An order is recieved for 6 g to be added to 250 mL of fluid for an iv infus

Ne4ueva [31]

9514 1404 393

Answer:

274 mL

Step-by-step explanation:

Often medical solutions expressed as a percentage are not really a percentage as such. A percentage is the ratio of two quantities with the same units.

Here, the context given by the problem suggests the "25%" solution is really (25 g)/(100 mL). That is, the units are grams and milliliters--different units.

With that assumption, we want to find the volume (v) of solution needed to deliver 6 g of medicine. An appropriate proportion* is ...

v/(6 g) = (100 mL)/(25 g)

v = (6 g)(100 mL)/(25 g) = 24 mL

So, the total volume of the infusion is ...

250 mL +24 mL = 274 mL

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* The concentration is given in terms of g/mL, but we have used a proportion that is mL/g. The reason for that is we want the variable to be in the numerator of the ratio. The variable here represents volume, so we have written the proportion with volumes in the numerators.

Having the variable in the numerator means the equation can be solved in one step--by multiplying by its denominator.

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3 years ago

Using traditional methods it takes 92 hours to receive an advanced flying license. A new training technique using Computer Aided

DanielleElmas [232]

Answer:

The value of the test statistic is z = 1.39.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

Step-by-step explanation:

Using traditional methods it takes 92 hours to receive an advanced flying license.

This means that at the null hypothesis, it is tested if the mean is of 92, that is:

$H_0: \mu = 92$

Test if there is evidence that the technique lengthens the training time

At the alternative hypothesis, it is tested if the mean is more than 92, that is:

$H_1: \mu > 92$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

92 is tested at the null hypothesis:

This means that $\mu = 92$

A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.

This means that $n = 70, X = 93, \sigma = \sqrt{36} = 6$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{93 - 92}{\frac{6}{\sqrt{70}}}$

$z = 1.39$

The value of the test statistic is z = 1.39.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.

Looking at the z-table, z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

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2 years ago