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3 years ago

Which of the following is a monomial? ОА. 20х9 – 7х Ов. 9/x Ос. 11х2 OD. 20x – 14

Mathematics

2 answers:

777dan777 [17]3 years ago

7 0

Answer:

C. 11x^2 is a monomial

Step-by-step explanation:

11x^2 is a monomial because it contains one term.

slava [35]3 years ago

3 0

Answer:

Step-by-step explanation:

11x² is monomial because it has only one term

9/x is not because power of x should be always whole number.

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The population standard deviation for the typing speeds for secretaries is 4 words per minute. If we want to be 90% confident th

Oksana_A [137]

Answer:

The minimum sample size that should be taken is 62.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

If we want to be 90% confident that the sample mean is within 1 word per minute of the true population mean, what is the minimum sample size that should be taken

This is n when $M = 1, \sigma = 4$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$1 = 1.96*\frac{4}{\sqrt{n}}$

$\sqrt{n} = 4*1.96$

$(\sqrt{n})^{2} = (4*1.96)^{2}$

$n = 61.5$

The minimum sample size that should be taken is 62.

8 0

3 years ago

Find the missing side and round to the nearest tenth

xxMikexx [17]

In this problem you use cosine because you know the hypotenuse and you want to know the adjacent side of the triangle. So in your calculator you would input cos(52). Then you would multiply that answer with the hypotenuse side. So your equation would be this: cos(52) x 13

5 0

3 years ago

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

1. A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin; 2. The radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball; 3. The volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

$\\ 1 \leq a < 10$ (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The La Plata river dolphin weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the scientific notation):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin.

<h3>Second case</h3>

Following the same reasoning, we can conclude that the radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball:

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$