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3 years ago

Convert 1.50 micrometers^2 to m^2. Show your solution.

Mathematics

1 answer:

Vadim26 [7]3 years ago

8 0

Here is the solution for the problem:

1m * 1m = 1m² 
1m = 1,000,000µm = 1 x 10^6µm 
1 x 10^6µm * 1 x 10^6µm = 1m² 
1x10^12µm² = 1m² (Now, divide both sides by 1x10^12). 
1x10^12µm² / (1 x 10^12) = 1m² / (1 x 10^12) 
1µm² = 0∙000 000 000 001 m² (x both sides by 1.5). 
1.5µm² = 0∙000 000 000 0015 m² 
1.5µm² = 1∙5x10^-12 m²

So 1.50 micrometers^2 is equal to 1.5 x 10^-12 m². I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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In order to study the relationship between playing football and traumatic brain injuries, researchers placed accelerometers in t

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Answer:

56.96% probability on a randomly selected team of 48 players that the average number of head hits per player is between 340 and 360 hits

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a single player:

$\mu = 355, \sigma = 80$

For the sample mean(of 48 playes).

$n = 48, s = \frac{80}{\sqrt{48}} = 11.547$

What is the probability on a randomly selected team of 48 players that the average number of head hits per player is between 340 and 360 hits?

This is the pvalue of Z when X = 360 subtracted by the pvalue of Z when X = 340. So

X = 360

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{360 - 355}{11.547}$

$Z = 0.43$

$Z = 0.43$ has a pvalue of 0.6664

X = 340

$Z = \frac{X - \mu}{s}$

$Z = \frac{340 - 355}{11.547}$

$Z = -1.30$

$Z = -1.30$ has a pvalue of 0.0968

0.6664 - 0.0968 = 0.5696

56.96% probability on a randomly selected team of 48 players that the average number of head hits per player is between 340 and 360 hits

4 0

2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

3 years ago

Find the Area of the figure below, composed of a parallelogram and one semicircle.

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Answer:

I Don't Know

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2 years ago

a rectangle has a length of 5 yards and a width of 3 yards. what is its perimeter? explain your answer

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