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3 years ago

What is a dense region of the sun with tightly packed atoms

Mathematics

1 answer:

zmey [24]3 years ago

5 0

Answer:

Well for one. This is the wrong section as its mathematics. I believe this is a science question. That being said. The answer here is the Suns Core.

Explanation:

The suns core has more than 15 million Kelvin and is tightly-packed and dense.

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Given A(3, 4), B(4, 3) and C(2, 1), what are the coordinates of C', the image of C under a rotation of 90 degrees about the orig

Aleks [24]

Given:

The coordinates of the points A, B and C are (3,4), (4,3) and (2,1)

The points are rotated 90° about the origin.

We need to determine the coordinates of the point C'.

Coordinates of the point C':

The general rule to rotate the point 90° about the origin is given by

$(x, y) \Rightarrow(-y, x)$

Substituting the coordinates of the point C in the above formula, we get;

$(2, 1) \Rightarrow(-1, 2)$

Therefore, the coordinates of the point C' is (-1,2)

5 0

3 years ago

When truckers are on long-haul drives, their driving logs must reflect their average speed. Average speed is the total distance

KengaRu [80]

Answer:

0.500 hour

Step-by-step explanation:

Givens

Distance

Total Distance on the highway = 3hr * 60 miles / hr = 180 miles
Distance in the city = 20 miles
Total Distance = 180 + 20 = 200 miles

Time

Time on the highway = 3 hours
Time in the city = x

Formula

d /t = r

Solution

Average rate = 57.14
Total Distance / Total Time = average rate
200 / (3 + x) = 57.14 Multiply both sides by 3 + x
200 = 57.14 * (3 + x) Remove the brackets
200 = 57.14*3 + 57.14x Multiply the factors on the right
200 = 171.42 + 57.14x Subtract 171.42 from both sides.
200 - 171.42 = 171.42 - 171.42 + 57.14x So the subtraction
28.58 = 57.14x Divide by 57.14
28.57/57.14 = x
x = 0.500 hour Answer

3 0

3 years ago

Read 2 more answers

SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as

vovangra [49]

Answer:

a) $P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

$P(Z>1.25)=1-P(Z$

b) $P(400$

$P(-1$

c) $z=-0.842$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the SAT scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(500,100)$

Where $\mu=500$ and $\sigma=100$

We are interested on this probability

$P(X>625)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

And we can find this probability using the complement rule and with the normal standard table or excel:

$P(Z>1.25)=1-P(Z$

Part b

We are interested on this probability

$P(400$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(400$

And we can find this probability with this difference:

$P(-1$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-1$

Part c

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.8$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

If we use condition (b) from previous we have this:

$P(X$