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3 years ago

What is the length of DD'?

Mathematics

2 answers:

Blababa [14]3 years ago

8 0

Answer:

The length of DD' is $\sqrt{29}$ or 5.39 units.

Step-by-step explanation:

The distance formula is

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

From the given figure it is clear that the coordinates of D are (2,0) and coordinates of D' are (7,2).

Using distance formula we get

$DD'=\sqrt{(7-2)^2+(2-0)^2}$

$DD'=\sqrt{25+4}$

$DD'=\sqrt{29}$

$DD'\approx 5.39$

Therefore the length of DD' is $\sqrt{29}$ or 5.39 units.

zavuch27 [327]3 years ago

6 0

The length of DD' is 5.39 units.

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Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi

ziro4ka [17]

Answer:

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

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".

Solution to the problem

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

$X \sim N(5.75241,0.06281)$

Where $\mu=5.75241$ and $\sigma=0.06281$

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

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

For the case when z =-2 we can do this:

$-2 = \frac{X-5.75241}{0.06281}$

And if we solve for X we got:

$X = 5.75241 -2*0.06281 =5.6121$

And for the other case when Z=2 we have:

$2 = \frac{X-5.75241}{0.06281}$

And if we solve for X we got:

$X = 5.75241 +2*0.06281 =5.8886$

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

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3 years ago

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

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Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

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According to this theorem,

$BC^2=BD\cdot AB.$