How can u enlarge or reduce a drawing proportinally

2 answers:

7 0

The process for dialating a figure by drawing parallel segments. ratio ... a number that describes by how much you have reduced or enlarged a map, ... if you can rotate and/or flip one of them so that you candilate it onto the other; two figures are similar if one is congruent to a dilation of the other. splits two sides proportinally.

Tresset [83]3 years ago

4 0

Maybe this website can help you with your question.
https://www.asu.edu/cfa/wwwcourses/art/SOACore/scalemain.htm

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Suppose that the distribution of IQ's of North Catalina State University's students can be approximated by a normal model with m

konstantin123 [22]

Answer:

The required probability is 0.6517

Step-by-step explanation:

Consider the provided information.

North Catalina State University's students can be approximated by a normal model with mean 130 and standard deviation 8 points.

μ₁ = 130 and σ₁ = 8

Chapel Mountain University's students can be approximated by a normal model with mean 120 and standard deviation 10 points.

μ₂ = 120 and σ₂ = 10

As both schools have IQ scores which is normally distributed, distribution of this difference will also be normal with a mean of μ₁-μ₂ and standard deviation will be $\sqrt{\sigma_1^2+\sigma_2^2}$

Therefore,

μ = 130-120=10

$\sigma = \sqrt{8^2+10^2}=12.806$

Now determine the probability of North Catalina State University student's IQ is at least 5 points higher than the Chapel Mountain University student's IQ:

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

$z=\frac{5-10}{12.806}\approx-0.39$

Now by using the z table we find the z- score of -0.39 is 0.6517.

Hence, the required probability is 0.6517

6 0

3 years ago

I will mark brainliest please help

amid [387]

DGE, it’s a vertical angle

4 0

3 years ago

Limitation of -4 to the left of the absolute value of x+4 divided by x+4

timama [110]

I believe you're asking about the one-sided limit,

$\displaystyle \lim_{x\to-4^-}\frac{|x+4|}{x+4}$

Recall the definition of absolute value:

• $|x| = x$ if $x\ge0$

• $|x| = -x$ if $x < 0$

Since we're approaching -4 from the left, we're effectively focusing on a domain of $x$ or $x+4$ . So, by the definition above, we have $|x+4| = -(x+4)$ . Then in the limit, we have