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Olenka [21]
3 years ago
11

In AMNP, what is sin M? 12 P M 5 13 N

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
4 0

Answer:

A

Step-by-step explanation:

triangle ez ez and yes i am smart big brain

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What is the area of the shaded portion of the circle? (16π – 32) in2 (16π – 8) in2 (64π – 32) in2 (64π – 8) in2
GrogVix [38]

Answer:

(16π - 32) in²

Step-by-step explanation:

In the picture attached, the circle is shown.

Area of one-quarter of circle = 1/4*Area of a circle = 1/4*π*radius² = 1/4*π*8² = 16π in²

Area of the triangle = 1/2*base*height = 1/2*8*8 = 32 in²

Shaded area = Area of one-quarter of circle - Area of the triangle = (16π - 32) in²

5 0
3 years ago
The average size of a new house built in a Hays county in 2010 was 1,872 square feet. A random sample of 40 new homes built in H
vlada-n [284]

Answer:

The test statistic is z = 2.7.

The pvalue of the test is 0.0035 < 0.05, which means that this sample provides enough evidence to conclude that the average house size has increased in the Hays County since 2010.

Step-by-step explanation:

The average size of a new house built in a Hays county in 2010 was 1,872 square feet. Evidence that it has increased?

This means that the null hypothesis is:

H_0: \mu = 1872

And the alternate hypothesis is:

H_a: \mu > 1872

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

1872 is tested at the null hypothesis:

This means that \mu = 1872

Sample of 40. The average square footage was 2,031 with a standard deviation of 373 square feet.

This means that n = 40, X = 2031, \sigma = 373

Value of the test-statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{2031 - 1872}{\frac{373}{\sqrt{40}}}

z = 2.7

Pvalue and decision:

The pvalue of the test is the probability of finding a sample mean above 2031. This is 1 subtracted by the pvalue of z = 2.7.

Looking at the z-table, z = 2.7 has a pvalue of 0.9965

1 - 0.9965 = 0.0035

The pvalue of the test is 0.0035 < 0.05, which means that this sample provides enough evidence to conclude that the average house size has increased in the Hays County since 2010.

7 0
3 years ago
If AD⎯⎯⎯⎯⎯⎯⎯⎯
kondor19780726 [428]

Answer:

I don't really understand the question

Step-by-step explanation:

AD

5 0
3 years ago
Disregard my work, but how do you get the answer? There is also a graph at the bottom which is optional to use.
kakasveta [241]
\bf ~~~~~~~~~~~~\textit{function transformations}&#10;\\\\\\&#10;% templates&#10;f(x)=  A(  Bx+  C)+  D&#10;\\\\&#10;~~~~y=  A(  Bx+  C)+  D&#10;\\\\&#10;f(x)=  A\sqrt{  Bx+  C}+  D&#10;\\\\&#10;f(x)=  A(\mathbb{R})^{  Bx+  C}+  D&#10;\\\\&#10;f(x)=  A sin\left( B x+  C  \right)+  D&#10;\\\\&#10;--------------------

\bf \bullet \textit{ stretches or shrinks horizontally by  }   A\cdot   B\\\\&#10;\bullet \textit{ flips it upside-down if }  A\textit{ is negative}\\&#10;~~~~~~\textit{reflection over the x-axis}&#10;\\\\&#10;\bullet \textit{ flips it sideways if }  B\textit{ is negative}\\&#10;~~~~~~\textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{  C}{  B}\\&#10;~~~~~~if\ \frac{  C}{  B}\textit{ is negative, to the right}\\\\&#10;~~~~~~if\ \frac{  C}{  B}\textit{ is positive, to the left}\\\\&#10;\bullet \textit{ vertical shift by }  D\\&#10;~~~~~~if\   D\textit{ is negative, downwards}\\\\&#10;~~~~~~if\   D\textit{ is positive, upwards}\\\\&#10;\bullet \textit{ period of }\frac{2\pi }{  B}

with that template in mind.

since the original or "parent" function is y = x²−4x+3, if we change the "x" argument to "x-2", we end up with a horizontal shift.

the x-2 part would be in the template the Bx+C part, with B = 1 and C = -2, or a horizontal shift to the right of 2/1 or 2 units.

since the parent function has a point at (2, -1), if we move that horizontally only to the right, we'd end up at (4, -1).
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4 years ago
Aaron Lloyd what is a?<br>​
algol [13]

Answer:

Rugby lawyer

Step-by-step explanation:

Aaron is a partner in the firm’s dispute resolution division. He advises clients on a range of litigious and risk related matters, with particular expertise in the areas of corporate misconduct, white collar criminal and regulatory affairs, sports law and employment law.  Aaron leads our sports law practice, and is a member of the firm’s health and safety, public law, and organisational integrity teams.

Well regarded by clients for his ability to analyse and strategise complex situations, Aaron is internationally recognised for his ability to implement pragmatic and commercial strategies to minimise risk and create opportunity. This ability has resulted in clients avoiding significant litigation and commercial consequences.

Aaron is an experienced advocate, having argued cases in the District Court, High Court, Employment Court, the Court of Appeal and Supreme Court of New Zealand, along with numerous tribunals.

He is recognised by international legal directories including by Chambers & Partners (Asia Pacific), Who’s Who Legal, Expert Guides, Best Lawyers and Doyles.

Before joining MinterEllisonRuddWatts Aaron practiced as a barrister with Paul Davison QC, and has lectured at the University of Auckland.

6 0
3 years ago
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