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anastassius [24]
3 years ago
9

25 POINTS AND BRAINLIEST 6. Consider the following figure.

Mathematics
2 answers:
avanturin [10]3 years ago
7 0

Answer:

Perimeter: 95+15*sqrt(3)=120.98

Area: ([45*sqrt(5)]/2) + 600=605.3

Nimfa-mama [501]3 years ago
3 0

Answer: The perimeter is 95 + 15 sqrt 3, and the area is 600 + 35 sqrt 3 / 2

Step-by-step explanation:

We can draw an imaginary line to form a 30 60 90 triangle. The ratio of side lengths in this special triangle is 1 sqrt 3 2. We are given that the side length opposite to 60 degrees is 15. 15 divided by sqrt 3 is equal to 5 sqrt 3. Now, to find the diagonal we can do 5 sqrt 3 * 2 = 10 sqrt 3. So now, we can find the perimeter. The perimeter is equal to 15 + 40 + 40 + 5 sqrt 3 + 10 sqrt 3 = 95 + 15 sqrt 3. Now, we can find the area. The area can be split into the rectangle's area and the triangle's area. The rectangle's area is 15 * 40 = 600. The triangle's area is 15 * 5 sqrt 3 / 2 = 35 sqrt 3 / 2. The total area is 600 + 35 sqrt 3 / 2.

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How many soulutions does the system of equations have x-3y=15. and 3x-9y=45
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<h2>Infinitely many solutions</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}x-3y=15\\3x-9y=45\end{array}\right\\\\\text{Divide both sides of the second equation by 3}\\\\\dfrac{3x}{3}-\dfrac{9y}{3}=\dfrac{45}{3}\\\\x-3y=15\\\\\text{We received the same equation as the first.}\\\\\bold{CONCLUSION}\\\\\text{The system of equations has infinitely many solutions.}

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onsider the following hypothesis test: H 0: 50 H a: &gt; 50 A sample of 50 is used and the population standard deviation is 6. U
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Answer:

a) z(e)  >  z(c)   2.94 > 1.64  we are in the rejection zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) z(e) < z(c)  1.18 < 1.64  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) 2.12  > 1.64 and we can conclude the same as in case a

Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point  z(c) ;  α = 0.05  fom z table w get   z(c) = 1.64 we need to compare values (between z(c)  and z(e) )

The test hypothesis is:  

a) H₀      ⇒      μ₀  = 50     a)  Hₐ    μ > 50   ;    for value 52.5

                                          b) Hₐ    μ > 50   ;     for value 51

                                          c) Hₐ    μ > 50   ;      for value 51.8

With value 52.5

The test statistic    z(e)  ??

a)  z(e) =  ( μ  -  μ₀ ) /( σ/√50)      z(e) = (2.5*√50 )/6   z(e) = 2.94

2.94 > 1.64  we are in the rejected zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) With value 51

z(e) =  ( μ  -  μ₀ ) /( σ/√50)    ⇒  z(e) =  √50/6    ⇒  z(e) = 1.18

z(e) < z(c)  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e)  =  ( μ  -  μ₀ ) /( σ/√50)    ⇒ z(e)  = (1.8*√50)/ 6   ⇒ z(e) = 2.12

2.12  > 1.64 and we can conclude the same as in case a

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