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mina [271]
3 years ago
15

True or False? If false provide a counterexample

Mathematics
1 answer:
Leto [7]3 years ago
3 0
Im gonna go with true on this one
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Solve the system by graphing or using a table 3x+y=5 x-y=7
Serggg [28]
<span>3x+y=5
x -y=7
------------add
4x = 12
  x = 3

</span>x-y=7
3 - y = 7
y = 3 -7
y = -4

answer
(3, -4)
5 0
4 years ago
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Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviati
Ilia_Sergeevich [38]

Answer:

a) \bf 0.3446^{25}=2.7095*10^{-12}

b) 3,624.25 pounds

Step-by-step explanation:

a)

Since 4300/25 = 172, in average every single person should weight 172 pounds to exceed the design limit.

The probability that one person weights 172 pounds or more is the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of 172.

<em>In Excel and OpenOffice Calc, this value is found with the formula </em>

<em>=1-NORMDIST(172;160;30;1) </em>

<em> (NORMDIST(172;160;30;1) gives the area to the left of 172, so 1-NORMDIST(172;160;30;1) gives the area to the right of 172) </em>

and equals 0.3446

(see picture 1)

Hence, the probability that all the 25 people exceeds 172 pounds equals

\bf 0.3446^{25}=2.7095*10^{-12}

b)

Similarly, we must find a weight w such that if p is the area to the left of w, then  

\bf p^{25}>0.0001  

so  

\bf p=\sqrt[25]{0.0001}=0.6918

w would be the point such that the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of w equals 0.6918.

<em>In Excel and OpenOffice Calc this is found with </em>

<em>=NORMINV(1-0.6918;160;30) </em>

and equals 144.97 pounds

(See picture 2)

and the design limit that is exceeded by 25 occupants with probability 0.0001 is 25*144.97 = 3,624.25 pounds

6 0
4 years ago
Are these triangles similar or no?
Marrrta [24]
Yes these triangles are similar

8 0
4 years ago
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GUYS CAN WE GET THE PERSON WHO ANSWERS THIS CORRECTLY 1,000 THANKS?
Alexeev081 [22]

Answer:

answer: 2

Heres the answer lol!

7 0
3 years ago
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A ship leaves port on a bearing of 34.0° and travels 10.4 miles. the ship then turns due east and travels 4.6 miles. how far is
maks197457 [2]

 is a clockwise angle measured from due North. This is a problem, because all of the trigonometric functions are referenced to a counterclockwise angle measured from East.

A bearing of <span>34∘</span> corresponds to a trigonometric angle of <span><span>θ1</span>=<span>90∘</span>−<span>34∘</span>=<span>56∘</span></span>

The (x,y) values for the position of the ship after completing its first heading are:

<span>x=<span>(10.4mi)</span><span>cos<span>(<span>56∘</span>)</span></span></span>
<span>y=<span>(10.4mi)</span><span>sin<span>(<span>56∘</span>)</span></span></span>

The trigonometric angle for the second heading is <span><span>θ2</span>=<span>90∘</span>−<span>90∘</span>=<span>0∘</span></span>

The (x,y) values for the position of the ship after completing its second heading is:

<span>x=<span>(10.4mi)</span><span>cos<span>(<span>56∘</span>)</span></span>+<span>(4.6mi)</span><span>cos<span>(<span>0∘</span>)</span></span>≈10.4mi</span>
<span>y=<span>(10.4mi)</span><span>sin<span>(<span>56∘</span>)</span></span>+<span>(4.6mi)</span><span>sin<span>(<span>0∘</span>)</span></span>≈8.6mi</span>

The distance from port is:

<span>d=<span>√<span><span><span>(10.4)</span>2</span>+<span><span>(8.6)</span>2</span></span></span>≈13.5mi</span>

Its trigonometric angle is:

<span>θ=<span><span>tan<span>−1</span></span><span>(<span>yx</span>)</span></span></span>

<span>θ=<span><span>tan<span>−1</span></span><span>(<span>8.610.4</span>)</span></span></span>

<span>θ≈<span>39.6∘</span></span>

The bearing angle is:

<span><span>θb</span>=<span>90∘</span>−<span>39.6∘</span>=<span>50.4<span>∘</span></span></span>

6 0
4 years ago
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