Variable c is 5 more than variable a. Variable c is also 3 less than variable a. Which pair of equations best models the relatio
2 answers:
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Diagram of two triangles
The two triangles are similar because they are both right triangles, meaning that one angle is 90° and the other two are acute (less than 90°).
The diagram on the left is missing its hypotenuse - the variable <em>c</em><em>.</em><em> </em>The hypotenuse is opposite of the right angle. The diagram on the right side is missing one of its legs.
Note: The hypotenuse is the <em>longest</em><em> </em>side of a right triangle.
Word Problen
The legs are 21 blocks and 20 blocks because they are by the right angle. Use the Pythagorean theorem to find the diagonal path's length, which is the hypotenuse.
Standard form of Pythagorean theorem.
Equation with the legs substituted and the missing hypotenuse value - <em> </em><em>c</em><em>.</em>
Square the legs and add.
Take the square root and simplify. The square root of 841 is 29 and -29, but distance is positive.
Thus the diagonal distance is 29 blocks.
Check by substituting.
Answer:
a) 0.1587 (or 15.87%)
b) 0.0478 (or 4.78%)
c) 0.7935 (or 79.35%)
Step-by-step explanation:
Find the following probabilities:
(a) the thickness is less than 3.0 mm
We need to find the area under the Normal curve with a mean of 4.5 mm and a standard deviation of 1.5 mm to the left of 3 mm
<h3>(See picture 1 attached) </h3>We can do that with the help of a calculator or a spreadsheet.
<em>In Excel use </em>
<em>NORMDIST(3,4.5,1.5,1) </em>
<em>In OpenOffice Calc use </em>
<em>NORMDIST(3;4.5;1.5;1) </em>
and we get the value 0.1587 (or 15.87%)
(b) the thickness is more than 7.0 mm
Now we need the area to the right of 7 (1 - the area to the left of 7)
<h3>(See picture 2 attached) </h3><em>In Excel use </em>
<em>1-NORMDIST(7,4.5,1.5,1) </em>
<em>In OpenOffice Calc use </em>
<em>1-NORMDIST(7;4.5;1.5;1) </em>
and we get the value 0.0478 (or 4.78%)
(c) the thickness is between 3.0 mm and 7.0 mm
We are looking for the area between 3 and 7
<h3>(See Picture 3) </h3>Since the area under the Normal equals 1, we have
Area to the left of 3 + Area between 3 and 7 + Area to the right of 7 = 1
Hence,
0.1587 + Area between 3 and 7 + 0.0478 = 1
and
Area between 3 and 7 = 1 - 0.1587 - 0.0478 = 0.7935 (or 79.35%)
