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Answer:
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ASA and AAS
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Step-by-step explanation:
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We do not know if these are right triangles; therefore we cannot use HL to prove congruence.
We do not have 2 or 3 sides marked congruent; therefore we cannot use SSS or SAS to prove congruence.
We are given that EF is parallel to HJ. This makes EJ a transversal. This also means that ∠HJG and ∠GEF are alternate interior angles and are therefore congruent. We also know that ∠EGF and ∠HGJ are vertical angles and are congruent. This gives us two angles and a non-included side, which is the AAS congruence theorem.
Since EF and HJ are parallel and EJ is a transversal, ∠JHG and ∠EFG are alternate interior angles and are congruent. Again we have that ∠EGF and ∠HGJ are vertical angles and are congruent; this gives us two angles and an included side, which is the ASA congruence theorem.
Answer:
The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.
Step-by-step explanation:
With the weekly average we can estimate the daily average for customers, assuming 7 days a week:
We can model this situation with a Poisson distribution, with parameter λ=108. But because the number of events is large, we use the normal aproximation:
Then we can calculate the z value for x=100:
Now we calculate the probability of x>100 as:
The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.
Answer:
(x,y)→(y,-x)
Step-by-step explanation:
Parallelogram ABCD:
A(2,5)
B(5,4)
C(5,2)
D(2,3)
Parallelogram A'B'C'D':
A'(5,-4)
B'(4,-5)
C'(2,-5)
D'(3,-2)
Rule:
A(2,5)→A'(5,-2)
B(5,4)→B'(4,-5)
C(5,2)→C'(2,-5)
D(2,3)→D'(3,-2)
so the rule is
(x,y)→(y,-x)
Answer:
4
Step-by-step explanation:
Let's set up an equation using the formula for the area of a triangle.
Hint #22 / 3
\begin{aligned} \text{Area of a triangle} &= \dfrac12 \cdot \text{base} \cdot \text{height}\\\\ 12&= \dfrac12 \cdot 6 \cdot x \\\\ 12&= 3x \\\\ \dfrac{12}{\blueD{3}}&= \dfrac{3x}{\blueD{3}} ~~~~~~~\text{divide both sides by } {\blueD{ 3}}\\\\ \dfrac{12}{\blueD{3}}&= \dfrac{\cancel{3}x}{\blueD{\cancel{3}}}\\\\ x &=\dfrac{12}{\blueD{3}}\\\\ x &=4\end{aligned}
Area of a triangle
12
12
3
12
3
12
x
x
=
2
1
⋅base⋅height
=
2
1
⋅6⋅x
=3x
=
3
3x
divide both sides by 3
=
3
3
x
=
3
12
=4
