Which pair or pairs of polygons are congruent? A. 3 and 4 B. 1 and 3 C. 1, 3, and 4 D. 2, 3, and 4 E. 1, 2, 3, and 4
2 answers:
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Answer:
$12159 per year.
Step-by-step explanation:
If I invest $x each year at the simple interest of 7.5%, then the first $x will grow for 35 years, the second $x will grow for 34 years and so on.
So, the total amount that will grow after 35 years by investing $x at the start of each year at the rate of 7.5% simple interest will be given by
=
=
{Since sum of n natural numbers is given by }
= 35x + 47.25x
= 82.25x
Now, given that the final amount will be i million dollars = $1000000
So, 82.25x = 1000000
⇒ x = $12,158. 05 ≈ $12159
Therefore. I have to invest $12159 per year. (Answer)
Answer:
Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))
1. R is reflexive if for each element a∈B, aRa.
2. R is symmetric if satisfies that if aRb then bRa.
3. R is transitive if satisfies that if aRb and bRc then aRc.
Then, our set B is .
a) We need to find a relation R reflexive and transitive that contain the relation
Then, we need:
1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,
2. Observe that
- 1R4 and 4R1, then 1 must be related with itself.
- 4R1 and 1R4, then 4 must be related with itself.
- 4R1 and 1R2, then 4 must be related with 2.
Therefore is the smallest relation containing the relation R1.
b) We need a new relation symmetric and transitive, then
- since 1R2, then 2 must be related with 1.
- since 1R4, 4 must be related with 1.
and the analysis for be transitive is the same that we did in a).
Observe that
- 1R2 and 2R1, then 1 must be related with itself.
- 4R1 and 1R4, then 4 must be related with itself.
- 2R1 and 1R4, then 2 must be related with 4.
- 4R1 and 1R2, then 4 must be related with 2.
- 2R4 and 4R2, then 2 must be related with itself
Therefore, the smallest relation containing R1 that is symmetric and transitive is
c) We need a new relation reflexive, symmetric and transitive containing R1.
For be reflexive
- 1 must be related with 1,
- 2 must be related with 2,
- 3 must be related with 3,
- 4 must be related with 4
For be symmetric
- since 1R2, 2 must be related with 1,
- since 1R4, 4 must be related with 1.
For be transitive
- Since 4R1 and 1R2, 4 must be related with 2,
- since 2R1 and 1R4, 2 must be related with 4.
Then, the smallest relation reflexive, symmetric and transitive containing R1 is
Answer:
Step-by-step explanation:
(a) H0: μ_D=0
Ha: μ_D ≠ 0
b) Find attached the solution
(c) By technology,
p - value = 0.4437
Hence, the p-value is 0.4437
