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bixtya [17]
3 years ago
13

A family of four is trying to choose health insurance that will cost the least based on their typical care needs from the prior

year.
Compare the plans listed here. Explain which
plan you would choose and why.

PLAN A-has a monthly premium of $300 and charges a $10 co-pay per
prescription, but all office visits and dental care are included.

PLAN B- has a monthly premium of $250 and charges a $15 copay per
prescription and 10% copay per office visit. Two dental cleanings are included per person, but additional dental work requires a 20% copay.
Mathematics
1 answer:
Tcecarenko [31]3 years ago
3 0
PLAN B SO THEN IT WOULD OF BE THEN PLAN B SO PLAN B JUST PUT PLAN B BECAUSE I SAID SO
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Answers and the work for them please
Anon25 [30]

Answer:

The screen is black? there is nothing there

Step-by-step explanation:

this is nonsense man

3 0
2 years ago
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

=k^1+2k+1+k+1

=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

3 0
3 years ago
Can someone please help me asap
attashe74 [19]

Answer:

y=3x-4

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Your gas bill went up from $120 per month to $150 per month.What is the percent increase of your gas bill
Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

7 0
3 years ago
If ORBIT is coded BEOVG, then how is PLANET coded? /
arlik [135]

Answer:

1. CYNARG

Step-by-step explanation:

This a example of Caesar's Cipher, in which each letter in the original word leads to a ciphered letter according to the following equation:

C = (P + o) \text{mod} 26

In which C is the index of the Ciphered letter in the alphabet, P is the index of the original letter and o is the offset.

Finding the offset:

O is coded B

O is the 15th letter in the alphabet, so P = 15.

B is the 2nd letter in the alphabet, so C = 2

C = (P + o) \text{mod} 26

2 = (15 + o) \text{mod} 26

|o = |2-15| \text{mod} 26

o = 13

So

C = (P + 13) \text{mod} 26

PLANET:

P

P is the 16th letter in the alphabet.

C = (16 + 13) \text{mod} 26 = 3

So P is coded C.

L

L is the 12th letter in the alphabet:

C = (12 + 13) \text{mod} 26 = 25

L is coded Y(25th letter in the alphabet)

A

A is the 1st letter in the alphabet

C = (1 + 13) \text{mod} 26 = 14

A is coded N

N

N is the 14th letter in the alphabet

C = (14 + 13) \text{mod} 26 = 1

N is coded A

E

E is the 5th letter in the alphabet

C = (5 + 13) \text{mod} 26 = 18

E is coded R

So the correct answer is:

1. CYNARG

4 0
3 years ago
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