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Rasek [7]
3 years ago
8

Find the product of the first 3 positive integers and then the first 5 negative integers.

Mathematics
1 answer:
Westkost [7]3 years ago
7 0

Answer:

6 and -120

Step-by-step explanation:

The first 3 positive integers are 1, 2 and 3 and their product is 6, the first 5 negative integers are -1, -2, -3, -4 and -5 and their product is -120.

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Ivan has a $500 bond with a 5.8% coupon. Ivan purchased this bond for $515.
svetoff [14.1K]

Answer:

The yield is 5.974%

Step-by-step explanation:

We proceed as follows ;

coupon rate = Annual coupon payment/bond face value.

The face value is the original amount which the bond was bought and that is $515 according to the question. While the coupon rate is 5.8%

mathematically, annual coupon payment = coupon rate * bond face value = 0.058 * 515 = $29.87

mathematically;

current yield = Annual coupon payment/bond price

current yield = 29.87/500

= 0.05974 or simply 5.974%

so the answer is c. 5.6%

Step-by-step explanation:

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3 years ago
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Simplify 6 to the 3rd
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Answer:

Isnt it 2?

Step-by-step explanation:

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3 years ago
Find area of rhombus . please./.
lilavasa [31]

Answer: 126

Step-by-step explanation:

6 0
2 years ago
Write a word phrase for -12t+2
Dmitriy789 [7]
Negative twelve time T plus two
3 0
3 years ago
a teacher and 10 students are to be seated along a bench in the bleachers at a basketball game. In how many ways can this be don
Veronika [31]

Wow !

OK.  The line-up on the bench has two "zones" ...

-- One zone, consisting of exactly two people, the teacher and the difficult student.
   Their identities don't change, and their arrangement doesn't change.

-- The other zone, consisting of the other 9 students.
   They can line up in any possible way.

How many ways can you line up 9 students ?

The first one can be any one of 9.   For each of these . . .
The second one can be any one of the remaining 8.  For each of these . . .
The third one can be any one of the remaining 7.  For each of these . . .
The fourth one can be any one of the remaining 6.  For each of these . . .
The fifth one can be any one of the remaining 5.  For each of these . . .
The sixth one can be any one of the remaining 4.  For each of these . . .
The seventh one can be any one of the remaining 3.  For each of these . . .
The eighth one can be either of the remaining 2.  For each of these . . .
The ninth one must be the only one remaining student.

     The total number of possible line-ups is 

               (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)  =  9!  =  362,880 .

But wait !  We're not done yet !

For each possible line-up, the teacher and the difficult student can sit

-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.

That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .

So the total total number of ways to do this is

           (362,880) x (10)  =  3,628,800  ways.

If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !

4 0
3 years ago
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