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AleksandrR [38]
3 years ago
7

Which of these is a trinomial?

Mathematics
2 answers:
7nadin3 [17]3 years ago
8 0
I’m thinking it’s D but i’m not too sure
Colt1911 [192]3 years ago
5 0
The answer is D I took the test
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Evaluate 20/6 - 2/3. 2 1/6, 2 2/3, 3 1/3, 3 2/3
N76 [4]

Answer:

Step-by-step explanation:

20/6 - 4/6= 16/6= 2 4/6= 2 2/3

7 0
2 years ago
Ank estimated that he drove about 450 miles in a week. He actually drove 485 miles. Find the percent error in Hank's estimate. R
yanalaym [24]

Answer:

7.2%

Step-by-step explanation:

Given data

Estimate=450miles

Actual= 485 miles

%error= actual-estimate/ actual*100

%error=35/ 485*100

%error=0.072*100

%error=7.2%

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2 years ago
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Hunter-Best [27]
the answer is A : m=-1
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2 years ago
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Cora uses a ratio of 18 cherries to 3 peaches when she makes a pie filling. Lena uses an equivalent ratio to Cora. If Lena uses
Lelechka [254]
Lena would have to use 54 cherries
3 0
2 years ago
Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h
Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

4 0
3 years ago
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