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insens350 [35]
3 years ago
10

lara has a tin of peanuts cashews and pecans she will pick a second nut without looking. how many outcomes are there for her two

picks?
Mathematics
2 answers:
irina1246 [14]3 years ago
7 0
There are six different possiblities of what she picks up.
KiRa [710]3 years ago
6 0
Possible procedures:

1). peanut, peanut
2). peanut, cashew
3). peanut, pecan
4). cashew, peanut
5). cashew, cashew
6). cashew, pecan
7). pecan, peanut
8). pecan, cashew
9). pecan, pecan.

Nine (9) possible procedures.
But ...
(2) and (4) produce the same final result.
(3) and (7) produce the same final result.
(6) and (8) produce the same final result.

So there are only <em><u>six (6)</u></em> possible different outcomes.
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Making Connections Read the word problem. Use the review vocabulary to describe what each number represents.​
Wittaler [7]

Answer:

dividend, divisor, quotient

Step-by-step explanation:

The answers are in the same order as the vocabulary words

8 0
2 years ago
15/8 + (-3 4/7) plz help me.
givi [52]
I think it’s -1.69642857143
4 0
4 years ago
Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m
strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

8 0
3 years ago
Help me with the answer! 14 points!!
guajiro [1.7K]

Answer:

hi :D

Step-by-step explanation:

8 0
3 years ago
Select the statement that correctly compares two numbers. (GIVING BRAINLIEST)
mars1129 [50]

Answer:

4.13=4.130

If there is a 0 at the end of a decimal then you can drop it.

6 0
3 years ago
Read 2 more answers
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