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vagabundo [1.1K]
3 years ago
9

There is 2 and 3/4 punds of roast beef and 1 and 7/8 pounds of turkey and 2 1/16 pounds of salami if you must put 1/4 of pound o

f meat on sandwiches how many whole sandwiches of each type can you make
Mathematics
1 answer:
rosijanka [135]3 years ago
8 0
<span>2 and 3/4 punds of roast beef   2  3/4 =  (4x2) + 3 /4=  11/4 roast beef

</span><span>1 and 7/8 pounds of turkey     1  7/8 =  15/8 </span>turkey 

2 1/16 pounds of salami     2 1/16      33/16 <span>salami

</span><span>if you  put 1/4 of pound of meat 
</span>then 11/4 roast beef   divided by 1/4 = 11/4 x4 = 11 sandwiches roast beef 

15/8 turkey<span> divided  1/4 = 15/8   x 4 = 15/2  = 7 sandwishes
</span>
 33/16 salami divided <span> 1/4  = 33/16 x 4 = 33/ 4 = 8 sandwiches</span>
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Abby is registering at a Web site and must select a six-character password.The password can contain either letters or digits.
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Answer:

a. 2176782336 with repeated characters. 1402410240 with no repeated characters.

b. 7964171460 when all letters can be repeated, but not numbers. 118813760 when only one number and letters can be repeated.

I believe that the password with all letters able to repeat and numbers not being able to is the most secure password, because if someone were to guess the password there is a 1/7964171460 chance of guessing the password.

Step-by-step explanation:

A. Since there are 26 different letters and 10 different numbers, there are 36 characters we can type. If characters can be repeated then, there is 2176782336 different passwords, since on all six number there are 36 possibilities each. So, 36 x 36 x 36 x 36 x 36 x 36 or 36^6 is to evaluated to find the answer. If characters are not to be repeated, there are 1402410240 different passwords, since on the first number there are 36, second has 35, third has 34, fourth has 33, fifth has 32, and sixth has 31. So, 36 x 35 x 34 x 33 x 32 x 31 is to be evaluated to solve this.

B. Since all characters that are letters can be repeated, then there are 26 letters to use forever and 10 numbers you can use with a limit. So, 36 x 35 x 34 x 33 x 32 x 31 which is to be solved if only numbers were to be used, which is 1402410240. Then, you add that with 36 x 35^5 and 36 x 35 x 34^4 and 36 x 35 x 34 x 33^3 and 36 x 35 x 34 x 33 x 32^2. which will be 7964171460. If the password must contain one digit, then you must multiply 10 with 26^5. Since, there is 10 different digits to use for the first number and 26 letters to choose from for the other five. So, it will be 26^5 times 10 which is 118813760.

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3 years ago
Which equation represents the relationship “4 less than a number is 10?”
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A stack of cards consists of seven red and four blue cards. A second stack of cards consists of eleven red cards. A stack is sel
Ann [662]

Answer:

0.1733 = 17.33% probability the first stack was selected.

Step-by-step explanation:

To solve this question, it is needed to understand conditional probability, and the hypergeometric distribution.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

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)!}

Conditional probability:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

Probability of all red cards for the first stack:

For this, we use the hypergeometric distribution, as the cards all chosen without replacement.

7 + 4 = 11 cards, which means that N = 11.

7 red, which means that k = 7

3 are chosen, which means that n = 3

We want all red, so we find P(X = 3).

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 3) = h(3,11,3,7) = \frac{C_{7,3}*C_{4,0}}{C_{11,3}} = 0.2121

Conditional probability:

Event A: All red

Event B: From the first stack.

Probability of all red cards:

0.2121 of 50%(first stack)

1 of 50%(second stack). So

P(A) = 0.2121*0.5 + 1*0.5 = 0.60605

Probability of all red cards and from the first stack:

0.21 of 0.5. So

P(A \cap B) = 0.21*0.5 = 0.105

What is the probability the first stack was selected?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.105}{0.60605} = 0.1733

0.1733 = 17.33% probability the first stack was selected.

4 0
3 years ago
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