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uranmaximum [27]
2 years ago
14

Please help! How many 3/8 pound burgers can be made from 3 pounds of ground turkey?

Mathematics
1 answer:
Finger [1]2 years ago
8 0
<h3>✽ - - - - - - - - - - - - - - -  ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Simply divide the two values:

3/(3/8) = 8

You can make 8 burgers.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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264/2200 = 0.12

0.12*100 = 12%
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Please help, I honestly have no clue if it's each or none.
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C there is no mode

Step-by-step explanation:

The mode is the number that appears most often.  Since there is no number that appears more than once, there is no mode

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3 years ago
PLZ HELP WILL GIVE BRAINLIEST IF ANSWER CORRECTLY
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Answer:

3 = y

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3 years ago
Last month, when Joe took his puppy to the veterinarian, the puppy weighed 4.6 pounds. This month, the puppy weighed 5.98 pounds
Karo-lina-s [1.5K]
Percent increase : (new number - original number) / (original number)....then multiply that by 100 to get ur percent.

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6 0
3 years ago
A student takes an exam containing 1414 multiple choice questions. The probability of choosing a correct answer by knowledgeable
Readme [11.4K]

Answer:

0.0082 = 0.82% probability that he will pass

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which 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)!}

And p is the probability of X happening.

In this problem we have that:

n = 14, p = 0.3.

If the student makes knowledgeable guesses, what is the probability that he will pass?

He needs to guess at least 9 answers correctly. So

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066

P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014

P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002

P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024

P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002

P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0

P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082

0.0082 = 0.82% probability that he will pass

6 0
3 years ago
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