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stich3 [128]
3 years ago
6

deja bought a bag of mixed nuts the bag has 1/3 pounds of nuts she and her three friends split the nuts equal how much did each

person get ?
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
6 0

Answer:

Each friend gets 1/12

Step-by-step explanation:

1/3 pounds of nuts in total

4 people

1/3 / 4

1/3 * 1/4 = 1/12

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the answer is 12 or 9

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1 Helen and Emma share some money in the ratio 2 : 5
ddd [48]

The amount of money Emma receives in this scenario is £35.

<h3>What is Ratio?</h3>

This is defined as a comparison of numbers that indicates their sizes in relation to each other.

Helen and Emma = 2 : 5

2 = 5

x  = x+21

We then cross multiply

5x = 2 ( x+21)

5x = 2x + 42

5x - 2x = 42

3x = 42

x = 42/3 = 14

Emma received x+21

= 14+21 = £35

Read more about Ratio here brainly.com/question/2328454

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2 years ago
Which of the following fractions is in simplest form?<br> 9/16<br> 8/14<br> 6/20<br> 15/35
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4 years ago
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A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa
UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so \mu = \frac{14}{60} = 0.2333

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0
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