All categories

4 years ago

ANSWER THIS PLEASDEEEEEE

Mathematics

2 answers:

DiKsa [7]4 years ago

8 0

The greatest number of boxes is 8. Vegetables Is 7 and fruit is 6.

Citrus2011 [14]4 years ago

5 0

The vegetables will have 7 and the fruit will have 6 and At most is 8

You might be interested in

5+(−4) What is answwer

shepuryov [24]

Answer:

Step-by-step explanation:

the way to solve is just too look at the bigger picture. -4+5 is another way to write it. This would simply to 5-4, which is 1.

ℍ

5 0

2 years ago

Read 2 more answers

Water flows from the bottom of a storage tank at a rate of

Tresset [83]

>-------------------------------<

6 0

4 years ago

When you owe money, you have a negative balance. You owe $200 to your bank. You pay $30 back. What is your new bank balance?

Pachacha [2.7K]

-170$
because -200+30=-170

5 0

3 years ago

Read 2 more answers

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

4 years ago

If jen has 2 meters of taffy and 3 friends but she wants to split her taffy equally amongst her friends how many pieces of taffy

Reptile [31]

Well 2 ÷ 3 = 0.66666667
Each friend would get 0.7 taffy

6 0

4 years ago