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GuDViN [60]
3 years ago
6

Serena wants to create snack bags for a trip she is going on. She has 6 granola bars and 10 pieces of dried fruit. If the snack

bags should be identified without any food leftover, what is the greatest number of snack bags Serena can make?
Mathematics
1 answer:
Rainbow [258]3 years ago
5 0

Answer:

2

Step-by-step explanation:

Given the question :

Serena wants to create snack bags for a trip she is going on. She has 6 granola bars and 10 pieces of dried fruit. If the snack bags should be identified without any food leftover, what is the greatest number of snack bags Serena can make?

Number of granolas = 6

Number of dried fruits = 10

Since the snackbag is to be designed in such a way that there should be no food leftover, the greatest number of snack bags Serena can make could be obtained by getting the highest common factor of (6 and 10)

____6____10

2___3____5

Here, the highest common factor of 6 and 10 is 2

Hence, the greatest number of snack bags she can make is 2.

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Answer:

240

Step-by-step explanation:

5x4x12=240

if you multiply the amount of sizes he can get times the amount of crust sizes he can get you get the pizza it's self (20)

now including the toppings (12) this means 20x12 gets you the amount of toppings on those induviduale pizza's.

Hope this helps!

Brainliest please!

Please fully rate!

Have a great day!

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2 years ago
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The number of typing errors made by a typist has a Poisson distribution with an average of three errors per page. If more than t
damaskus [11]

Answer:

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

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.

Poisson distribution with an average of three errors per page

This means that \mu = 3

What is the probability that a randomly selected page does not need to be retyped?

Probability of at most 3 errors, so:

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

In which

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

P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240

Then

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0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

3 0
3 years ago
Break apart the addends to find the sum second grade
Leviafan [203]
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4 0
4 years ago
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Answer:

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Step-by-step explanation:

8 0
4 years ago
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