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harkovskaia [24]
3 years ago
6

James has 21 cups of flour he will pour the flour equally into 4 container how many cups of flour will be in each container

Mathematics
1 answer:
Degger [83]3 years ago
8 0

Answer:

5 1/4 cups of flour will be in each container

Step-by-step explanation:

21 ÷ 4= 5.25

5.25= 5 1/4

hope this helps babe <3

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A restaurant offers three sizes of coffee and four differerſt types of coffee.
Kruka [31]
<h3>Answer:  864</h3>

=======================================================

Work Shown:

There are,

  • 3 sizes of coffee
  • 4 types of coffee
  • 2 choices for cream (you pick it or you leave it out)
  • 2 choices for sugar (same idea as the cream)

This means there are 3*4*2*2 = 12*4 = 48 different coffees. We'll use this value later, so let A = 48.

There are 6 bagel options. Also, there are 3 choices in terms of if you order the bagel plain, with butter, or with cream cheese. This leads to 6*3 = 18 different ways to order a bagel. Let B = 18.

Multiply the values of A and B to get the final answer

A*B = 48*18 = 864

There are 864 ways to order a coffee and bagel at this restaurant.

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If you're curious why you multiply the values out, consider this smaller example.

Let's say you had 3 choices of coffee and 2 choices for a bagel. Form a table with 3 rows and 2 columns. Place the different coffee choices along the left to form each row. Along the top, we'll have the two different bagel choices (one for each column).

This 3 by 2 table leads to 3*2 = 6 individual table cells inside. Each cell in the table represents a coffee+bagel combo. This idea is applied to the section above, but we have a lot more options.

3 0
3 years ago
In a list of households, own homes and do not own homes. Five households are randomly selected from these households. Find the p
hichkok12 [17]

Answer:

The probability of success for this case would be:

p =\frac{9}{15}= 0.6 representing the proportion of homes that are own homes

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=15, p=0.6)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

And we want this probability:

P(X=3)

And uing the probability mass function we got:

P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165

Step-by-step explanation:

Adduming the following info: In a list of 15 households, 9 own homes and 6 do not own homes. Five households are randomly selected from these 15 households. Find the probability that the number of households in these 5 own homes is exactly 3.

Round your answer to four decimal places

P (exactly 3)=

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

The probability of success for this case would be:

p =\frac{9}{15}= 0.6 representing the proportion of homes that are own homes

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=15, p=0.6)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

And we want this probability:

P(X=3)

And uing the probability mass function we got:

P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165

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45÷50=90%
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