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ziro4ka [17]
2 years ago
6

Denzel has 3 2/5 boxes of party favors. One full box contains 15 bags of favors and each bag has 3 favors and it in addition to

this he has 7 xtra party favors that are not in bags or boxes. Select the correct operation two model a numerical expression for the total number of party favors
Mathematics
2 answers:
gladu [14]2 years ago
4 0

Answer:

You got 174 flavors

Easy peasy

:)


ddd [48]2 years ago
3 0

Given:

3 2/5 boxes of party favors.

1 full box contains 15 bags

each bag has 3 favors in it.

7 extra party bags no in boxes.


3boxes * 15 bags/box= 45 bags

2/5 boxes * 15 bags/box = (2*15)/5 = 30/5 = 6 bags

Additional 7 extra party bags.


Total number of bags:

45 + 6 + 7 = 58 party bags.


Total number of favors, 3 favors per bag.

58 party bags * 3 favors/bag = 174 favors.



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Answer:

C

Step-by-step explanation:

In this question, we are interested in calculating the z-score of a company employee.

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z-score = (x- mean)/SD

where in this case;

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Plugging these values into the equation above, we have;

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Answer:

The probabilities of each outcome are the following:

for X = 1 is 0.6

for X = 2 is 0.35

for X = 3 is 0.049

and for X = 4 is 0.001

Step-by-step explanation:

Let's consider X as the random variable for the sum of outcomes "S" exceeds 3, this is: S\geq 4

Let's now analyze and consider the ways that X equals the different values:

X = 1: I throw the dice and the result is: 4, 5 and 6.

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X = 2: The result after following the dice two times can be:

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2 and 2, 2 and 3... and so on until 2 and 6

3 and 1... and so on until 3 a 6

Then P(X=2) = P(1)xP(3,4,5,6) + P(2)xP(2....6) + P(3)x(1.....6)

Theory of Probability: Sum of all possible outcomes P(1)+P(2).......P(6) = 1

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X = 3: The result can be

1 and 1 and 2, 1 and 1 and 3.... until 1 and 1 and 6

1 and 2 and 1, 1 and 2 and 2..... until 1 and 2 and 6

2 and 1 and 1, 2 and 1 and 2.... until 2 and 1 and 6

Then P(X=3) = P(1)xP(1)xP(2....6) + P(1)xP(2)xP(1.....6) + P(1)xP(2)xP(1.....6)

P(X=3) = 0.1 x 0.1 x (1-0.1) + 0.1 x 0.2 x 1 + 0.2 x 0.1 x 1 = 0.01 x 0.9 + 0.2 + 0.2 = 0.049

Thus P(X=3) = 0.049

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1 and 1 and 1 and 1.... until 1 and 1 and 1 and 6

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