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4vir4ik [10]
2 years ago
6

Jan has 35 teaspoons of chocolate cocoa mix and 45 teaspoons of french vanilla cocoa mix. She wants tomput the same amount of mi

x into each jar , and she only wants one flavor mix in each jar. She wants to fill as many jars as possible. How many jars of french vanilla cocoa mix wil Jan fill?
Mathematics
1 answer:
Alexxx [7]2 years ago
5 0

Answer:

9

the largest number 45tsp can be divided by is 9 so 45tsp/9 =5tsp per jar in 9 jars

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Question:

What are the solution(s) to the quadratic equation 50 – x2 = 0?

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D) no real solution

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C) x = ±5Plus or minus 5 StartRoot 2 EndRoot

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2 years ago
Sten sat on a bench and read a book while Manu was at soccer practice. He read for 35 minutes and stopped at 6:10 p.m.
Dahasolnce [82]

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1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,
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Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

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PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

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Therefore,

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