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Orlov [11]
2 years ago
12

Julie used her gift card for the local coffee shop to buy iced teas for herself and five friends. After she and one friend place

d their orders, the balance on Julie’s gift card was $14.85. After all six members of the group got their iced teas, she had a balance of $3.97 on her gift card. Determine the cost for one glass of iced tea.
Mathematics
1 answer:
valentina_108 [34]2 years ago
4 0

Answer:

One glass of iced tea cost $2.72

Step-by-step explanation:

Let,

x be the amount of gift card.

y be the amount of one glass of iced tea.

According to given statement;

Julie and her friend bought tea and remaining balance was $14.85

x - 2y = $14.85     Eqn 1

All the six members bought tea and remaining balance was $3.97

x - 6y = $3.97      Eqn 2

Subtracting Eqn 2 from Eqn 1

(x-2y)-(x-6y)=14.85-3.97\\x-2y-x+6y=10.88\\4y=10.88

Dividing both sides by 4

\frac{4y}{4}=\frac{10.88}{4}\\y=2.72

Hence,

One glass of iced tea cost $2.72

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Twelve people join hands for a circle dance.In how many ways can they do this? Suppose six of these people are men, and the other six are women. In how many ways can they join hands for a circle dance, assuming they alternate in gender around the circle

Answer:

86400 ways

Step-by-step explanation:

Since the circle can be rotated, the number of ways to arrange a distinct number of n objects in a circle will be (n−1)!.

Now, if we rotate the circle with the six women, we will see that there are 5! ways with which they can be placed in the circle.

After picking the places for the women, we will now fill each gap between two women with a man.

We have 6 men. Thus, number of ways to arrange the men is 6!

Thus,number of ways they can join hands for a circle dance, assuming they alternate in gender around the circle = 5! × 6! = 86400 ways

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If a- 2b +3c =0 prove that a³ -8b³+ 27c²= - 18abc​
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Step-by-step explanation:

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Answer:

2=173

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A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d
cestrela7 [59]

Answer:

(10) Person B

(11) Person B

(12) P(5\ or\ 6) = 60\%

(13) Person B

Step-by-step explanation:

Given

Person A \to 5 coins (records the outcome of Heads)

Person \to Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}

n(Head) = 32

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}

n(Dice) = 30

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Head)}

Pr(5) = \frac{1}{32}

Pr(5) = 0.03125

From person B list of outcomes, the proportion of 5 is:

Pr(5) = \frac{n(5)}{n(Dice)}

Pr(5) = \frac{8}{30}

Pr(5) = 0.267

<em>From the above calculations: </em>0.267 > 0.03125<em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

Median = \frac{n(Head) + 1}{2}th

Median = \frac{32 + 1}{2}th

Median = \frac{33}{2}th

Median = 16.5th

This means that the median is the mean of the 16th and the 17th item

So,

Median = \frac{3+2}{2}

Median = \frac{5}{2}

Median = 2.5

For person B

Median = \frac{n(Dice) + 1}{2}th

Median = \frac{30 + 1}{2}th

Median = \frac{31}{2}th

Median = 15.5th

This means that the median is the mean of the 15th and the 16th item. So,

Median = \frac{5+5}{2}

Median = \frac{10}{2}

Median = 5

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}

From the sample space of person B, we have:

n(5\ or\ 6) =n(5) + n(6)

n(5\ or\ 6) =8+10

n(5\ or\ 6) = 18

So, we have:

P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}

P(5\ or\ 6) = \frac{18}{30}

P(5\ or\ 6) = 0.60

P(5\ or\ 6) = 60\%

Question 13: Person with higher probability of 3 or more

Person A

n(3\ or\ more) = 16

So:

P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}

P(3\ or\ more) = \frac{16}{32}

P(3\ or\ more) = 0.50

P(3\ or\ more) = 50\%

Person B

n(3\ or\ more) = 28

So:

P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}

P(3\ or\ more) = \frac{28}{30}

P(3\ or\ more) = 0.933

P(3\ or\ more) = 93.3\%

By comparison:

93.3\% > 50\%

Hence, person B has a higher probability of 3 or more

7 0
2 years ago
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