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Setler79 [48]
3 years ago
6

Shuping and joshua had the same amount of money. Joshua paid $9.10 for a bag and had $16.25 left. Shuping bought a pen and had $

19.60 left. How much did the pen cost
Mathematics
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

$5.75

Step-by-step explanation:

Joshua paid $9.10 for a bag and has $16.25 left

Now let's find out how much Joshua had before spending on a bag. To do this, we simply sum up the cost of the bag and how much Joshua had lefyt

9.10 + 16.25

=$25.35

Since Joshua and Shopping had the same amount, Shuping also had $25.35

To get the cost of the pen, we subtract the amount Shuping had left after purchasing the pen

This gives;

25.35 - 19.60

=$5.75

Aleonysh [2.5K]3 years ago
7 0

Answer: the cost of the pen is $5.75

Step-by-step explanation:

Let x represent the cost of the pen.

Shuping and joshua had the same amount of money. Joshua paid $9.10 for a bag and had $16.25 left. This means that the amount of money that Joshua had initially was

16.25 + 9.1 = $25.35

Shuping bought a pen and had $19.60 left. Since they have the same amount of money initially, it means that

x + 19.6 = 25.35

x = 25.35 - 19.6

x = $5.75

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Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


And since there are three of these combinations, The answer is


3\frac{5^2}{6^3}


Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

4 0
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