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Eva8 [605]
4 years ago
6

Mitchell wants to order CDs on the Internet. Each CD costs $15.99, and shipping for the entire order is $9.99. Mitchell has no m

ore than $100 to spend. What is the maximum number of CDs that Mitchell can order?
Mathematics
2 answers:
katovenus [111]4 years ago
8 0

Answer:

maximum 5 CDs

Step-by-step explanation:

Let Mitchell can order a maximum of x CDs

It has been given that each CD costs $15.99, and shipping for the entire order is $9.99.

Thus, we have the total cost for x CDs

15.99x+9.99

Now, Mitchell has no more than $100 to spend. It means

15.99x+9.99

Subtract 9.99 to both sides

15.99x

Divide both sides by 15.99

x

Hence, Mitchell can order maximum 5 CDs

gogolik [260]4 years ago
4 0
<h2>Hello!</h2>

The answer is: Mitchell can order a maximum of 5 CDs.

<h2>Why?</h2>

Let's make the entire shipping cost apart to calculate the actual money amount Mitchell could spend just buying CDs,

Money To Spend = CDsPrice+ShippingPrice\\100=CDsPrice+9,99\\CDsPrice=100-9,99=90,1

So, he will be able to spend 90,1 $

In order to calculate how many CDs Mitchell could buy with 90,1$, she/he should divide the total available amount by the CDs price which is 15,99$

Doing the calculation we have:

\frac{90,1}{15,99} = 5,62

So, since a CD is an indivisible object,  Mitchell can order a maximum of 5 CDs.

Have a nice day!

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Answer:

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\frac{y}{1} :\frac{408}{9.5}

y × 9.5 = 408 × 1

9.5y = 408

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y=42\frac{18}{19}

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skelet666 [1.2K]

Answer:

<h2>1/4 Chances</h2><h2>25% Chances</h2><h2>0.25 Chances (out of 1)</h2>

Step-by-step explanation:

Two methods to answer the question.

Here are presented to show the advantage in using the product rule given above.

<h2>Method 1:Using the sample space</h2>

The sample space S of the experiment of tossing a coin twice is given by the tree diagram shown below

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From the three diagrams, we can deduce the sample space S set as follows

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tree diagram in tossing a coin twice

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<h2>Method 2: Use the product rule of two independent event</h2>

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with the probabilities of each event A and B given by

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Event E occurring may now be considered as events A and B occurring. Events A and B are independent and therefore the product rule may be used as follows

        P(E)=P(A and B)=P(A∩B)=P(A)⋅P(B)=12⋅12=14

NOTE If you toss a coin a large number of times, the sample space will have a large number of elements and therefore method 2 is much more practical to use than method 1 where you have a large number of outcomes.

We now present more examples and questions on how the product rule of independent events is used to solve probability questions.

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