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Tanya [424]
3 years ago
14

Noah bought 15 baseball cards for $9. At this rate, what is the cost of 1 baseball card?

Mathematics
2 answers:
Paha777 [63]3 years ago
4 0

Answer:

$0.60

Step-by-step explanation:

9 divided by 15 = 0.6 so that would be $0.60, if you want to check if it's accurate then do 0.60x15 and you get 9

pychu [463]3 years ago
3 0
You do nine divided by fifteen which =$0.6
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1. Howard's CD House sells new CDs for $15 each and used CDs for $6 each. Bre'anna bought 51 CDs and spent a
Yanka [14]

The numbers of new CDs is 30 and the old CDs is 21.

To solve this problem, we have to write a system of equations in which we would know the numbers of each types of CD she bought.

<h3>System of Equations</h3>

This is used to solve a series or complex of word problems in which we can represent in a mathematical statements;

  • Let n represents numbers of new CDs
  • Let u represents numbers of old CDs

The equations can be represented as

n + u = 51 ...equation(i)\\15n + 6u = 576 ... equation(ii)

Let's take equation (i)

Making n the subject of formula

n + u = 51\\n = 51 - u...equation(iii)

we can put equation (iii) into equation (ii)

15n + 6u =576\\n = 51 - u\\15(51 - u) + 6u = 576\\765-15u + 6u = 576\\collect like terms\\765-576=15u - 6u\\189=9u\\9u/9 = 189/9\\u = 21

Let's put the value of u into equation (i)

n + u = 51\\u = 21\\n + 21 = 51\\n = 51 - 21\\n = 30

From the calculations above, the numbers of new CDs is 30 and the old CDs is 21.

Learn more on system of equations here;

brainly.com/question/13729904

4 0
3 years ago
What is the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34 Round yo
Galina-37 [17]

We are asked to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34.

The probability of a data score between two z-scores is given by formula P(a.

Using above formula, we will get:

P(-1.32

Now we will use normal distribution table to find probability corresponding to both z-scores as:

P(-1.32

P(-1.32

Now we will convert 0.27351 into percentage as:

0.27351\times 100\%=27.351\%

Upon rounding to nearest tenth of percent, we will get:

27.351\%\approx 27.4\%

Therefore, our required probability is 27.4% and option C is the correct choice.

5 0
3 years ago
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