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Oksanka [162]
3 years ago
9

Ben and Carla each have $10 to spend. They purchase 6 packs of baseball cards for $0.79 each and two balls that cost $5.95 each.

if they split the cost evenly, how much change will they receive?
Expression: 10 - 6(0.79) + 2(5.95) /2
First step: Multiply the parentheses.
Second step: Rewrite the equation: 10 - 4.47 + 11.90 /2
Third step: Add 4.47 and 11.90. Then rewrite the expression.
Expression: 10 - 16.64/2
Fourth step: Divide 16.64 and 2.
Rewrite the expression: 10 - 8.32.
Ben and Carla will each have $1.68 left in change after their purchase.
Hope I helped!
Mathematics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

ok thank you

Step-by-step explanation:

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The question "What is the LCM and GCF of 36 and 81?" can be split into two questions: "What is the LCM of 36 and 81?" and "What is the GCF of 36 and 81?"

In the question "What is the LCM and GCF of 36 and 81?", LCM is the abbreviation of Least Common Multiple and GCF is the abbreviation of Greatest Common Factor.

To find the LCM, we first list the multiples of 36 and 81 and then we find the smallest multiple they have in common. To find the multiples of any number, you simply multiply the number by 1, then by 2, then by 3 and so on. Here is the beginning list of multiples of 36 and 81:

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Multiples of 81: 81, 162, 243, 324, 405, 486, etc.

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You flip a coin and then roll a 6-sided number cube (a die).
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Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

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