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Scrat [10]
3 years ago
15

At a pet store, Ms. Pike paid $34.75 for 5 dog bowls and 7 bones. Mr. Moore paid $20.50 for 3 dog bowls and 4 bones. Determine t

he cost of one dog bowl(x) and one bone(y).
Mathematics
1 answer:
lianna [129]3 years ago
3 0

Answer:

Step-by-step explanation:

t a pet store, Ms. Pike paid $34.75 for 5 dog bowls and 7 bones. Mr. Moore paid $20.50 for 3 dog bowls and 4 bones. Determine the cost of one dog bowl(x) and one bone(y).

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89.6994g

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M = DV

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   = 89.6994g

Hope this helps!

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What is the total cost?
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$0.75

Step-by-step explanation:

if Felix wants 1/4 kilogram of marmalade and it cost $3 per kilogram, you divide 3/4= 0.75

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Which of these strategies would eliminate a variable in the system of equations?
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because would die of boredom .and i wouldn't have Netflix or Hulu.and i would have nothing to look forward. to when i come home from school i would have to just go and do homework. and without TV you would not have a news channel so you wouldn't know things to are important to life.

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5 0
3 years ago
The manager of a music store has kept records of
Marysya12 [62]

Answer:

(a) P(x\le 3) = 0.75

(b) P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

(c) P(x \ge 5) = 0.10

(d) P(x = 1\ or\ 2) = 0.55

(e) P(x > 2) = 0.45

Step-by-step explanation:

Given

\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

Solving (b): Probability of at most 3 CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x \ge 5) = P(5) + P(6\ or\ more)

This gives:

P(x\ge 5) = 0.05 + 0.05

P(x \ge 5) = 0.10

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}

This probability is calculated as:

P(x = 1\ or\ 2) = P(1) + P(2)

This gives:

P(x = 1\ or\ 2) = 0.30 + 0.25

P(x = 1\ or\ 2) = 0.55

Solving (e): Probability of more than 2 CDs

Here, we consider:

\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)

This gives:

P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05

P(x > 2) = 0.45

3 0
3 years ago
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