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vaieri [72.5K]
3 years ago
13

Mickey used these calculations to find how much he would spend on 7 cartons of eggs, if 12 cartons of eggs cost $22.20. Describe

his error.
12 cartons(÷22.20)
$22.22(÷22.20)


Unit price = $0.54
$0.54(7) = $3.78
Answer ; Mickey set up the equivalent rate wrong. The cartons of eggs should be in the denominator, because you need to find the unit price for 1 carton of eggs.
Mathematics
2 answers:
IgorLugansk [536]3 years ago
5 0

Answer: you would divide the total price of $22.20 by the amount of cartons, which is 12, giving you the unit rate of $1.85 per carton. the unit rate, $1.85 multiplied by 7 is $12.95, so it would cost $12.95 for 7 cartons of eggs

Step-by-step explanation:

minnie2 years ago
0 0

Sample student response: Mickey set up the equivalent rate wrong. The cartons of eggs should be in the denominator, because you need to find the unit price for 1 carton of eggs.

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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

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Answer:

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Step-by-step explanation:

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4 years ago
A line passes through (-1,7)and(2, 10)
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The equation is 
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Answer:

179.50 in^{3}

Step-by-step explanation:

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7 0
4 years ago
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