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oksano4ka [1.4K]
3 years ago
15

Can someone show how to solve this step by steps

Mathematics
1 answer:
horrorfan [7]3 years ago
3 0

Answer:

$133.33

Step-by-step explanation:

When dealing with such a problem we can use the Rule of Three. This rule is basically used in order to find the missing value when dealing with a ratio. Like so...

12% of final price  <======>   $16

100% of final price   <=====>   x

Now we multiply the two available diagonal values together and divide by the last value in order to get the value of the variable, which in this case would be the original price in dollars

(100 * 16) / 12 = x

1600 / 12 = x

133.33 = x

Finally, we can see that the original price of the item was $133.33

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At the beginning of the week, a puppy weighed 5 1/4 pounds. During the week, the puppy gains 8 ounces.
lys-0071 [83]

Answer:

5 3/4

Step-by-step explanation:

8 ounces is half (1/2) of a pound

1/2 is equivalent to 2/4

So I added 5 1/4 and 2/4

Which gives us 5 3/4

5 0
3 years ago
Combine like terms: -21a+16a
Zinaida [17]
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4 0
4 years ago
Yall can anyone help pls ? I'll give brainliest
viva [34]

Answer

5 1/3

Use Pemdas

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M-multiplication

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7 0
3 years ago
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Sue wants to increase 160 by 4%
ivann1987 [24]

Answer:

a) To increase 160 by 4% we multiply by 1.04

b) To decrease 160 by 6% we multiply by 0.94

Step-by-step explanation:

a) Here we have to increase 160 by 4% we are to note that;

4 \% = \frac{4}{100} = 0.04

Therefore, increasing 160 by 4% is to add additional 0.04×160 to 160

That is 160 + 0.04×160 = 160×(1 + 0.04) = 160×1.04

Therefore Sue should have multiplied 160 by 1.04 to increase 160 by 4%.

b) Here to decrease 160 by 6%, we subtract 6% of 160 from 160 that is;

Decrease 160 by 6% = 160 - 160×0.06 = 160×(1 - 0.06) = 160 × 0.94 = 150.4

Hence to decrease 160 by 6% we multiply by 0.94.

8 0
4 years ago
Find the greatest possible error for the measurement 0.991 g.
Veronika [31]
Explanation:

Any calibration scale consists of markings indicating calibrated values. The ‘space’ between the marks (lines) is the area of uncertainty with respect to the calibration.

Thus, the possible error is always one-half of the value between the markings, because ON either one you have a calibrated value. In between, no matter how close you think you can “judge” the distance, there is no calibrated reference point, so the ‘error’ of stating a value is +/- the value of half of the calibration accuracy. 0.991 is accurate (assuming that is the calibration limit), and 0.992 or 0.990 would also be “accurate”. The possible error is the +/- 0.0005 beyond that third digit that might be more to one side or the other.

That means the measured value of 0.991g could be between 0.9905g and 0.9915g.

4 0
3 years ago
Read 2 more answers
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