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erastova
1 month ago
14

A customer wants you to enlarge a photo to 2 3 4 234 its current height. The photo's current height is 3 1 4 314 inches. What sh

ould its enlarged height be, in inches? 6 6 3 16 6316 6 11 16 61116 7 8 15 16
Mathematics
1 answer:
zzz [12.3K]1 month ago
4 0

Answer:

The question lacks clarity, so here is a more clearly articulated question along with the response:

A customer requests to scale up a photo to 2 3/4 times its existing height. The current height of the photo stands at 3 1/4 inches. What will the new height be after enlarging, in inches?

Answer:

Enlarged\ height = 8\frac{15}{16}\ inches

Step-by-step explanation:

current\ height\ =\ 3\frac{1}{4} = \frac{13}{4} \\ magnification\ factor = 2\frac{3}{4} = \frac{11}{4} \\Magnification (enlargement)\ =\ (magnification\ factor) \times (current\ height)\\Magnification\ =\ \frac{11}{4} \times \frac{13}{4}\\ Magnification\ =\ \frac{143}{16}

Next, we convert the improper fraction into a mixed number:

When 143 is divided by 16, the result is 8 whole numbers with a remainder of 15, thus the mixed number is:

\frac{143}{16} = 8\frac{15}{16} (A scientific calculator can also be utilized)

This means the enlarged height will be = 8\frac{15}{16}\ inches

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The result is 75,905.
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The measures in the table describe the weights of animals that visited a vet on one day, in pounds. Mean Median Mode Mean Absolu
lawyer [12517]

Answer: On average, a pet's weight during this vet visit is approximately 2.4 pounds away from 12.9 pounds.

If the MAD for another day's weights was 1.5, then that day’s weights would be less variable compared to the weights of pets seen today.

Step-by-step explanation:

Given: The data in the table outlines the weights of animals visiting a vet one day, in pounds.

Mean = 12.9

Median= 12.0

Mode = 12.0

Mean Absolute Deviation = 2.4

It’s understood that the mean absolute deviation (MAD) of a dataset indicates the average distance between each data point and the mean. It reflects the variation present in the dataset.

Therefore, the average weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.

Moreover, if another day had a MAD of 1.5, and since 1.5 < 2.4,[TAG_42]]

it implies that the weights on that day would be less variable compared to those of pets encountered on this day.

8 0
1 month ago
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Sunita bought a 10 kg box of grapes from the market. She gave 3 1/2 kg of her grapes to her friend Reena and 2 1/4 kg to Anita.
Leona [12618]

Answer:

17/4 kg

Step-by-step explanation:

Sunita initially has = 10kg of grapes

Amount distributed:

To Reena =3 1/2 kg

7/2 kg

To Anita =2 1/4 kg

9/4 kg

Total grapes given away

7/2+9/4

(14+9)/4

23/4 kg

The remaining grapes after distribution = Total grapes first minus grapes given away

10/1 - 23/4

(40-23)/4

= 17/4 or 4 1/4

Thus, she retains 17/4 kg of grapes

5 0
1 month ago
Profit is the difference between revenue and cost. The revenue, in dollars, of a company that manufactures cell phones can be mo
zzz [12365]

Answer:

  • The profit can be expressed as 70x + 50
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Detailed explanation:

The revenue function R(x) is provided as R(x)=2x^2+55x+10

The cost function C(x) is given as C(x)=2x^2-15x-40


Profit is calculated by subtracting cost from revenue, expressed as Profit = Revenue − Cost

We substitute the given revenue and cost functions into this formula.

Denote the profit function as P(x).

P(x)=R(x)-C(x)\\P(x)=2x^2+55x+10-(2x^2-15x-40)\\P(x)=2x^2+55x+10-2x^2+15x+40\\P(x)=70x+50

Therefore, the profit function formula simplifies to 70x + 50


Since x denotes the quantity of phones sold, to find the profit when 240 phones are sold, we substitute x = 240 into the profit expression above.

P(x)=70x+50\\P(240)=70(240)+50\\P(240)=16,850

Hence, selling 240 phones yields a profit of $16,850

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2 months ago
Gary has 32 ounces of soda. He shares it with 3 of his friends., splitting the soda evenly into 4 cups. He drinks 8 ounces of so
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0 ounces remain; since he distributed it evenly among four individuals, each cup contained 8 ounces, meaning he consumed the entire cup.

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