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Sindrei [870]
1 year ago
7

A pet store donated 50 pounds of food for adult dogs, puppies, and cats to an animal shelter. 19 and 3/4 pounds was adult dog fo

od and 18 and 7/8 pounds was puppy food. How many pounds of cat food did the pet store donate?
Mathematics
2 answers:
Sati [7]1 year ago
8 0
So first we will add together the fractions. 3/4 is equal to 6/8, so we will add 6/8 to 7/8 and get 13/8, or 1 5/8.  Then we can add together the whole numbers, 19 and 18, to get 37. Then add 37 + 1 5/8 and get 38 5/8. Lastly, we subtract that amount from 50 to get a finale number of 11 3/8. The pet store donated 11 3/8 pounds of cat food! Hope this helps!
madam [21]1 year ago
3 0
50 - (19 3/4 + 18 7/8) = the cat food
50 - (19 6/8 + 18 7/8) 
50 - (37 13/8) =
50 - (38 5/8) = 
11 3/8 lbs of cat food <==
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Each Friday, the school prints 400 copies of the school newsletter. The equation c = 400w models the relationship between the nu
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THE SECOND ONE IS ANY WHOLE NUMBER I JUST TOOK THE QUIZ

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The manager at Gabriela's Furniture Store is trying to figure out how much to charge for a book shelf that just arrived. The boo
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thy should sell it for $235.2 because in the shop they sell everything for 60% the original price for which they bought it

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2 years ago
give an example on an addition problem in which you would and would not group the addends differently to add
Murrr4er [49]
Addends are any of the numbers added together in an equation. 

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1 year ago
a car travels 30 1/2 miles in 2/3 of an hour. what is the average speed, in miles per hour, of the car ?
Contact [7]

Answer:

speed=225.75 miles per hour

Step-by-step explanation:

given:

s=301/2

t=2/3

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6 0
1 year ago
In high-school 135 freshmen were interviewed.
timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

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1 year ago
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