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otez555 [7]
3 years ago
15

Janice spent $54 to buy some pairs of pants each pair of pants cost the same whole dollar amount how many pairs of pants could s

he have bought
Mathematics
1 answer:
miv72 [106K]3 years ago
6 0
Think of factor pairs. She could buy 1 pair for $54, 54 pairs for $1, 2 pairs for $27, 27 pairs for $2, 3 pairs for $18, 18 for $3, 6 pairs for $9, or 9 pairs for $6
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Sanitizing with 171 degree water for how many seconds
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3600 seconds.
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Graph the quadratic function <br> F(x)=3x^2+6x-24
Oliga [24]
For this case we have the following function:
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7 0
3 years ago
A 16 ounce package of hamburger meat has 3% fat how much of the total package is fat
Oxana [17]
0.48 ounces

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8 0
3 years ago
A skier has decided that on each trip down a slope, she will do 2 more jumps than before. On her first trip she did 6 jumps. Der
Lady_Fox [76]
<span><span>∑<span>i=2</span><span>i=10</span></span>(4+2n)=</span><span> You can do it by hand, w/o much work . plug in n=2 into (4+2n) then plug in n=3 into (4+2n) then plug in n=4 into (4+2n) etc until you get to n=10. Add them all up.
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5 0
3 years ago
There are 30 students in a class. 10 students have a pet dog, 13 students have a pet cat, and 7 students have a pet fish. 4 stud
tankabanditka [31]

Answer:

9 students have pets

Step-by-step explanation:

From the above question, we are given the following information

Total number of students = 30

Let Pet Dog = D

Pet Cat = C

Pet Fish = F

Number is students that have pet dog

(D) = 10 students

Number of students that have pet cat (C) = 13 students

Number of students that have pet fish (F) = 7 students

Number of students that have Pet dog and cat ( D and C) = 4 students

Number of students that have Pet cat and fish (C and F) = 6 students

Number of student that has pet dog and pet fish (D and F) = 2 students

1 student has all three = ( D and F and C)

Number of student that have a pet Dog only

= n(D) - [n( D and C) + n( D and F) - n(D and C and F)]

= 10 -( 4+ 2 -1)

= 10 - 5

= 5

Number of student that have Pet cat only

= n(C) -[ n( D and C) + n( C and F) - n( D and C and F)]

= 13 -( 4 + 6 - 1)

= 13 - 9

= 4

Number is student that have a pet fish only

= n(F) - [n (C and F) + n( D and F) - n( D and C and F)]

= 7 - [6 + 2 - 1]

= 7 - 7

= 0

The number of students that have pets is calculated as:

(Number of students that have dogs only + Number of student that have cats only + Number of students that have fish only)

= 5 + 4 + 0

= 9

Therefore only 9 students have pets.

4 0
3 years ago
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