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Rasek [7]
3 years ago
14

Linda is training for a half marathon. she ran 3 3/4 miles monday, 3 miles tuesday, and 2 1/2 miles wednesday. how many miles di

d she run in those three days
Mathematics
1 answer:
GenaCL600 [577]3 years ago
4 0
She would've ran a total of 8 1/4 miles.
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What is the slope of the line that passes through the pair of points (-5/3, -1) and (-2, 9/2)?
vovangra [49]
Slope = (9/2 + 1) / (-2 + 5/3)
     
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= ----------
      -1/3
= 11/2 (-3/1)
= -33/2
or
= -16.5
7 0
3 years ago
35 x 30 = 35 x ____tens, <br> ____=____tens, <br> ___=____.
JulsSmile [24]

3 tens , 105 tens , 1050

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3 years ago
Jerome will also order t-shirt for the volunteers he will order both adults and child size t shirt
RideAnS [48]

a >= 200

5 a + 2.5 c <=1500

Step-by-step explanation:

Step 1 :

Let c be the number of child t shirt and a be the number of adult t-shirts

Given the cost of adult t-shirt is $5 and the cost of child t-shirt is $2.50

Step 2 :

Inequality 1

Given that Jerome needs at least 200 adult t-shirts .

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Step 3 :

Inequality 2

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5 a + 2.5 c <=1500

5 0
3 years ago
Read 2 more answers
What belongs in the box marked with the question mark in the proof?
koban [17]
<span>the blank boxes are for you to plug in x=20 to prove its right. so it would be 3(20)-4=2(20+8) 60-4=2(28) 56=56 so its true!</span>
6 0
3 years ago
A committee of 5 people is to be chosen from a group of 8 women and 10 men. How many diffferent committees are possible? How man
Jet001 [13]
A) 5 to be chosen among a Total : 10 Men + 8 Women

¹⁸C₅ = (18!)/(5!)(13!) = 8,568 groups of five

b) A must to have men and women. If so we have to deduct all groups of 5 that are all men and all group of 5 that are all women
Groups of 5 with only men: ¹⁰C₅ = 252
Groups of 5 with only women: ⁸C₅ = 56

So number of committees of 5 men and women mixed =
8568 - 252 - 56 = 8,260 committees 

c) 3 Women and 2 Men:

⁸C₃ x ¹⁰C₂ = 2,520 groups of 3 W and 2 M

d) More women than men, it means:
  3 W + 2 M OR (we have found it in c) = 2,520)
  4 W + 1 M OR   ⁸C₄ x ¹⁰C₁   →→→→ =    700
  5 W + 0 M OR   ⁸C₅ x ¹⁰C₀   →→→→  =     56

Total where W>M = 3,276 groups of 5 where women are at least 3


3 0
3 years ago
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