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creativ13 [48]
3 years ago
12

Okay, After this i'm done asking for help. I hope..

Mathematics
2 answers:
frozen [14]3 years ago
6 0

Answer:

Hehe boi

Step-by-step explanation:

nalin [4]3 years ago
6 0
C
(Heheheh 20 characters)
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HELP PLS I NEED THIS >3
pychu [463]

Answer:

56

Step-by-step explanation:

Here, n = 8, r = 3

\huge \orange{ \because \: ^nC_r =  \frac{n!}{r!(n - r)!} } \\  \\  \huge \therefore \: ^8C_3 =  \frac{8!}{3!(8 - 3)!} \\  \\  \huge \therefore \: ^8C_3 =  \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!}  \\  \\   \huge \red{ \boxed{\therefore \:^8C_3 =   \purple{56}}}

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3 years ago
Question 1(Multiple Choice Worth 1 points)
tester [92]

The correct answer is d15

7 0
3 years ago
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andrey2020 [161]

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3 years ago
Write a Real World Problem for the equation x - 100 = 40. Then solve the equation
kenny6666 [7]
My uncle jimbob took 100 dollars away from me, now i only have 40

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7 0
4 years ago
Read 2 more answers
To 1 significant digit, how much Coumarin (in grams) would a 70 kg (154lb) student have to ingest to reach the LD50?
timofeeve [1]

Answer:

2 × 10 g

Step-by-step explanation:

The lethal dose 50 (LD50) of Coumarin is 293 mg/kg body mass. The amount of Coumarin that a 70 kg student would have to ingest to reach the LD50 is:

70 kg body mass × (293 mg/kg body mass) = 20510 mg

1 gram is equal to 1000 milligrams. The mass (in grams) corresponding to 20510 milligrams is:

20510 mg × (1 g/ 1000 mg) = 20.51 g ≈ 2 × 10 g

6 0
4 years ago
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