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Gnesinka [82]
3 years ago
15

Please help this is algebra 1 problem is in the picture

Mathematics
1 answer:
Naddika [18.5K]3 years ago
7 0

Answer:

0,1,2,3,4,5

Step-by-step explanation:

You can not buy more than 5 books because you'd have a negative amount of money. So 0,1,2,3,4,5 are the possible values for b.

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B+7=10 what is the sum of b?
juin [17]
B = 3. Hope it helps!
5 0
3 years ago
Can someone please help my with my work.
Aleksandr [31]

Answer:

X+0.375=y

Step-by-step explanation:

6 0
3 years ago
A cold tablet contains the following amounts of active ingredients: acetaminophen 325 mg, chlorpheniramine maleate 2 mg, and dex
kakasveta [241]

Answer:

The number of tablets that can be prepared is 3076.

Step-by-step explanation:

The total amount of active ingredients in the tablet is the sum of the amounts provided in the formula:

325 mg + 2mg+15 mg=342 mg

The percentages of each component in the formula are:

Acetaminophen:\frac{325mg*100}{342mg}=95.03%

Chlorpheniramine maleate:\frac{2mg*100}{342mg} =0.58%

Dextromethorphan hydrobromide:\frac{15mg*100}{342mg}=4.39%

If 1 Kg=10^{6} mg of acetaminophen is used, the needed amount of chlorpheniramine maleate would be:

\frac{10^{6} mg *0.58}{95.03}=6153.85 mg

Since there are 125 g = 125000 mg of chlorpheniramine maleate, there is enough of these ingredient to run the available acetaminophen out. Thus, the total amount of active ingredients that can be prepared with 1 kg of acetaminophen is:

\frac{10^{6}mg*100}{95.03}=1052307.7mg

Since each tablet weighs 342 mg, the number of tablets that can be prepared is:

\frac{1052307.7mg}{342mg}=3076.923

Which means that 3076 tablets can be prepared and a there will be a remanent of 0.923*342 mg = 315.69 mg of active ingredients.

7 0
3 years ago
What is the probability of drawing a 3 or a 4 or a heart from a deck of cards?
Effectus [21]
\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}

Answer:\ \dfrac{19}{52}
5 0
3 years ago
HELPPPPPPPPPPPPP 30 points!!!!!
Maru [420]

The answer is 2 and would you like an explanation

8 0
3 years ago
Read 2 more answers
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