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Crazy boy [7]
3 years ago
6

Martin has 64 sticks of gum altogether. How many packs of gum does he have?

Mathematics
1 answer:
klasskru [66]3 years ago
4 0
Doesn’t make sense.. how much can fit into one packet
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The sum of the ages of Mr Lin and his son Ethan is 38 years. In three years' time, Mr Lin will be three times as old as his son.
Sati [7]

Answer:

Mr.Lin's present age is 30 years now. His son is 8 years old now.

Step-by-step explanation:

Let L = Mr.Lin's present age,

   S = his son's present age.

First equation  is

L + S = 38.       (1)

The second equation is  

L + 3 = 3*(S+3)   (2)   ("In three years' time, Mr.Lin will be three times as old as his son.")

From (1), express S = 38-L and substitute it into (2). You  will get

L + 3 = 3*((38-L)+3),   or

L + 3 = 114 - 3L + 9,

4L = 114 + 9 - 3,  

4L = 120,

L = 30.

8 0
3 years ago
|2x + 5| = 13 please show work
gtnhenbr [62]

Answer:

x=4

Step-by-step explanation:

|2x + 5| = 13

      -5      -5

|2x| = 8

2x/2  8/2

|x| = 4

3 0
3 years ago
Daddy chill 6969696969
iren2701 [21]

Answer:

I've never been to waffle house

8 0
3 years ago
Read 2 more answers
A low-strength children’s/adult chewable aspirin tablet contains 81 mg of aspirin per tablet. How many tablets may be prepared f
Svetlanka [38]

Answer:

12,345 tablets may be prepared from 1 kg of aspirin.

Step-by-step explanation:

The problem states that low-strength children’s/adult chewable aspirin tablets contains 81 mg of aspirin per tablet. And asks how many tablets may be prepared from 1 kg of aspirin.

Since the problem measures the weight of a tablet in kg, the first step is the conversion of 81mg to kg.

Each kg has 1,000,000mg. So

1kg - 1,000,000mg

xkg - 81mg.

1,000,000x = 81

x = \frac{81}{1,000,000}

x = 0.000081kg

Each tablet generally contains 0.000081kg of aspirin. How many such tablets may be prepared from 1 kg of aspirin?

1 tablet - 0.000081kg

x tablets - 1kg

0.000081x = 1

x = \frac{1}{0.000081}

x = 12,345 tablets

12,345 tablets may be prepared from 1 kg of aspirin.

4 0
3 years ago
There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student
Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0
3 years ago
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