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dimaraw [331]
3 years ago
6

3m-7=8 solve please NEED HELP

Mathematics
2 answers:
Airida [17]3 years ago
8 0

m=5, simplify the equation to the form then move all terms containing m to the left and all other terms to the right after doing that simplify left and right side of the equation and lastly divide both sides of the equation by 3 to get m.

Charra [1.4K]3 years ago
7 0

3m-7=8

Add 7 to both sides

3m=15

Divide by 3.

M=5

Your answer is 5.

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Vera_Pavlovna [14]
It’s easy just make a line and put numbers each line to solve the linear equation and than solve it simple
4 0
3 years ago
Read 2 more answers
Cassidy says that capacity is the same as the amount. Do you agree? Explain why or why not.
telo118 [61]
The capacity is the same as the amount  because it refers to the amount of what is inside the object. Lets say you have a bag of chips it refers to the amount inside the bag of chips.
6 0
3 years ago
Paul buys 5 bags of chips and 9 bags of pretzels for $16.95. Ian buys 1 bag of chips and 1 bag of pretzels for $2.55. Find the c
egoroff_w [7]

Answer: The cost of one bag of chips is $1.5 and cost of one bag of pretzels is $1.05.

Step-by-step explanation:

Let x= Cost of one bag of chips

y= Cost of one bag of pretzels.

As per given, we have

5x+9y=16.95        ...(i)

x+y=2.55            ...(ii)

Multiply 5 to (ii), we get

5x+5y=12.75     ...(iii)

Eliminate (iii) from (i), we get

4y=4.2\\\\\Rightarrow\ y=1.05

Put value of y in (ii), we get

x+1.05=2.55\\\\\Rightarrow\ x=2.55-1.05=1.5

Hence, the cost of one bag of chips is $1.5 and cost of one bag of pretzels is $1.05.

3 0
3 years ago
The principal $3000 is accumulated with 3% interest, compounded semiannually for 6 years.
timurjin [86]
The formula is
A=p (1+r/k)^kt
A accumulated amount?
P principle 3000
R interest rate 0.03
K compounded semiannually 2
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A=3,586.85
7 0
3 years ago
Identify the class​ width, class​ midpoints, and class boundaries for the given frequency distribution.
Crank

Answer:

a. Class width=4

b.

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

c.

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

a)

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

b)

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

M=\frac{lower class limit+upper class limit}{2}

Class

Interval Midpoints

45-48 \frac{45+48}{2}=46.5

49-52 \frac{49+52}{2}=50.5

53-56 \frac{53+56}{2}=54.5

57-60 \frac{57+60}{2}=58.5

61-64 \frac{61+64}{2}=62.5

65-68 \frac{65+68}{2}=66.5

69-72 \frac{69+72}{2}=70.5

c)

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

7 0
3 years ago
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