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Fynjy0 [20]
3 years ago
15

Solve the equation. -1/3 + w = 5 3/8 what is w

Mathematics
1 answer:
Kruka [31]3 years ago
5 0
-1/3+w=5 3/8. First isolate the variable. w=5 3/8+1/3= 5 17/24
w=5 17/24
I hope this helps.
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PLZ HELP ASAP DUE RN
matrenka [14]

Answer:

The first and third options are correct so choose both

Step-by-step explanation:

4 0
2 years ago
Who made an error? <br> The correct scale factor= ?
Ostrovityanka [42]

Answer:

reggie made an error

the correct scale factor is 2/3

Step-by-step explanation:

we want to get from A to A', B to B', and ultimately C to C'

to get there, we must multiply each value in each point by the scale factor.

let's start out with reggie's scale factor. he multiplies each value in C by 3/2 to get to C'. we can try this out with one point, e.g. A

for A: 3/2 * (-12, -6) = (-18, -9). this is not A' = (-8, -4)! thus, 3/2 cannot be the scale factor

now, onto hillary's scale factor of 2/3

for A: 2/3 * (-12, -6) = (-8, -4). this is A'! thus, hillary is correct and reggie made an error

the correct scale factor is thus hillary's: 2/3

4 0
2 years ago
Which number satisfies the inequality 9h + 15 &gt; 55?
telo118 [61]
9h + 15 > 55
     - 15    - 15
9h > 40
h > 40/9
h > 4.44444.
any number greater than 4.44444
4 0
3 years ago
Read 2 more answers
The equation y = 35.25x + 40 gives the labor cost y of repairing a car if takes x hours to repair the car.
Lady_Fox [76]
Y = 32.25 x + 40
If it takes 6 hours to repair a car:
y = 32.25 * 6 + 40 = 193.5 + 40 = $233.50
The closest are $251.50 and $211.50.
$251.50 - $233.50 = $18
$233.50 - $211.50 = $22
$18 < $22
Answer:
The closest estimate is A ) $251.50
3 0
3 years ago
Read 2 more answers
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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