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LekaFEV [45]
3 years ago
5

Ten tiles numbered 1 through 10 are placed in a bag. You randomly choose one tile. Without replacing the first tile, you randoml

y choose a second tile. Find the probability of choosing the 4 and then a number less than 5. Write your answer as a fraction in simplest form.
Mathematics
1 answer:
ruslelena [56]3 years ago
7 0
You would have one tenths (1/10) of a chance to get a 4 on the first go. you would have four ninths (4/9) of a chance to get a number less than 5 out of the bag on the second go.

the overall probability would be two forty-fifths (2/45)
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Line equations from point/slope
babunello [35]

Answer:

The answer is

y =  \frac{1}{4} x + 3

Step-by-step explanation:

Plug in and solve the equation. We are trying to find the y-intercept:

y=  \frac{1}{4}x  + b \\  2 =  \frac{1}{4} ( - 4) + b \\ 2 =  - 1 + b \\  \frac{ + 1 =  + 1  \:  \:  \:  \:  \:  \:  \:  \: }{3 = b}

We found the y-intercept, so now we plug it in it's right full spot in the formula.

y =  \frac{1}{4} x + 3

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2 years ago
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Aleksandr-060686 [28]

Answer:

Just say you have no friends you will beat the system.

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3 years ago
Cara grew 4 inches in a second grade and 3in in a third grade. If cara was 44 inches tall at the start of second grade, how tall
ivann1987 [24]

Answer:

51 inches (or 4 feet 2 inches)

Step-by-step explanation:

if she was 44 inches at the start of second grade, grew 4 inches through second, then grew 3 more inches through third grade, she would be 51 inches (Or 4 feet 2 inches)

5 0
3 years ago
A coach is ordering shirts for a team
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Answer:

Step-by-step explanation:

Yes its B C=8k+24

5 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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