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alekssr [168]
2 years ago
6

Circumference of a circular disc is 157 cm, find its radius​

Mathematics
2 answers:
Zielflug [23.3K]2 years ago
7 0

Answer:

Owwwww would be a good time

just olya [345]2 years ago
7 0
Circumference = 2πr

Radius:

157 = 2πr

157/2π = r

r = 157/2π or 24.99 (nearest hundredth)
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Please show me how to solve this equation step by step!<br> 1/25(20-t)=4/25t-3/5
rjkz [21]

Remark

Multiply both sides of the equation by 25. Then you have an equation that you have already done probably.

Step One

Multiply both sides of the equation by 25

1 * (20 - t) = 4t - (3/5) * 25    Every part of the equation must be multiplied by 25

(20 - t) = 4t - 75/5                 3*25/5 = 75/5

20 - t = 4t - 15                       Add t to both sides

20 - t + t =4t + t  - 15             Combine like terms

20 = 5t - 15                            Add 15 to both sides

20 + 15 = 5t - 15 + 15             Combine like terms  

35 = 5t                                    Divide by 5

35/5 = 5t/5                             Switch sides

t = 7  Answer

5 0
2 years ago
Find the probability for the experiment of tossing a coin three times. Use the sample space S = {HHH, HHT, HTH, HTT, THH, THT, T
myrzilka [38]

Answer:

1) 0.375

2) 0.375

3) 0.5

4) 0.5

5) 0.875

6) 0.5                          

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375

2. The probability of getting exactly two tails

P(Exactly two tail)

Favorable outcomes ={ HTT,THT, TTH}

\text{P(Exactly two tail)} = \dfrac{3}{8} = 0.375

3. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

4. The probability of getting a tail on the last toss

P(tail on the last toss)

Favorable outcomes ={HHT,HTT,THT,TTT}

\text{P(tail on the last toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

5. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

\text{P(at least one head)} = \dfrac{7}{8} = 0.875

6. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

3 0
2 years ago
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