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olga_2 [115]
3 years ago
6

(3x+2y)(3y+2x)hiihhhhhhhhhhhhhhhh

Mathematics
1 answer:
Alik [6]3 years ago
4 0

Answer:

The answer is  <u>x = 2y</u>

Step-by-step explanation:

(3x + 2y) (3y + 2 y)

3x = 3y + 2y + 2y

3x = 7y

x = 7 ÷ 3

x = 2y

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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
3 years ago
A recipe called for the ratio of sugar to flour to be 7 : 2. If you used 63 ounce of sugar, how many ounces
lora16 [44]

Answer:

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63 ÷ 7 = 9

So one share is 9

9 × 2 = 18 ounces of flour

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