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Oxana [17]
3 years ago
10

If 3a + 2b = 7 and 2a + 2b = 9 , what is the value of

Mathematics
1 answer:
Simora [160]3 years ago
3 0
Hey there :)

We have two equations:
3a + 2b = 7 
2a + 2b = 9

We need to solve simultaneously to find the values of a and b

eq.1    3a + 2b = 7
eq.2 ( 2a + 2b = 9 ) x -1  ) multiply by -1 to cancel 2b

  3a + 2b = 7
- 2a - 2b = -9         ( Add both together )
-------------------
   a = - 2 

Substitute the value you found for a in a in order to find b

3( - 2 ) + 2b = 7                                                     2( - 2 ) + 2b = 9 
- 6 + 2b = 7                                    OR                  - 4 + 2b = 9
2b = 13                                                                   2b = 13        
b = \frac{13}{2}                                                                 b = \frac{13}{2} 

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Ludmilka [50]
1.) 52:97.5
2.) 1:1.875

Hope this helps!
5 0
3 years ago
If you get both of these questions right I will give you 50 points
Stella [2.4K]

Answer:

1) c 226

2)

1. E

2. D

3. B

4. A

5. C

Step-by-step explanation:

6^3=216

5x(4-2)

5x2=10

216+10=226

8 0
3 years ago
The mass, m kilograms, of a horse is 429 kg, correct to the nearest kilogram.
Liula [17]

Answer:

400 < m< 430 kg

Step-by-step explanation:

the mass of horse is 429 kg

estimated to the nearest hundred is 400  kg

estimated to the nearest  tens is 430 kg

thus the value of m mass lies between 400 to 430

therefore

400 < m< 430 kg

6 0
1 year ago
Read 2 more answers
Help me with this two questions
Anni [7]
First question: 0

Second question: 0.94
5 0
3 years ago
Suppose each of 12 players rolls a pair of dice 3 times. Find the probability that at least 4 of the players will roll doubles a
polet [3.4K]

Answer:

Our answer is 0.8172

Step-by-step explanation:

P(doubles on a single roll of pair of dice) =(6/36) =1/6

therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)

=1-(1-1/6)3 =91/216

for 12 players this follows binomial distribution with parameter n=12 and p=91/216

probability that at least 4 of the players will get “doubles” at least once =P(X>=4)

=1-(P(X<=3)

=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)

=1-0.1828

=0.8172

7 0
3 years ago
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