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Mkey [24]
4 years ago
5

Use the elimination method to solve the system of equations.

Mathematics
1 answer:
ExtremeBDS [4]4 years ago
4 0

Answer:

Option D. (8, – 4)

Step-by-step explanation:

3x + 4y = 8 ..... (1)

x – y = 12.... (2)

To solve the above equation by elimination method, do the following:

Step 1:

Multiply equation 1 by the coefficient of x in equation 2 i.e 1.

Multiply equation 2 by the coefficient of x in equation 1 i.e 3. This is illustrated below:

1 × Equation 1

1 × (3x + 4y = 8)

3x + 4y = 8 ...... (3)

3 × Equation 2

3 × ( x – y = 12)

3x – 3y = 36......(4)

Step 2:

Subtract equation 3 from equation 4. This is illustrated below:

. 3x – 3y = 36

– (3x + 4y = 8)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

– 7y = 28

Divide both side by the coefficient of y i.e –7

y = 28/–7

y = – 4

Step 3:

Substitute the value of y into any of the equation to obtain the value of x. In this case, we shall substitute the value of y into equation 2 as shown below:

x – y = 12

y = –4

x – (–4) = 12

x + 4 = 12

x = 12 – 4

x = 8

Therefore, the solution to the equation above is (8, – 4)

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zysi [14]

Answer:  a) 0.0792   b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)   (1)

Similarly ,

P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

5 0
3 years ago
Solve for x: m(fx+a)=z+g
Inga [223]
X= -a/f = z/mf + g/mf
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MrRissso [65]

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Answer:

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Step-by-step explanation:

First, we need to know the concentration of NaCl in a normal saline solution, this is by definition 0.9%, meaning we have 0.9g of NaCl per 100ml of solution, we want to know how much NaCl we have in 3L (3000ml):

3000ml*\frac{0.9g}{100ml}=27g=27000mg

So, we have 27000mg in 3L of normal saline solution.

Now, acording to our milliequivalent (mEq) equation (mEq=\frac{mg}{pE}) where pE is de molecular mass of NaCl divided by their charges, in this case 1:

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Finally we substitute in the mEq formula:

mEq=\frac{mg}{pE}=\frac{27000}{58}=466mEq

I hope you find this information useful! Good luck!

4 0
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Answer:

Right Angles: C, F

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Obtuse Angles are angles that are bigger than 90 degrees.

Acute Angles are angles that are smaller than 90 degrees.

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