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lions [1.4K]
3 years ago
5

What is the solution to the system of equations x+y=10 and x+2y=4 using the linear combination method?

Mathematics
1 answer:
Elis [28]3 years ago
3 0

Answer:

The solution:

X = 16 and Y = -6

Step-by-step explanation:

The equations to be solved are:

x+y = 10 ------- equation 1

x+2y = 4 ----------- equation 2

we can multiply equation 1 by -1 to make the value of x and y negative.

This will give us

-x- y = - 10 ------- equation 3

x+2y = 4 ----------- equation 2

We will now add equations 3 and 2 together so that x will cancel itself out.

this will give us

y = -10 +4 = -6

hence, we have the value of y as -6.

To get the value of x, we can put this value of y into any of the equations above.  (I will use equation 1)

x - 6 = 10

from this, we have that x = 4

Therefore, we have our answer as

X = 16 and Y = -6

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3 years ago
An industry representative claims that 10 percent of all satellite dish owners subscribe to at least one premium movie channel.
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Answer: 1) 0.6561    2) 0.0037

Step-by-step explanation:

We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

P(X=x)=^nC_xp^x(1-p)^{n-x}

, where p =Probability of getting success in each trial.

As per given , we have

The probability that any satellite dish owners subscribe to at least one premium movie channel.  : p=0.10

Sample size : n= 4

Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.

1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}

=(1)(0.90)^4=0.6561

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.

2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

= P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037

∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.

8 0
3 years ago
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