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saw5 [17]
3 years ago
15

Which ordered pair is the solution to the system of equations?

Mathematics
2 answers:
nata0808 [166]3 years ago
4 0
The answer is C(3, -9)
Setler [38]3 years ago
4 0

Answer:

Option C is correct

The ordered pair which is the solution of the system of given equations is (3, -9)

Step-by-step explanation:

Given the system of equation:

y =x -12                    ......[1]

-5x - 3y =12                ......[2]

Substitute equation [1] in [2] we have;

-5x-3(x-12)=12

Using distributive property : a \cdot (b+c) =a\cdot b+a\cdot c

-5x - 3x+36 = 12

Combine like terms;

-8x +36 = 12

Subtract 36 to both sides of an equation; we get

-8x +36 -36 = 12 -36

Simplify:

-8x = -24

Divide by -8 to both sides we get;

\frac{-8x}{-8} =\frac{-24}{-8}

Simplify:

x = 3

Substitute the value of x = 3 in equation [1] to solve for y;

y = 3 -12 = -9

or y = -9

Therefore, the only ordered pair (x, y)  the solution of the system of equation is ( 3, -9).



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What is the value of the digit 2 in the number 425,611?
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6 0
3 years ago
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

\mu_s=np=500*0.90=450

The standard deviation will be:

\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

\mu=np=300*0.932=279.6

The standard deviation will be:

\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

P(270\leq x\leq280)=P(269.5

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

P(X=280)=P(279.5

8 0
3 years ago
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