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Vlada [557]
3 years ago
13

How many Solutions does this system have? (1 point)

Mathematics
1 answer:
mixas84 [53]3 years ago
5 0

The given system of equation that is 2x+y=3 and 6x=9-3y has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}

Let us first bring the equation in standard form for comparison

\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}

\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

To check how many solutions are there for system of equations a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0, we need to compare ratios of \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}

In our case,  

a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3

a_{2}  = 6, b_{2} = 3,\text{ and }c_{2} = -9

\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}

As \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}, so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

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A simple random sample of 10 paired values (x,y) yields the following statistical calculations: ∑x=108, ∑y=138, ∑(x2) =1249, ∑(y
lora16 [44]

The linear correlation coefficient is r=1.054

Explanation:

It is given that $\Sigma x=108$,  $\Sigma y=138$ , $\Sigma x^{2} =1249$ , $\Sigma y^{2} =2280$ and $\Sigma(x y)=1676$

Also, the random sample is n=10

The formula to determine the correlation coefficient is given by

$r=\frac{n\left(\sum x y\right)-\left(\sum x\right)(\Sigma y)}{\sqrt{\left[n \sum x^{2}-\left(\sum x\right)^{2}\right]\left[n \Sigma y^{2}-(\Sigma y)^{2}\right]}}$

Substituting the values in the formula, we have,

$r=\frac{10(1676)-(108)(138)}{\sqrt{\left[10(1249)-(108)^{2}\right]\left[10(2280)-(138)^{2}\right]}}$

Simplifying the values, we get,

$r=\frac{16760-14904}{\sqrt{\left[12490-11664\right]\left[22800-19044\right]}}$

Subtracting the values in both numerator and denominator, we have,

$r=\frac{1856}{\sqrt{\left[826\right]\left[3756\right]}}$

Multiplying the denominator,

$r=\frac{1856}{\sqrt{3102456}}$

Simplifying, we have,

$r=\frac{1856}{1761.4}$

Dividing, we get,

r=1.054

Thus, the linear correlation coefficient is r=1.054

7 0
3 years ago
A cyclist rides a distance of 36 km at an average speed of 16 km/h.
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Answer:

2.25 hours

Step-by-step explanation:

Given data

distance= 36 km

speed= 16 km/h

Speed= distance/time

substitute

16= 36/t

t= 36/16

t= 2.25 hours

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663 round nearest ten and hundred
In-s [12.5K]


663 rounded to the nearest ten is 660

663 rounded to the nearest hundred is 700

4 0
3 years ago
Read 2 more answers
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