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3 years ago

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Put this equation into function notation: y=(1200-2x)/3

1 answer:

Arte-miy333 [17]3 years ago

6 0

Y=mx+b
y=(1200-2x)/3
y=1200/3-2/3x
y=400-2/3x
y=-2/3x+400

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You have been assigned to determine whether more people prefer Coke or Pepsi. Assume that roughly half the population prefers Co

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Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

For this case we have the following info given:

$ME=0.03$ the margin of error desired

$Conf= 0.95$ the level of confidence given

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

the critical value for 95% of confidence is $z=1.96$

We can use as estimator for the population of interest $\hat p=0.5$ . And on this case we have that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

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let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

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3 years ago