First find the slope of the line so change in y over change in x so 4-2/1-2 which is 2/-1 so -2 is your slope then plug that as well as a point from above to find your y intercept so y=mx+b 2=-2(2)+ b then you have 2=-4+b then add 4 to each side and you find b is six then you make ur equation from your slope and y int(b) y=-2x+6

6 0

2 years ago

Let W be the subspace of R5 spanned by the vectors w1, w2, w3, w4, w5, where

Gwar [14]

To find W⊥, you can use the Gram-Schmidt process using the usual inner-product and the given 5 independent set of vectors.

Define projection of v on u as 
p(u,v)=u*(u.v)/(u.u) 
we need to proceed and determine u1...u5 as: 
u1=w1 
u2=w2-p(u1,w2) 
u3=w3-p(u1,w3)-p(u2,w3) 
u4=w4-p(u1,w4)-p(u2,w4)-p(u3,w4) 
u5=w5-p(u4,w5)-p(u2,w5)-p(u3,w5)-p(u4,w5) 

so that u1...u5 will be the new basis of an orthogonal set of inner space. 

However, the given set of vectors is not independent, since 
w1+w2=w3, 
therefore an orthogonal basis cannot be found.

7 0

3 years ago

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by: