What must you do if the base is not a triangle or rectangle?

HELP ME PLZ AND I WILL SHOW YOU MY KITTTTY

oksian1 [2.3K]

Welp looks like I won't be seeing that cat, darn it

6 0

3 years ago

Let a=(1,2,3,4), b=(4,3,2,1) and c=(1,1,1,1) be vectors in R4. Part (a) [4 points]: Find (a⋅2c)b+||−3c||a. Part (b) [6 points]:

love history [14]

Solution :

Given :

a = (1, 2, 3, 4) , b = ( 4, 3, 2, 1), c = (1, 1, 1, 1) ∈ $R^4$

a). (a.2c)b + ||-3c||a

Now,

(a.2c) = (1, 2, 3, 4). 2 (1, 1, 1, 1)

= (2 + 4 + 6 + 6)

= 20

-3c = -3 (1, 1, 1, 1)

= (-3, -3, -3, -3)

||-3c|| = $\sqrt{(-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 }$

$=\sqrt{9+9+9+9}$

$=\sqrt{36}$

= 6

Therefore,

(a.2c)b + ||-3c||a = (20)(4, 3, 2, 1) + 6(1, 2, 3, 4)

= (80, 60, 40, 20) + (6, 12, 18, 24)

= (86, 72, 58, 44)

b). two vectors $\vec A$ and $\vec B$ are parallel to each other if they are scalar multiple of each other.

i.e., $\vec A=r \vec B$ for the same scalar r.

Given $\vec p$ is parallel to $\vec a$ , for the same scalar r, we have

$\vec p = r (1,2,3,4)$

$\vec p = (r,2r,3r,4r)$ ......(1)

Let $\vec q = (q_1,q_2,q_3,q_4)$ ......(2)

Now given $\vec p$ and $\vec q$ are perpendicular vectors, that is dot product of $\vec p$ and $\vec q$ is zero.

$q_1r + 2q_2r + 3q_3r + 4q_4r = 0$

$q_1 + 2q_2 + 3q_3 + 4q_4 = 0$ .......(3)

Also given the sum of $\vec p$ and $\vec q$ is equal to $\vec b$ . So

$\vec p + \vec q = \vec b$

$(r,2r,3r,4r) + (q_1+q_2+q_3+q_4)=(4, 3,2,1)$

∴ $q_1 = 4-r , \ q_2=3-2r, \ q_3 = 2-3r, \ q_4=1-4r$ ....(4)

Putting the values of $q_1,q_2,q_3,q_4$ in (3),we get

$r=\frac{2}{3}$

So putting this value of r in (4), we get

$\vec p =\left( \frac{2}{3}, \frac{4}{3}, 2, \frac{8}{3} \right)$

$\vec q =\left( \frac{10}{3}, \frac{5}{3}, 0, \frac{-5}{3} \right)$

These two vectors are perpendicular and satisfies the given condition.

c). Given terminal point is $\vec a$ is (-1, 1, 2, -2)

We know that,

Position vector = terminal point - initial point

Initial point = terminal point - position point

= (-1, 1, 2, -2) - (1, 2, 3, 4)

= (-2, -1, -1, -6)

d). $\vec b = (4,3,2,1)$

Let us say a vector $\vec d = (d_1, d_2,d_3,d_4)$ is perpendicular to $\vec b.$

Then, $\vec b.\vec d = 0$

$4d_1+3d_2+2d_3+d_4=0$

$d_4=-4d_1-3d_2-2d_3$

There are infinitely many vectors which satisfies this condition.

Let us choose arbitrary $d_1=1, d_2=1, d_3=2$

Therefore, $d_4=-4(-1)-3(1)-2(2)$

= -3

The vector is (-1, 1, 2, -3) perpendicular to given $\vec b.$

6 0

2 years ago

A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true

zloy xaker [14]

Answer:

The probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.009.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

As the sample size is large, i.e. <em>n</em> = 492 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by the normal distribution.

The mean and standard deviation of the sampling distribution of sample proportion are:

$\mu_{\hat p}=p=0.07\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.07(1-0.07)}{492}}=0.012$