All categories

3 years ago

Three vectors A, B, C in three-dimensional space satisfy the following properties

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

Answer:

$\frac{7}{8}\pi$

Step-by-step explanation:

observe

||a–b+c|| = ||a+b+c||

(a-b+c)² = (a+b+c)²

(a+b+c)² – (a-b+c)² = 0

((a+b+c)+(a-b+c))((a+b+c)–(a-b+c)) = 0

(2a+2c)(2b) = 0

(a+c)b = 0

a•b + c•b = 0

||a||×||b||×cos(π/8) + ||c||×||b||×cos(θ) = 0

$\cos(\theta)=-\frac{||a||\times ||b|| \times \cos(\frac{\pi}{8})}{||c||\times ||b||}=-\cos(\frac{\pi}{8})\\ \theta=\frac{7}{8}\pi$

You might be interested in

5x – 2x + 7 – x = ?x + ?

fomenos

Step-by-step explanation:

5x – 2x + 7 – x = ?x + ?

I need the answer for "no solution" and "infinite solutions".

NO SOLUTION:

no solution because any value you give to x is not solvable, it remains +7

5x – 2x + 7 – x = ?x + ?

2x = 2x + 7

INFINITE SOLUTION

infinite solutions because any value you give to x is solvable

5x – 2x + 7 – x = ?x + ?

5x – 2x + 7 – x = 2x + 7

2x + 7 = 2x + 7

8 0

2 years ago

Leena knits 10 rows in 12 min. At this pace, how long will it take her to knit the next 25 rows?

tatiyna

For 10 rows, she takes = 12 min.
So, for 1 row, she'll take = 12/10

Now, for 15 rows, it would be: 12/10 * 25 = 12/2 * 5 = 60/2 = 30

So, Your Final Answer would be 30 minutes

Hope this helps!

7 0

3 years ago

Read 2 more answers

Select the correct answer.

gizmo_the_mogwai [7]

Answer:

Factors are (x-1), (x+1), and (x+2).

Step-by-step explanation:

$g(x)=x^{3}+2x^{2} -x-2\\g(x)=(x - 1) (x + 1) (x + 2)$

7 0

3 years ago

0.3(n-5)=0.4-0.2 solve for n!

mariarad [96]

Answer:

$n = 5.67$

Step-by-step explanation:

0.3(n-5)=0.4-0.2

0.3(n-5) = 0.2

Dividing 0.3 both side

$\frac{0.3(n-5) }{0.3} = \frac{0.2}{0.3}$

$n-5 = 0.67$

$n = 0.67+5$

$n = 5.67$

4 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago