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3 years ago

What is the component form and magnitude of the vector shown?

Mathematics

2 answers:

Blababa [14]3 years ago

8 0

Answer:

Step-by-step explanation:

Solnce55 [7]3 years ago

8 0

Answer:

Step-by-step explanation:

edge 2020

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Assume that we have two events, and , that are mutually exclusive. Assume further that we know and . If an amount is zero, enter

sweet [91]

Answer:

Explained below.

Step-by-step explanation:

The complete question is:

Assume that we have two events, A and B, that are mutually exclusive. Assume further that we know P(A) = 0.30 and P(B) =0.40.

What is P(A and B)?

What is P(A | B)?

Is P(A | B) equal to P(A)?

Are events A and B dependent or independent?

A student in statistics argues that the concepts of mutually exclusive events and independent events are really the same, and that if events are mutually exclusive they must be independent. Is this statement accurate?

What general conclusion would you make about mutually exclusive and independent events given the results of this problem?

Solution:

The probability of the two events A and B are:

P(A) = 0.30 and P(B) =0.40

(a)

Compute the value of P (A ∩ B) as follows:

$P(A\cap B)=0$

This is because mutually exclusive events are those events that cannot occur together.

(b)

Compute the value of P (A | B) as follows:

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0}{0.40}=0$

Thus, the value of P (A | B) is 0.

(c)

No, P (A | B) is not equal to the P (A).

(d)

As mentioned in part (a), mutually exclusive events are those events which cannot occur together.

That is, $P(A\cap B)=0$ .

Events A and B are independent if the chance of their concurrent happening is equivalent to the multiplication of their distinct probabilities.

That is, $P(A\cap B)=P(A)\times P(B)$ .

The concepts of mutually exclusive events and independent events are not the same.

(e)

As the it is provided that A and B are mutually exclusive events, we know that $P(A\cap B)=0$ .

Now compute the value of $P(A)\times P(B)$ as follows:

$P(A)\times P(B)=0.30\times 0.40=0.12\neq 0$

Thus, the events A and B are not independent.

Thus, if two events are mutually exclusive events they cannot be independent.

4 0

3 years ago

In what quadrant is the terminal side of -323°?

dexar [7]

Answer:

Quadrant 4 i believe plz tell me if im wrong

Step-by-step explanatio

6 0

3 years ago

Read 2 more answers

What is the volume of the right prism?

Serggg [28]

We know that:
$V = l \times w \times h$

Substituting known values, we have:
$V= 4ft \times 4ft \times 15ft$
$V= 240ft^{3}$

6 0

3 years ago

Shen ran for president of the chess club, and he received 14 votes. There were 35 members in the club. What percentage of the cl

RUDIKE [14]

Answer:

40%

Step-by-step explanation:

14/35 members voted him. Lets convert this into a percentage:

14/35 * 100% = 14/35 *100/100 = 40%

I hope this helps! :)

3 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

8 0

3 years ago