Ask question

Ask question

All categories

2 years ago

14

Can somebody please help me with the question below and please explain how to find the answer? plz, I'm studying for a test plzz

zz help!!!

1 answer:

blagie [28]2 years ago

3 0

Answer:

-10/9 < -2/3 < 4/6 < 6/5 < 5/4

(hope this helps!)

Step-by-step explanation:

You might be interested in

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

Triangle ABC is such that AB = 3cm , BC = 4cm , angle ABC = 120 and BAC = theta : Write down in terms of theta , and expression

lakkis [162]

1)inside the triangule is 180º .There are 3 angules, one with 120º, theta and
angle ACB, lets call it x.

x + theta + 120 = 180
x + theta= 180-120
x + theta= 60
x= 60- theta
angle ACB= 60- theta

8 0

3 years ago

1. Integers are not closed under which operation?

velikii [3]

Answer:

Integers are not closed under the DIVISION operation

Example:

(9 ÷ 2 = 4½)

4 0

3 years ago

4.) Which equations below have no solution? Select all that apply, and justify your response with work. a.) x - 9 = 2(x - 3) + 1

Sophie [7]

Answer: Equations (b) and (e) do not have solutions

Step-by-step explanation: b) The x cancels out (-10x + 35) = -10x + 318; 35 = 318???? NO. e) 3(x + 2) + 1 = x + 2(4 + x); 3x + 7 = 3x + 8. The x cancels out again.

The other s can be solved for x.

3 0

2 years ago

Read 2 more answers