Ask question

Ask question

All categories

3 years ago

12

2) If events A and B are DEPENDENT, then A) A and B must occur together. B) A and B cannot occur together. C) A's occurrence can

affect the probability of B's occurrence. D) A's occurrence cannot affect the probability of B's occurrence.

2 answers:

jonny [76]3 years ago

6 0

Answer:

by 5

Step-by-step explanation: because tgey wou;d nred help

Vilka [71]3 years ago

3 0

Answer:

C.) If events A and B are DEPENDENT, then A's occurrence can affect the probability of B's occurrence. For example, when two cards are chosen from a deck without replacement, the possibilities change for the second card.

You might be interested in

Consider the differential equation x^2 y''-xy'-3y=0. If y1=x3 is one solution use redution of order formula to find a second lin

Anastasy [175]

Suppose $y_2(x)=y_1(x)v(x)$ is another solution. Then

$\begin{cases}y_2=vx^3\\{y_2}'=v'x^3+3vx^2//{y_2}''=v''x^3+6v'x^2+6vx\end{cases}$

Substituting these derivatives into the ODE gives

$x^2(v''x^3+6v'x^2+6vx)-x(v'x^3+3vx^2)-3vx^3=0$

$x^5v''+5x^4v'=0$

Let $u(x)=v'(x)$ , so that

$\begin{cases}u=v'\\u'=v''\end{cases}$

Then the ODE becomes

$x^5u'+5x^4u=0$

and we can condense the left hand side as a derivative of a product,

$\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C$

$x^5u=C\implies u=Cx^{-5}$

Solve for $v$ :

$v'=Cx^{-5}\implies v=-\dfrac{C_1}4x^{-4}+C_2$

Solve for $y_2$ :

$\dfrac{y_2}{x^3}=-\dfrac{C_1}4x^{-4}+C_2\implies y_2=C_2x^3-\dfrac{C_1}{4x}$

So another linearly independent solution is $y_2=\dfrac1x$ .

3 0

3 years ago

Which equation can be used to find 30 percent of 600

agasfer [191]

A/b *100
For example,
30/600.
30/600=0.05
0.05*100=5%

5 0

3 years ago

If the sides of a right triangle is 10 cm, and the other side is 8 cm, what is the hypotenuse? Round to the nearest tenth.

Advocard [28]

Answer:

Step-by-step explanation:

Pythagorean theorem,

hypotenuse² = Base² + altitude²

= 8² + 10²

= 64 + 100

= 164

hypotenuse = √164 = 12.8 cm

5 0

3 years ago

Read 2 more answers

Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos

crimeas [40]

Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

There is a 19.58% probability that exactly one is diet and exactly one is regular.

8 0

3 years ago

n a large high school, 37% of the teachers believe that five minutes is not enough time for students to change classes. However,

timurjin [86]

The difference (faculty-student) in the sample proportions of those who believe that five minutes is not enough time for students to change classes typically varies about 0.096 from the true difference in proportions. Option D is correct.

<h3>What is the standard deviation?</h3>

It is a measurement of statistical data dispersion. The degree to which the value varies is known as standard deviation.

The standard deviation of the sampling distribution is found as;

$\rm \sigma_F-\sigma_s= \sqrt{\frac{0.37 \times (1-0.37)}{28}+\frac{0.89 \times (1-0.89)}{100} } \\\\ \sigma_F-\sigma_s=0.096$

The difference in the sample proportions of those who think five minutes is not enough time for students to switch classrooms generally differs by roughly 0.096 from the actual difference in proportions.

Hence option D is correct.

To learn more about the standard deviation, refer to: brainly.com/question/16555520.

#SPJ1

6 0

2 years ago