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3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

Mathematics

2 answers:

Lisa [10]3 years ago

4 0

<h3>Answer: 2.1</h3>

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Work Shown:

event A = getting an integer between 1 and 8

event A = getting 3 points awarded

event B = getting an integer between 9 and 15

event B = getting 2 points awarded

event C = getting an integer between 16 and 19

event C = getting 1 point awarded

event D = getting the integer 20

event D = getting 0 points awarded

P(x) = probability of event x

V(x) = net value of event x

P(A) = 8/20 = 2/5

V(A) = 3

P(B) = (15-9+1)/20 = 7/20

V(B) = 2

P(C) = (19-16+1)/20 = 4/20 = 1/5

V(C) = 1

P(D) = 1/20

V(D) = 0

-------------

P(A)*V(A) = (2/5)*3 = 6/5 = 1.2

P(B)*V(B) = (7/20)*2 = 14/20 = 7/10 = 0.7

P(C)*V(C) = (1/5)*1 = 0.2

P(D)*V(D) = (1/20)*0 = 0

-------------

E[x] = expected value

E[x] = P(A)*V(A)+P(B)*V(B)+P(C)*V(C)+P(D)*V(D)

E[x] = 1.2+0.7+0.2+0

E[x] = 2.1

arsen [322]3 years ago

4 0

Answer:idk

Step-by-step explanation:

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With the aid of an illustrative example, discuss the relationship between the area of a region and the definite integral. *

melisa1 [442]

Answer:

The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the a and b above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral.

Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, who is often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S, suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval, \left[a,b\right]. The numbers a and b are x-values and are called the limits of integration; specifically, a is the lower limit and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as n\to \infty . Second, the boundaries of the region are called the limits of integration.

We call the function f(x) the integrand, and the dx indicates that f(x) is a function with respect to x, called the variable of integration. Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on the computation of the integral.

his leads to the following theorem, which we state without proof.

Step-by-step explanation:

8 0

3 years ago

Please help. Easy graph question.

Dmitrij [34]

Answer:B

Step-by-step explanation:

6 0

3 years ago

PLS HELP asap!! The graph of an equation is shown below. Which of the following points is a solution to the equation?

Nady [450]

Your answer would be A.(0,1)

5 0

3 years ago

Solve 22 = - 3d + 1. Show your steps.

sattari [20]

22 - 1 = -3d
21= -3d
21/-3= d
-7 = d

Bring the 1 to the other side by doing the opposite sign, so subtract 1. Solve the left side.

Bring the 3 to the other side by doing the opposite sign, so divide 21 by -3.

End with D equaling -7

4 0

3 years ago

Read 2 more answers

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

4 years ago