En el juego de domino son 28 fichas con la cara hacia abajo cual es la probabilidad de sacar la ficha 6-7

a) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve a

Neporo4naja [7]

Answer:

Tangent at (1,2) has value $\frac{15}{4}$ . And (-2, 4) is the point of intersection of horizontal tangent.

Step-by-step explanation:

Given curve equation,

$y^2=x^3+3x^2\hfill (1)$

To find tangent at (x,y)=(1,2) and point of intersection of horizontal tangent, differentiate (1) withrespect to x we get,

$2y\frac{dy}{dx}=x^3+3x^2$

$\implies \frac{dy}{dx}=\frac{3(x^2+2x)}{2y}$

At (1, 2), tangent line is,

$\frac{dy}{dx}|_{(1,2)}}=\frac{15}{4}$

To find point of intersection of horizontal tangent we have to do,

$\frac{dy}{dx}=0$

$\implies x(x+2)=0\implies x=0 or -2$

Thus,

At x=0, y=0

but snce,

$\lim_{y\to 0}\frac{dy}{dx}\to \infty$

at (0,0) there exist a vertical tangent. And,

At x=-2, y=4.

Thus (-2, 4) is the point of intersection of horizontal tangent.

6 0

2 years ago

In a large population, 61 % of the people have been vaccinated. if 4 people are randomly selected, what is the probability that

muminat

In a large population, 61% of the people are vaccinated, meaning there are 39% who are not. The problem asks for the probability that out of the 4 randomly selected people, at least one of them has been vaccinated. Therefore, we need to add all the possibilities that there could be one, two, three or four randomly selected persons who were vaccinated.

For only one person, we use P(1), same reasoning should hold for other subscripts.

P(1) = (61/100)(39/100)(39/100)(39/100) = 0.03618459
P(2) = (61/100)(61/100)(39/100)(39/100) = 0.05659641
P(3) = (61/100)(61/100)(61/100)(39/100) = 0.08852259
P(4) = (61/100)(61/100)(61/100)(61/100) = 0.13845841

Adding these probabilities, we have 0.319761. Therefore the probability of at least one person has been vaccinated out of 4 persons randomly selected is 0.32 or 32%, rounded off to the nearest hundredths.

8 0

1 year ago

There are 345 students at a college who have taken a course in calculus, 212 who have taken a course in discrete mathematics, an

ollegr [7]

Answer:

369 students have taken a course in either calculus or discrete mathematics

Step-by-step explanation:

I am going to build the Venn's diagram of these values.

I am going to say that:

A is the number of students who have taken a course in calculus.

B is the number of students who have taken a course in discrete mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the number of students who have taken a course in calculus but not in discrete mathematics and $A \cap B$ is the number of students who have taken a course in both calculus and discrete mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

188 who have taken courses in both calculus and discrete mathematics.

This means that $A \cap B = 188$

212 who have taken a course in discrete mathematics

This means that $B = 212$

345 students at a college who have taken a course in calculus

This means that $A = 345$

How many students have taken a course in either calculus or discrete mathematics

$(A \cup B) = A + B - (A \cap B) = 345 + 212 - 188 = 369$

369 students have taken a course in either calculus or discrete mathematics

4 0

2 years ago

Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll

JulijaS [17]

Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

$Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So, the possibility of getting a total of 7 = $\frac{6}{36}=\frac{1}{6}$

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = $\frac{2}{36}$ = $\frac{1}{18}$

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = $\frac{28}{36}$ = $\frac{7}{9}$

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

$\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11$

$\frac{11}{3}+\frac{11}{3}-\frac{77}{9}$

$\frac{22}{3}-\frac{77}{9}$

$\frac{66-77}{9}$

$-\frac{11}{9}$

$-1.22$

Therefore the expected payoff for this game is -$1.22.

4 0

2 years ago

Tom supports North Storm Football Club. At the match between North Storm and West Stanford there were 71,167 spectators. If 41,1

dangina [55]

Answer:
30,058 spectator

Explanation:
The total number of spectators is equal to the sum of West Stanford's and North Storm's supporters.
We are given that:
Total number of spectators = <span>71,167 spectator
North Storm spectators = </span><span>41,109 spectator
So, to get the number of West Stanford spectators, all we have to do is subtract North Storm spectators from the total spectators as follows:
West Stanford spectators = </span>71,167 - 41,109 = 30,058 spectator

Hope this helps :)

8 0

2 years ago