MATHEMATICS COLLEGE

At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a research paper and a final exam. Let F be the even that a course has a final exam. Let R be the event that a course requires a research paper. (a) Find the probability that a course has a final exam or a research paper. Your answer is : (b) Find the probability that a course has NEITHER of these two requirements. Your answer is :

Answers

Answer 1

Answer:

Answer:

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

Related Questions

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 8.8 years, and standard deviation of 2.2 years.If 5 items are picked at random, 5% of the time their mean life will be less than how many years?
In the equation (y-7)=(3)/(4)(x-5) a point on that line is1.(-5,-7)2.(7,5)3.(-7,-5)4.(5,7)
What is 2/3 - 1/2 = in simplest form
Find the first derivative for y = f(x). fox ) 3x² -5x-1 at a Pocat where a = 4
2,4000,000 in standard form

234 base five
Writen in base 10

Answers

$234_5=2*5^2+3*5^1+4*5^0=50+15+4=69_(10)$

Consider the following function.Step 2 of 2 : Find two points on the line to graph the function.

Answers

The twο pοints οn the line are:

A. (0, 3)

B. (3,−2)

What exactly dοes functiοn mean?

Functiοn is wοrk expressiοn the cοnnectiοns between variοus cοmpοnents that wοrk tοgether tο prοduce the same result. A utility is made up οf a variety οf distinctive cοmpοnents that cοοperate tο create distinct results fοr each input.

Here,

$$\mathrm{r(x)=3 - \frac53 (x)}$

This is a linear equatiοn.

Any twο pοint οf this linear equatiοn gives a unique line οn the graph.

when x = 0, r(x )= 3

when x = 3, r(x) = −2

The twο pοints οn the line are:

A. (0, 3)

B. (3,−2)

Using A. (0, 3) and B. (3,−2), Let us plοt a straight line.

(see the attachment belοw fοr the graph)

To know more about function visit:

brainly.com/question/28193995

#SPJ1

Tony worked 9 hours Monday, 9.5 hours Tuesday, 8.75 hours Wednesday,9.25 hours Thursday 8 hours Friday and 6 hours Sunday. How many hours
did he work?

Answers

The answer is 50.5 hours

all you have to do is add all the hours up to get the grand total.

Hope this helps

have a great day!

Solve the equation using a graphical method.
t/3-1/2=t+3/9

Answers

Answer:

-1.25

Step-by-step explanation:

t÷3-1÷2 = t÷1+3÷9

2t-3÷6 = 9t+3=9

18t - 27 = 54 + 18

18t -54t = 18 + 27

-36t = 45

Divide through by -36

t = -1.25

Compute the area of the region D bounded by xy=1, xy=16, xy2=1, xy2=36 in the first quadrant of the xy-plane. Using the non-linear change of variables u=xy and v=xy2, find x and y as functions of u and v.x=x(u,v)= ?

y=y(u,v)=?

Find the determinant of the Jacobian for this change of variables.

∣∣∣∂(x,y)/∂(u,v)∣∣∣=det=?

Using the change of variables, set up a double integral for calculating the area of the region D.

∫∫Ddxdy=?

Evaluate the double integral and compute the area of the region D.

Area =

Answers

Answer:

53.7528

Step-by-step explanation:

Notice that when

$xy = 1 ,\,\,\, xy = 16 , \,\,\, xy^2 = 1 \,\,\,, xy^2 = 36 \n\n$

If you set

$u = xy , v = xy^2$

as they suggest, then

$y = (v)/(u)} \,\,\,\, \text{and} \,\,\,\, \n\n{\displaystyle x = (u)/(y) = (u)/(v/u) = (u^2)/(v)$

Then

${\diplaystyle (\partial(x,y))/(\partial(u,v))} =\det \begin{pmatrix} 2u/v && -u^2/v^2 \n -v/u^2 && 1/u \end{pmatrix} = (1)/(v) }$

Therefore

$\iint\limits_(D) dx\,dy = \int\limits_(1)^(36)\int\limits_(1)^(16) (1)/(v) \, du \, dv = 15 \ln(36) = 53.7528$

A Jacobian matrix is formed by the first partial derivatives of a multivariate function that utilizes a training algorithm, and further calculation as follows:

Jacobian:

To evaluate the integral, cover the bounds, the integrand, and the differential area dA.

Transform the four equations in terms of u and v, notice that $u= xy \ \ and \ \ xy = 1, xy = 16$

implies that $1\leq u \leq 16.$

Similarly, $v= xy^2\ \ and\ \ xy^2= 1 , xy^2= 25$ implies that $1 \leq v \leq 25$

so write this integration region as $S= {(u,v) |1 \leq u \leq 18, 1 \leq v \leq 25}.$

Translate the equations from uv - plane to xy- plane. It is obtained by solving,

$u= xy, y= xy^2 \n\n\left.\begin{matrix}u=xy & \n v=xy^2& \end{matrix}\right\} \to \left.\begin{matrix}u^2=x^2y^2 & \n v=xy^2& \end{matrix}\right\} \n\n\to x=(u^2)/(v), y=(v)/(u)$

Convert dA part of the integral , using is $dA= |(\partial (x,y))/(\partial(u,v))| dudv.$

That is, $dA= \begin{vmatrix}(\partial x)/(\partial u) & (\partial x)/(\partial v)\n (\partial y)/(\partial u) & (\partial y)/(\partial v) \end{vmatrix} \ du dv \n\n$

Sampule the partial derivatives to find the Jacobian.

$dA=\begin{vmatrix}(2u)/(v) &-(u^2)/(v^2) \n -(v)/(u^2) &(1)/(u) \end{vmatrix} \ dudv\n\n=[((2u)/(v)) ((1)/(u)) -(- (u^2)/(v^2))(-(v)/(u^2))]\ du dv\n\n=((2)/(v)- (1)/(v)) \ dudv\n\n=(1)/(v)\ du dv\n\n$

The Jacobian the transformation is $dA= (1)/(v)dudv$

The region is $S={(u,v) |1\leq u \leq 16, 1\leq v\leq 25}.$

Rewrite the integral, using the transformation: $S,\ x=(u^2)/(v) =, y=(v)/(u) \ \ and\ \ dA=(1)/(v) dudv\n\n\int\int_R 1dA =\int \int_S (1)/(v)\ dudv= \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

Evaluate the inner integral with respect to u.

$\to \int\int_R 1dA = \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

by solving the value we get

$= 30 \ ln (5) \approx 48.28$

Find out more about the Jacobians here:

brainly.com/question/9381576

The sum of two rational numbers is (A: Rational) (B: Irrational)
number.

The sum of a rational number and an irrational number is
an (A:a rational) or (B: a irrational)
number.

~~~ZoomZoom44~~~~

Answers

Answer:

Its B

Step-by-step explanation:

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