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Question Number 213428 Answers: 0 Comments: 0

Question Number 212926 Answers: 2 Comments: 0

find integers x,y such that (x/(x−3)) −(4/(y^2 −45)) = (1/(100))

$$\:\mathrm{find}\:\mathrm{integers}\:\mathrm{x},\mathrm{y}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{3}}\:−\frac{\mathrm{4}}{\mathrm{y}^{\mathrm{2}} −\mathrm{45}}\:=\:\frac{\mathrm{1}}{\mathrm{100}} \\ $$

Question Number 210832 Answers: 2 Comments: 1

If the probability of A solving a question is 1/2 and the probability of B solving the question is 2/3 then the probability of the question being solved is

$$ \\ $$If the probability of A solving a question is 1/2 and the probability of B solving the question is 2/3 then the probability of the question being solved is

Question Number 209732 Answers: 1 Comments: 0

Q) The collection A={12,13,15,18,23,24,25,26}& B⊆A if m,M ∈B ; m=min & M =max & nm=10k which number of B : 1)59 2)60 3)61 4)62

$$\left.{Q}\right)\:{The}\:{collection}\:{A}=\left\{\mathrm{12},\mathrm{13},\mathrm{15},\mathrm{18},\mathrm{23},\mathrm{24},\mathrm{25},\mathrm{26}\right\}\&\:{B}\subseteq{A} \\ $$$${if}\:\:{m},{M}\:\in{B}\:\:;\:{m}={min}\:\&\:{M}\:={max}\:\&\:\:{nm}=\mathrm{10}{k} \\ $$$${which}\:{number}\:{of}\:\:{B}\:: \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{5}\left.\mathrm{9}\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{60}\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{61}\:\:\:\:\:\:\mathrm{4}\right)\mathrm{62} \\ $$$$ \\ $$

Question Number 209510 Answers: 0 Comments: 0

three points are randomly selected on a circle to form a triangle. 1) find the probability that the center of the circle lies inside the triangle. 2) find the probability that the triangle is an acute triangle.

$${three}\:{points}\:{are}\:{randomly}\:{selected} \\ $$$${on}\:{a}\:{circle}\:{to}\:{form}\:{a}\:{triangle}.\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{probability}\:{that}\:{the}\:{center} \\ $$$${of}\:{the}\:{circle}\:{lies}\:{inside}\:{the}\:{triangle}. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{probability}\:{that}\:{the}\: \\ $$$${triangle}\:{is}\:{an}\:{acute}\:{triangle}. \\ $$

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Question Number 207787 Answers: 1 Comments: 0

n married couples are invited to a dance party. for the first dance n paires are radomly selected. what′s the probability that no woman dances with her own husband? 1) if a pair must be of different genders. 2) if a pair can also be of the same gender.

$$\boldsymbol{{n}}\:{married}\:{couples}\:{are}\:{invited}\:{to} \\ $$$${a}\:{dance}\:{party}.\:{for}\:{the}\:{first}\:{dance} \\ $$$$\boldsymbol{{n}}\:{paires}\:{are}\:{radomly}\:{selected}.\: \\ $$$${what}'{s}\:{the}\:{probability}\:{that}\:{no}\:{woman} \\ $$$${dances}\:{with}\:{her}\:{own}\:{husband}? \\ $$$$\left.\mathrm{1}\right)\:{if}\:{a}\:{pair}\:{must}\:{be}\:{of}\:{different} \\ $$$$\:\:\:\:\:{genders}. \\ $$$$\left.\mathrm{2}\right)\:{if}\:{a}\:{pair}\:{can}\:{also}\:{be}\:{of}\:{the}\:{same}\: \\ $$$$\:\:\:\:\:{gender}. \\ $$

Question Number 207752 Answers: 1 Comments: 3

Two ships have the same berth in a port. It is known that the arrival times of the two ships are independent and have the same probability of docking on a Sunday (00.00−24.00) If the berth time of the first ship is 2 hours and the berth time of the second ship is 4 hours, the probability that one ship will have to wait until the berth can be used is □ ((67)/(144)) □ ((67)/(288)) □ (1/4) □((33)/(144))

$$\:\mathrm{Two}\:\mathrm{ships}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{berth}\: \\ $$$$\:\mathrm{in}\:\mathrm{a}\:\mathrm{port}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\:\mathrm{arrival}\:\mathrm{times}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{ships}\: \\ $$$$\:\mathrm{are}\:\mathrm{independent}\:\mathrm{and}\:\mathrm{have}\:\mathrm{the}\: \\ $$$$\:\mathrm{same}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{docking}\: \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{Sunday}\:\left(\mathrm{00}.\mathrm{00}−\mathrm{24}.\mathrm{00}\right) \\ $$$$\:\mathrm{If}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{ship} \\ $$$$\:\mathrm{is}\:\mathrm{2}\:\mathrm{hours}\:\mathrm{and}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time} \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{second}\:\mathrm{ship}\:\mathrm{is}\:\mathrm{4}\:\mathrm{hours},\: \\ $$$$\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{one}\:\mathrm{ship} \\ $$$$\:\mathrm{will}\:\mathrm{have}\:\mathrm{to}\:\mathrm{wait}\:\mathrm{until}\:\mathrm{the} \\ $$$$\:\mathrm{berth}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}\:\mathrm{is}\: \\ $$$$\:\Box\:\frac{\mathrm{67}}{\mathrm{144}}\:\:\:\:\:\Box\:\frac{\mathrm{67}}{\mathrm{288}}\:\:\:\:\Box\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\Box\frac{\mathrm{33}}{\mathrm{144}} \\ $$

Question Number 207751 Answers: 0 Comments: 0

It is known that a balanced 6−sided dice originally had 2,3,4,5,6 and 7. The dice wre thrown once and the result was observed. If an odd numbers appears, than the number is replaced with the number 8. However, if an even number appears , the number is replaced with the number 1. Then the dice whose dice have been replaced are thrown again, the probability of an odd dice odd dice appearing is □ (1/3) □ (2/3) □ (1/2) □ 1

$$\:\:\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{a}\:\mathrm{balanced}\:\mathrm{6}−\mathrm{sided}\: \\ $$$$\:\mathrm{dice}\:\mathrm{originally}\:\mathrm{had}\:\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\:\mathrm{and}\:\mathrm{7}. \\ $$$$\:\mathrm{The}\:\mathrm{dice}\:\mathrm{wre}\:\mathrm{thrown}\:\mathrm{once}\:\mathrm{and}\: \\ $$$$\:\mathrm{the}\:\mathrm{result}\:\mathrm{was}\:\mathrm{observed}.\:\mathrm{If}\:\mathrm{an}\: \\ $$$$\mathrm{odd}\:\mathrm{numbers}\:\mathrm{appears},\:\mathrm{than}\:\mathrm{the}\: \\ $$$$\:\mathrm{number}\:\mathrm{is}\:\mathrm{replaced}\:\mathrm{with}\:\mathrm{the}\: \\ $$$$\:\mathrm{number}\:\mathrm{8}.\:\mathrm{However},\:\mathrm{if}\:\mathrm{an}\:\mathrm{even}\: \\ $$$$\:\mathrm{number}\:\mathrm{appears}\:,\:\mathrm{the}\:\mathrm{number} \\ $$$$\:\mathrm{is}\:\mathrm{replaced}\:\mathrm{with}\:\mathrm{the}\:\mathrm{number}\:\mathrm{1}. \\ $$$$\:\mathrm{Then}\:\mathrm{the}\:\mathrm{dice}\:\mathrm{whose}\:\mathrm{dice}\:\mathrm{have}\: \\ $$$$\:\mathrm{been}\:\mathrm{replaced}\:\mathrm{are}\:\mathrm{thrown}\:\mathrm{again},\: \\ $$$$\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{dice}\: \\ $$$$\:\mathrm{odd}\:\mathrm{dice}\:\mathrm{appearing}\:\mathrm{is}\: \\ $$$$\:\:\Box\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\Box\:\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\:\:\Box\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\Box\:\mathrm{1}\: \\ $$

Question Number 207665 Answers: 3 Comments: 0

(1/1) (((20)),(( 0)) ) +(1/2) (((20)),(( 1)) ) +(1/3) (((20)),(( 2)) ) +...+(1/(21)) (((20)),((20)) ) =?

$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{0}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{1}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{2}}\end{pmatrix}\:+...+\frac{\mathrm{1}}{\mathrm{21}}\:\begin{pmatrix}{\mathrm{20}}\\{\mathrm{20}}\end{pmatrix}\:=? \\ $$

Question Number 207387 Answers: 1 Comments: 0

Let cardE=n , and the set of parts S={(A,B)∈P(E)×P(E) / A∩B=∅} Show that cardS= 3^n

$${Let}\:\:{cardE}={n}\:,\:{and}\:\:{the}\:{set}\:{of}\:{parts} \\ $$$${S}=\left\{\left({A},{B}\right)\in{P}\left({E}\right)×{P}\left({E}\right)\:/\:\:{A}\cap{B}=\varnothing\right\} \\ $$$${Show}\:{that}\:\:{cardS}=\:\mathrm{3}^{{n}} \\ $$

Question Number 207374 Answers: 2 Comments: 0

Show that Σ_(k=0) ^n (C_n ^k )^2 =C_(2n) ^n

$${Show}\:{that}\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$

Question Number 206514 Answers: 0 Comments: 0

In this covid −19 pandemic, it is known that are 5,667,355 confirmed cases out of 273,500,000 in X country population based WHO. One of the equipment to test the covid−19 is GeNose C19−S developed by UGM. GeNose C19−S is a rapid screening equipment for Sars−CoV2 virus infection through the breath of Covid −19 patient . It is claim that the sensivity of the test is 0,90 that is, if a person has the disease, then the probability that the diagnostic blood test comes back positive is 0,90. In addition , the specificity of the test is 0,95, i.e if a person is free the disease, then the probability that the diagnostic test comes back negative is 0,95. Let D and H is the event that a randomly selected individual has the disease and disease−free (healty), respectively. a. What is the positive predictive value of the GeNose C19−S test? That is, given that the blood test is positive for disease, what is the probability that person actually has the disease? b. If the doctor perfoms the test for the second time , taking P(D) equals to the value of probability you obtained from part a), determine the update positive predictive value of the test.

$$\:{In}\:{this}\:{covid}\:−\mathrm{19}\:{pandemic},\:{it}\:{is}\: \\ $$$$\:{known}\:{that}\:{are}\:\mathrm{5},\mathrm{667},\mathrm{355}\:{confirmed}\: \\ $$$$\:{cases}\:{out}\:{of}\:\mathrm{273},\mathrm{500},\mathrm{000}\:{in}\:{X}\:{country} \\ $$$$\:{population}\:{based}\:{WHO}.\: \\ $$$$\:{One}\:{of}\:{the}\:{equipment}\:{to}\:{test}\:{the} \\ $$$$\:{covid}−\mathrm{19}\:{is}\:{GeNose}\:{C}\mathrm{19}−{S}\:{developed} \\ $$$$\:{by}\:{UGM}.\:{GeNose}\:{C}\mathrm{19}−{S}\:{is}\:{a}\:{rapid} \\ $$$$\:{screening}\:{equipment}\:{for}\:{Sars}−{CoV}\mathrm{2} \\ $$$$\:{virus}\:{infection}\:{through}\:{the}\:{breath} \\ $$$$\:{of}\:{Covid}\:−\mathrm{19}\:{patient}\:.\:{It}\:{is}\:{claim} \\ $$$$\:{that}\:{the}\:{sensivity}\:{of}\:{the}\:{test}\:{is}\:\mathrm{0},\mathrm{90} \\ $$$$\:{that}\:{is},\:{if}\:{a}\:{person}\:{has}\:{the}\:{disease},\: \\ $$$$\:{then}\:{the}\:{probability}\:{that}\:{the}\:{diagnostic} \\ $$$$\:{blood}\:{test}\:{comes}\:{back}\:{positive} \\ $$$$\:{is}\:\mathrm{0},\mathrm{90}.\:{In}\:{addition}\:,\:{the}\:{specificity} \\ $$$$\:{of}\:{the}\:{test}\:{is}\:\mathrm{0},\mathrm{95},\:{i}.{e}\:{if}\:{a}\:{person}\: \\ $$$$\:{is}\:{free}\:{the}\:{disease},\:{then}\:{the}\:{probability} \\ $$$$\:{that}\:{the}\:{diagnostic}\:{test}\:{comes}\:{back} \\ $$$$\:{negative}\:{is}\:\mathrm{0},\mathrm{95}.\: \\ $$$$\:{Let}\:{D}\:{and}\:{H}\:{is}\:{the}\:{event}\:{that}\:{a}\: \\ $$$$\:{randomly}\:{selected}\:{individual}\:{has} \\ $$$$\:{the}\:{disease}\:{and}\:{disease}−{free}\: \\ $$$$\:\left({healty}\right),\:{respectively}. \\ $$$$\:{a}.\:{What}\:{is}\:{the}\:{positive}\:{predictive}\: \\ $$$$\:\:{value}\:{of}\:{the}\:{GeNose}\:{C}\mathrm{19}−{S}\:{test}? \\ $$$$\:\:{That}\:{is},\:{given}\:{that}\:{the}\:{blood}\:{test}\: \\ $$$$\:{is}\:{positive}\:{for}\:{disease},\:{what}\:{is}\:{the} \\ $$$$\:{probability}\:{that}\:{person}\:{actually}\:{has} \\ $$$$\:\:{the}\:{disease}?\: \\ $$$$\: \\ $$$$\:{b}.\:{If}\:{the}\:{doctor}\:{perfoms}\:{the}\:{test}\:{for} \\ $$$$\:{the}\:{second}\:{time}\:,\:{taking}\:{P}\left({D}\right)\:{equals} \\ $$$$\:{to}\:{the}\:{value}\:{of}\:{probability}\:{you} \\ $$$$\left.\:{obtained}\:{from}\:{part}\:{a}\right),\:{determine}\: \\ $$$$\:{the}\:{update}\:{positive}\:{predictive}\: \\ $$$$\:\:{value}\:{of}\:{the}\:{test}.\: \\ $$

Question Number 205409 Answers: 0 Comments: 0

let x, y, z be random numbers from 0 to 10 where x,y,z∈R what is the probability that a) all the following is satisfied ∣x−y∣≥2 ∣x−z∣≥2 ∣y−z∣≥2 b) the probability that one or two of them are not satisfied c) the probability that all of them are not satisfied

$$\mathrm{let}\:{x},\:{y},\:{z}\:\mathrm{be}\:\mathrm{random}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{10} \\ $$$$\mathrm{where}\:{x},{y},{z}\in\mathbb{R} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\left.{a}\right)\:\mathrm{all}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{satisfied} \\ $$$$\mid{x}−{y}\mid\geqslant\mathrm{2} \\ $$$$\mid{x}−{z}\mid\geqslant\mathrm{2} \\ $$$$\mid{y}−{z}\mid\geqslant\mathrm{2} \\ $$$$\left.{b}\right)\:\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\:\mathrm{one}\:\mathrm{or}\:\mathrm{two}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{not}\:\mathrm{satisfied} \\ $$$$\left.{c}\right)\:\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\mathrm{all}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{not}\:\mathrm{satisfied} \\ $$$$ \\ $$

Question Number 205394 Answers: 1 Comments: 0

let x and y be random numbers from 0 to 10 where x,y∈R ∣x−y∣≥d what is the probability that their sum is less than 10 in the following cases a) d=0 b) d=1 c) d=2

$$\mathrm{let}\:{x}\:\mathrm{and}\:{y}\:\mathrm{be}\:\mathrm{random}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{10} \\ $$$$\mathrm{where}\:{x},{y}\in\mathbb{R} \\ $$$$\mid{x}−{y}\mid\geqslant{d} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{10} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\left.{a}\right)\:{d}=\mathrm{0} \\ $$$$\left.{b}\right)\:{d}=\mathrm{1} \\ $$$$\left.{c}\right)\:{d}=\mathrm{2} \\ $$

Question Number 205230 Answers: 0 Comments: 1

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A dice is cast twice, and the sum of the appearing numbers is 10. The probability that the number 5 has appeared at least once is.

$$\mathrm{A}\:\mathrm{dice}\:\mathrm{is}\:\mathrm{cast}\:\mathrm{twice},\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{appearing}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{10}. \\ $$$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\:\mathrm{5}\:\mathrm{has}\: \\ $$$$\mathrm{appeared}\:\mathrm{at}\:\mathrm{least}\:\mathrm{once}\:\mathrm{is}. \\ $$

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There is a field. Everyday kids throw some balls on the field. At night the farmer goes and place the bucket in a place where it will contain the most amount of balls. the field can be represented as a line of length 10. the bucket can be represented as a line of length 2. If the kids have thrown 3 balls into the field, what is the probability that the bucket will contain 2 balls how a sample looks like Where the blue line represents the field. and the red line represents the bucket. and the dots are the balls

$$ \\ $$$$ \\ $$There is a field. Everyday kids throw some balls on the field. At night the farmer goes and place the bucket in a place where it will contain the most amount of balls. the field can be represented as a line of length 10. the bucket can be represented as a line of length 2. If the kids have thrown 3 balls into the field, what is the probability that the bucket will contain 2 balls $$\mathrm{how}\:\mathrm{a}\:\mathrm{sample}\:\mathrm{looks}\:\mathrm{like} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$Where the blue line represents the field. and the red line represents the bucket. and the dots are the balls $$ \\ $$

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