If P (event E) = 0.47, then find P(not E). -Maths 9th

If P (event E) = 0.47, then find P(not E). -Maths 9th
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1 Answer

Answer :

P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53
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Related questions

If P (event E) = 0.47, then find P(not E). -Maths 9th

Description : If P (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

IfP (event E) = 0.47, then find P(not E). -Maths 9th

Description : IfP (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Description : Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Last Answer : Let x = 0.7Bar ⇒ x = 0.777.......... .........(1) Multiplying (1) by 10 ⇒ 10x = 7.7...... = 7.777 ...........(2) Subtracting (1) from (2) ⇒ 10x - x = 9x ⇒ 7.777 - 0.777 = 7 ... and then solving it, we get. ⇒ (594 + 770 + 470)/990 ⇒ 1834/990 ⇒ 917/495 Hence, 0.6 + 0.7Bar + 0.47Bar = 917/495

Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Description : Express 0.6bar +0.7bar+0.47 bar in the form p/q where p and q are integers and q is not equal to 0 -Maths 9th

Last Answer : Let x=0.666....... (1) Multiply equation (1 by 10 10x = 6.666....... (2) Subtract equation (1) from (2) x=6/9 Similarly 0.7bar =7/9 and 0.47bar = 47/99. 6/9+7/9+47/99=190/99

Teacher asked the students 'Can we write 0.47(recurring) -Maths 9th

Description : Teacher asked the students 'Can we write 0.47(recurring) -Maths 9th

Last Answer : Yes, Let x = 0.477777.... ....(i) 10x = 4.77777... .....(ii) Subtracting (i) from (ii), we get 9x = 4.3 or x = 43/90 Scientific temper, knowledge, curosity.

Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Description : Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct mean. -Maths 9th

Last Answer : Mean of 36 observations = 12 Total of 36 observations = 36 x 12 = 432 Correct sum of 36 observations = 432 – 74 + 47 = 405 Correct mean of 36 observations = 405/ 36 =11.25

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

Can the experimental probability of an event be a negative number? -Maths 9th

Description : Can the experimental probability of an event be a negative number? -Maths 9th

Last Answer : No, because the number of trials in which the event can happen cannot be negative and the total number of trials is always positive.

Can the experimental probability of an event be greater than 1? -Maths 9th

Description : Can the experimental probability of an event be greater than 1? -Maths 9th

Last Answer : No, as the number of trials can't be greater than the total number of trials.

What do you mean by Event ? -Maths 9th

Description : What do you mean by Event ? -Maths 9th

Last Answer : It is something that happens, whose probability we may want to measure, as drawing a spade from a pack of 52 cards.

What to you mean by Compound or Composite Event? -Maths 9th

Description : What to you mean by Compound or Composite Event? -Maths 9th

Last Answer : An event containing more than one sample point is called a compound or composite event. It is a subset of the sample spaces. For example, (i) If a die is thrown, then the statement getting an odd number on the top is ... If A be the event that one head and one tail turn up, then A = {(HT), (TH)}

What do you mean by Impossible Event? -Maths 9th

Description : What do you mean by Impossible Event? -Maths 9th

Last Answer : If we assume that S is the sample space of a random experiment then ϕ (null set) is a subset of S. As no outcome of the experiment belong to ϕ, the event ϕ can never occur when the experiment is performed. Hence ϕ is an impossible event.

What do you mean by sure event? -Maths 9th

Description : What do you mean by sure event? -Maths 9th

Last Answer : Let S be the sample space of a random experiment since S is a subset of itself, it also represents an event. Since all the outcomes of a random experiment belong to sample space S, the event S represents a sure event.

What do you mean by complementary event? -Maths 9th

Description : What do you mean by complementary event? -Maths 9th

Last Answer : Let S be the sample space associated with a random experiment and let E be an event. The event of non-occurrence of E is called the complementary event of E. It is represented by \(\bar{E}\) or E′. \(\bar{E}\) is also ... the complementary event of A. If A = {2, 3, 5}, then \(\bar{A}\) = {1, 4, 6}

The odds against certain event are 5:2 and the odds in favour of another in dependent event are 6:5. The probability that at least one of -Maths 9th

Description : The odds against certain event are 5:2 and the odds in favour of another in dependent event are 6:5. The probability that at least one of -Maths 9th

Last Answer : (c) \(rac{52}{77}\)Given, odds against Event 1 = 5 : 2⇒ P(Event 1 not happening) = \(rac{5}{5+2}\) = \(rac{5}{7}\)Odds in favour of Event 2 = 6 : 5⇒ P(Event 2 happens) = \(rac{6}{6+5}\) = \( ... }\)(∵ Both event are independent)⇒ P(At least one event happens) = 1 - \(rac{25}{77}\) = \(rac{52}{77}\).

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

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The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

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Description : If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

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If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

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If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

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If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

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If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

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If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Description : If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Last Answer : (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

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Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : x = 1 y = -2 2x-y = p Therefore, p = 2(1)-(-2) = 2 + 2 = 4

if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : 2x-y=p put x-1,y=-2 =2(1)-(-2)=p p=4

If the coordinates of two points are P( -2,3) and Q ( -3, 5) then find (abscissa of P)–(abscissa of Q) -Maths 9th

Description : If the coordinates of two points are P( -2,3) and Q ( -3, 5) then find (abscissa of P)–(abscissa of Q) -Maths 9th

Last Answer : Abscissa of P – Abscissa of Q = (–2) – (–3) = –2 + 3 = 1.

If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Description : If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Last Answer : Solution :-

If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Description : If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Last Answer : THINGS WHICH ARE COINCIDE WITH EACH OTHER ARE EQUAL TO ONE ANOTHER PO+OR=PR 2PO=PR PO=OR PO=1/2PR HENCE PROVED

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

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If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Description : If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Last Answer : (a) 0Let h be the first term and k be the common ratio of a GP, then a = hkp - 1, b = hkq - 1, c = hkr - 1∴ (q - r) log a + (r - p) log b + (p - q) log c = log [hkp -1]q - r + log [hkq -1]r - p + log[hkr -1]p - ... r + r - p + p - q) (kp - 1)q - r (kq -1)r - p (kr -1)p - q = log(ho ko) = log 1 = 0.

If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th

Description : If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th

Last Answer : (d) log3\(\big(rac{3}{4}\big)\) 1, log9 (31 - x + 2), log3 (4.3x -1) are in A.P. ⇒ log33 , log3 (31- x + 2)1/2, log3 (4.3x - 1) are in A.P. ⇒ 3, (31 - x + 2)1/2, (4.3x - 1) are in G.P.\ ... (rac{1}{3}\), \(rac{3}{4}\)∴ Rejecting the negative value, we have 3x = \(rac{3}{4}\)⇒ x = log3\(rac{3}{4}.\)

If a, b, c are positive real numbers such that a + b + c = p, then 1/a+1/b+1/c is greater than -Maths 9th

Description : If a, b, c are positive real numbers such that a + b + c = p, then 1/a+1/b+1/c is greater than -Maths 9th

Last Answer : answer:

For three distinct positive numbers p, q and r, if p + q + r = a, then -Maths 9th

Description : For three distinct positive numbers p, q and r, if p + q + r = a, then -Maths 9th

Last Answer : answer:

If cos2B = (cos(A+C))/(cos(A-C)) , then show that tan A, tan B and tan C are in G.P. -Maths 9th

Description : If cos2B = (cos(A+C))/(cos(A-C)) , then show that tan A, tan B and tan C are in G.P. -Maths 9th

Last Answer : answer:

If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

Description : If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

Last Answer : answer: