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trapecia [35]
3 years ago
14

In a multiple choice examination of 20 questions, four marks are given for each correct answers and one mark is deducted for eac

h wrong answer. There is no penalty for not attempting a question. a candidate attempts "a" questions and get "c" correct. Write down, and simplify, an expression for the candidate's total mark in terms of "a" and "c".​
Mathematics
1 answer:
finlep [7]3 years ago
8 0

Answer:

Total Mark = 5c - a

Step-by-step explanation:

Total Mark = 4c - (a - c) = 5c - a

Marks for correct questions = 4c

Marks deducted for incorrect questions = a - c

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Square lMNO is shown in the diagram below what are the coordinates of the midpoint of a diagonal LN
sashaice [31]

Answer:

The coordinates of the midpoint of LN are (-2\frac{1}{2},4\frac{1}{2})

answer (4)

Step-by-step explanation:

* Lets explain how to find the midpoint of a line

- The coordinates of the midpoint of a line whose endpoints are (x1 , y1)

 and (x2 , y2) are x=\frac{x_{1}+x_{2}}{2},y=\frac{y_{1}+y_{2}}{2}

∵ LMNO is a square

∵ The coordinates of point L are (-6 , 1)

∵ The coordinates of point N are (1 , 8)

- Let the coordinates of point L are (x1 , y1) , the coordinates of point

 N are (x2 , y2) and the coordinates of the midpoint of LN are (x , y)

∴ x1 = -6 , x2 = 1 and y1 = 1 , y2 = 8

- Use the rule of the midpoint above to find the midpoint (x , y)

∵ x=\frac{-6+1}{2}=\frac{-5}{2}=-2\frac{1}{2}

∵ y=\frac{1+8}{2}=\frac{9}{4}=4\frac{1}{2}

∴ The coordinates of the midpoint are (-2\frac{1}{2},4\frac{1}{2})

* The coordinates of the midpoint of LN are (-2\frac{1}{2},4\frac{1}{2})

8 0
2 years ago
Find the 14th term of the geometric sequence 5 -10 20
Anna [14]

Answer:

  a14 = -40,960

Step-by-step explanation:

The sequence has a first term of 5 and a common ratio of -10/5 = -2, so the n-th term is given by ...

  an = a1·r^(n-1)

  an = 5·(-2)^(n-1)

The 14th term is then ...

  a14 = 5·(-2)^(14-1) = -5·2^13 = -5·8192

  a14 = -40,960

7 0
3 years ago
Read 2 more answers
work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company deci
NISA [10]

Answer:

Step-by-step explanation:

Here is the complete question.

Dominique, Marco, Roberto , and John work for a publishing company. The company wants to send two employees to a statistics conference. To be fair, the company decides that the two individuals who get to attend will have their names drawn from a hat. This is like obtaining a simple random sample of size 2. (a) Determine the sample space of the experiment. That is, list all possible simple random samples of size n = 2. (b) What is the probability that Dominique and Marco attend the conference? (c) What is the probability that John attends the conference?  ​(d) What is the probability that John stays​ home?

Since there are four employees to select the two to send from. then for us to get the sample space for the experiment we need to merge all possible two employees and represent them as set.

Let D = Dominique, M = Marco, R = Roberto and J = John

a) The sample space for the experiment is the total number of possible outcomes that we can have. It is as given below

S = 4C2 = 4!/(4-2)!2! (Selecting 2 out of 4 employees)

Total sample space = 4!/2!2!

Total sample space = 4*3*2!/2!2

Total sample space = 12/2 = 6

The sample space are S = {DM, DR, DJ, MR, MJ, RJ}

b) Probability is the ratio of number of event to the sample space.

P = n(E)/n(S)

Given n(S) = 6

n(E) is the event of Dominique and Marco attending the conference.

E = {DM}

n(E) = 1

P(D and M) = 1/6

Hence  the probability that Dominique and Marco attend the conference is 1/6

c) For John to attend the conference, the event outcome will be given as;

E = {DJ, MJ, RJ}

n(E) = 3

n(S) = 6

Probability for John to attend the conference is 3/6 = 1/2

d) Probability that John stays at home = 1 - Prob (John attends the conference)

Probability that John stays at home = 1 - 1/2

Probability that John stays at home = 1/2

7 0
3 years ago
H.C.F of 672 and 120 is ​
Vlada [557]

Answer:

24

Step-by-step explanation:

Check attachment for explanation... hope it helps :)

8 0
2 years ago
Please help quick for 25 points with explanations. Any silly and absurd answers will be reported.
amid [387]

Answer:

Note that the tangents to the circles at A and B intersect at a point Z on XY by radical center. Then, since∠ZAB=∠ZQA and ∠ZBA=∠ZQB. ∠AZB+∠AQB=∠AZB+∠ZAB+∠ZBA=180°. ∴ZAQB is cyclic.But if O is the center of w, clearly ZAOB is cyclic with diameter ZO, so ∠ZQO is 90° ⇒Q is the mid-point of XY.Then, by Power of a Point, PY· PX = PA · PB = 15 and it is given that PY+PX = 11. Thus,PX=(11±\sqrt{61})/2. So, PQ=\frac{\sqrt{61} }{2}, PQ²=\frac{61}{4}. Thus, the answer is 61+4=65.

Step-by-step explanation:

4 0
3 years ago
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