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Dvinal [7]
3 years ago
14

What is the average of 4 1/2, 3 1/3, 2, 2 1/6? I will give brainleist.

Mathematics
1 answer:
kotykmax [81]3 years ago
7 0

Answer:

3

Step-by-step explanation:

  4 1/2 + 3 1/3 + 2 + 2 1/6

= 4 3/6 + 3 2/6 + 2 + 2 1/6

= 4 + 3 + 2 + 2 + 3/6 + 2/6 + 1/6

= 11 + 6/6

= 11 + 1

= 12

There are 4 numbers, therefore average = 12 ÷ 4 = 3

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Part A: Plot the points A(-8,7) and B(6,-9). Mark the halfway point on AB and label it point M. What are the coordinates of M? P
vredina [299]

Answer:

a. M(x,y) = (-1,-1)

b. D(-4,-9)

Step-by-step explanation:

Given

A(-8,7) and B(6,-9).

Solving (a):

Determine the Midpoint M;

This is calculated as follows;

M(x,y) = (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Where

(x_1,y_1) = (-8,7)

(x_2,y_2) = (6,-9)

Substitute these values in the formula

M(x,y) = (\frac{-8+6}{2},\frac{7-9}{2})

M(x,y) = (\frac{-2}{2},\frac{-2}{2})

M(x,y) = (-1,-1)

<em>Hence; the midpoint is (-1,-1)</em>

Solving (a):

Here, we have

C = (2,7)

M; Midpoint = (-1,-1)

Required: Determine D

This is calculated as follows;

M(x,y) = (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Where

(x_1,y_1) = (2,7)

(x,y) = (-1,-1)

Substitute these values in the formula

(-1,-1) = (\frac{2+x_2}{2},\frac{7+y_2}{2})

Solving for x2

-1 = \frac{2 + x_2}{2}

Multiply both sides by 2

-2 = 2 + x_2

Subtract 2 from both sides

x_2 = -2 - 2

x_2 = -4

Solving for y2

-1 = \frac{7 + y_2}{2}

Multiply both sides by 2

-2 = 7 + y_2

Subtract 7 from both sides

y_2 = -2 - 7

y_2 = -9

Hence, the coordinates of D is

D(-4,-9)

7 0
4 years ago
How is 35 to 10 proportional to 7 to 5
KiRa [710]
If you simplify 35/10 both side divide by 7 is equal to 7/5

6 0
3 years ago
Solve by the addition method: 2x-5y=18 ,3x+4y=4
alexgriva [62]
2 x - 5 y = 18  / * 4
3 x + 4 y = 4   / * 5  ( we will use the addition method )
--------------------------
 8 x - 20 y = 72
+
15 x + 20 y = 20
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23 x = 92
x = 92 : 23
x = 4
8 - 5 y = 18
- 5 y = 18 - 8
- 5 y = 10
y = 10  : ( - 5 ) 
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( x, y ) = ( 4, - 2 ) 
8 0
4 years ago
Two boys, starting at the same point, walk-in opposite directions for x meters, turn right and walk another y meters, and turn r
Alecsey [184]

Answer:

2y meters

Step-by-step explanation:

The first x meters are cancelled with the last x meters because after turning right two times they walk in the opposite direction that each of them started to walk.

The y meters each of them walk are also in the opposite direction, so given that each of them walked y meters, the distance between them is 2y meters.

5 0
3 years ago
Suppose n people, n ≥ 3, play "odd person out" to decide who will buy the next round of refreshments. The n people each flip a f
blondinia [14]

Answer:

Assume that all the coins involved here are fair coins.

a) Probability of finding the "odd" person in one round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}.

b) Probability of finding the "odd" person in the kth round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}.

c) Expected number of rounds: \displaystyle \frac{2^{n - 1}}{n}.

Step-by-step explanation:

<h3>a)</h3>

To decide the "odd" person, either of the following must happen:

  • There are (n - 1) heads and 1 tail, or
  • There are 1 head and (n - 1) tails.

Assume that the coins here all are all fair. In other words, each has a 50\,\% chance of landing on the head and a

The binomial distribution can model the outcome of n coin-tosses. The chance of getting x heads out of

  • The chance of getting (n - 1) heads (and consequently, 1 tail) would be \displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n.
  • The chance of getting 1 heads (and consequently, (n - 1) tails) would be \displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n.

These two events are mutually-exclusive. \displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n  = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1} would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.

<h3>b)</h3>

Since the coins here are all fair, the chance of determining the "odd" person would be \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} in all rounds.

When the chance p of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the kth round: (1 - p)^{k - 1} \cdot p. That's the same as the probability of getting one success after (k - 1) unsuccessful attempts.

In this case, \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}. Therefore, the probability of succeeding on round k round would be

\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}.

<h3>c)</h3>

Let p is the chance of success on each round in a geometric distribution. The expected value of that distribution would be \displaystyle \frac{1}{p}.

In this case, since \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}, the expected value would be \displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}.

7 0
3 years ago
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