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Andrews [41]
3 years ago
10

Which distribution has the greatest spread

Mathematics
1 answer:
Sauron [17]3 years ago
4 0

For solving this question we will use the relationship between the range of data and the spread of data. The relationship is that "Higher the Range of the Data, Higher is the Spread and lower the range of the data then lower is the spread".

So, keeping in mind this definition let us proceed and check each option and measure their range.

We know that Range is measured as

Range=(Highest Value of the Data)-(Lowest Value of the Data)

a. 13.7-3.1=10.6

b. 423.6-1.3=422.3

c. 38.4-7.2=31.2

d. 231.4-5.5=225.9

As we can see Option b. has the largest range and as range is a measure of spread then option b is the distribution which has the largest spread.

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If p and q are both true, then which of the following statements has the same truth-value as ~p → q?
olganol [36]

If p and q are both true, then

\neg p \implies q

is an implication of the form

F \implies T

which is true, because every implication starting with false is true, i.e.

F \implies T = T,\quad F \implies F = T

So, we're looking for an expression evaluating to true. Let's see what we have:

A) is an AND proposition. Logical AND is true only if both parts are true. So, you have

\neg P \land \neg Q = F \land F = F

So it's not the right option.

B) is an OR proposition. Logical OR is true whenever one of the two parts is true. So, you have

\neg P \lor\neg Q = F \lor F = F

So it's not the right option.

C) is again an AND proposition. You have

P \land \neg Q = T \land F = F

So this is not the right option.

D) Finally, the last one is again an implication, and again it starts with false:

\neg Q \implies P = F \implies T = T

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8 0
2 years ago
Four darts are thrown at this dartboard. If all four darts hit the board, how many different point totals are possible. (Dartboa
jonny [76]

Answer:

Different point total are possible = 497

Step-by-step explanation:

Given - Four darts are thrown at this dartboard. Dartboard regions are  

             1, 4, 7, 10 points

To find - If all four darts hit the board, how many different point totals are

              possible.

Proof -

Are given there are 4 regions -

1st region - 1 point

2nd region - 4 points

3rd region - 7 points

4th region - 10 points

Case I :

If all the darts hit the different region -

Total points possible are - 1 + 4 + 7 + 10 = 22 points

Case II :

If all the darts hit the same region -

It means either they hit 1 region or 2nd region or 3rd region or 4th region

If they all 4 hit first region , points are - 1+1+1+1 = 4

If they all 4 hit second region , points are - 4+4+4+4= 16

If they all 4 hit third region , points are - 7+7+7+7 = 28

If they all 4 hit fourth region , points are - 10+10+10+10 = 40

So,

the Total points possible are - 4 + 16 + 28 + 40 = 88

Case III :

If 3 darts hit the same region -

Sub-case 1 :

If 1 dart hit 1st region, other 3 dart hit 2nd region

Points are - 1 + 4 + 4+ 4 = 13

Sub-case 2 :

If 1 dart hit 1st region , other 3 dart hit 3rd region

Points are - 1 + 7 + 7 + 7 = 22

Sub-case 3 :

If 1 dart hit 1st region , other 3 dart hit 4th region

Points are - 1 + 10 + 10 + 10 = 31

Sub-case 4 :

If 1 dart hit 2nd region , other 3 dart hit 1st region

Points are - 4 + 1 + 1 + 1 = 7

Sub-case 5 :

If 1 dart hit 2nd region , other 3 dart hit 3rd region

Points are - 4 + 7 + 7 + 7 = 25

Sub-case 6 :

If 1 dart hit 2nd region , other 3 dart hit 4th region

Points are - 4 + 10 + 10 + 10 = 34

Sub-case 7 :

If 1 dart hit 3rd region , other 3 dart hit 1st region

Points are - 7 + 1 + 1 + 1 = 10

Sub-case 8 :

If 1 dart hit 3rd region , other 3 dart hit 2nd region

Points are - 7 + 4 + 4 + 4 = 19

Sub-case 9 :

If 1 dart hit 3rd region , other 3 dart hit 4th region

Points are - 7 + 10 + 10 + 10 = 37

Sub-case 10 :

If 1 dart hit 4th region , other 3 dart hit 1st region

Points are - 10 + 1 + 1 + 1 = 13

Sub-case 11 :

If 1 dart hit 4th region , other 3 dart hit 2nd region

Points are - 10 + 4 + 4 + 4 = 22

Sub-case 12 :

If 1 dart hit 4th region , other 3 dart hit 3rd region

Points are - 10 + 7 + 7 + 7 = 31

So,

Total points possible are - 13+22+21+7+25+34+10+19+37+13+22+31 = 254

Case IV :

If 2 darts hit the same region -

Sub-case 1:

If 2 darts hits 1st region , other 2 darts hit 2nd region

Points are - 1 + 1 + 4 + 4 = 10

Sub-case 2:

If 2 darts hits 1st region , other 2 darts hit 3rd region

Points are - 1 + 1 + 7 + 7 = 16

Sub-case 3:

If 2 darts hits 1st region , other 2 darts hit 4th region

Points are - 1 + 1 + 10 + 10 = 23

Sub-case 4:

If 2 darts hits 2nd region , other 2 darts hit 3rd region

Points are - 4 + 4 + 7 + 7 = 22

Sub-case 5:

If 2 darts hits 2nd region , other 2 darts hit 4th region

Points are - 4 + 4 + 10 + 10 = 28

Sub-case 6:

If 2 darts hits 3rd region , other 2 darts hit 4th region

Points are - 7 + 7 + 10 + 10 = 34

So,

Total points possible are - 10 + 16 + 23 + 22 + 28 + 34 = 133

Case V :

If 1 darts hit the same region -

This case is include in the Case III.

∴ we get

Different point total are possible = Points in Case I +  II +  III +  IV

                                                       = 22 + 88 + 254 + 133

                                                       = 497

⇒Different point total are possible = 497

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Answer:

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Step-by-step explanation:

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