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1 year ago

15

chegg tell whether each sequence is arithmetic. if the sequence os arithmetic, identify the common difference and find the indic

ated term 3,5/2,3/2,-3/2,...

1 answer:

spin [16.1K]1 year ago

5 0

No, the given sequence is not an arithmetic sequence.

What is Arithmetic Sequence ?

An arithmetic sequence is a list of numbers with a definite pattern. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence.

In the above question,

The sequence is 3,5/2,3/2,-3/2,...

Take the 2nd term and minus the 1st term.

Now take the 3rd term and minus the 2nd term.

We can clearly notice that the differences are not same. Hence there is no common difference and therefore it's not an arithmetic sequence

To read about arithmetic sequence click here :

brainly.com/question/27079755

#SPJ4

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The numbers are 5,2,65

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A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

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Answer:

(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

Given

Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

$Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}$

$n(Head) = 32$

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

$n(Dice) = 30$

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

<em>From the above calculations: </em> $0.267 > 0.03125$ <em> Hence, person B is more likely to get 5</em>

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

$Median = 16.5th$

This means that the median is the mean of the 16th and the 17th item

So,

$Median = \frac{3+2}{2}$

$Median = \frac{5}{2}$

$Median = 2.5$

For person B

$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

$Median = \frac{31}{2}th$

$Median = 15.5th$

This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

<em>Person B has a greater median of 5</em>

Question 12: Probability that B gets 5 or 6

This is calculated as:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

So, we have:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

$P(5\ or\ 6) = \frac{18}{30}$

$P(5\ or\ 6) = 0.60$

$P(5\ or\ 6) = 60\%$

Question 13: Person with higher probability of 3 or more

Person A

$n(3\ or\ more) = 16$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}$

$P(3\ or\ more) = \frac{16}{32}$

$P(3\ or\ more) = 0.50$

$P(3\ or\ more) = 50\%$

Person B

$n(3\ or\ more) = 28$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}$

$P(3\ or\ more) = \frac{28}{30}$

$P(3\ or\ more) = 0.933$

$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

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2 years ago

What is the Value of x+y ? PLZ Help ASAP

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ANSWER

$x + y = 60$

EXPLANATION

Method 1

Vertically opposite angles are equal, therefore

$3x + 55 = 85$

This implies that,

$3x = 85 - 55$

We simplify to get,

$3x = 30$

Dividing through by 3 gives,

$x = 10$

Also,

$2y - 5 = 95$

This implies that,

$2x = 95 + 5$

We simplify the right hand side to get,

$2x = 100$

We divide both sides both sides by 2 to get,

$x = 50$

$x + y = 50 + 10$

$x + y = 60$

Method 2

Angles on straight line sums to 180°.

$2y-5+85=180$

This implies that,

$2y+80=180$

$2y=180-80$

$2y=100$

$y=50$

Also,

$3x+55+95=180$

$3x+150=180$

$3x=180-150$

$3x=30$

$x=10$

$x+y=50+10$

$x+y=60$

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