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sladkih [1.3K]
3 years ago
10

What is the answer to this question?

Mathematics
1 answer:
Vinil7 [7]3 years ago
8 0

Answer:

I think it's A

Step-by-step explanation:

It's opposites 6 goes over 15 and 15 goes over 6

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What is (x^-2 y^10)^2/x^5 y^-3 using only positive exponents
Eddi Din [679]

Answer:

\frac{(x^{ - 2}{y}^{10})^{2}  }{ {x}^{5}  {y}^{ - 3} }  \\   \frac{ {x}^{ - 4}  {y}^{20} }{ {x}^{5}  {y}^{ - 3  } }  \\   {x}^{ - 9}  {y}^{23} \\  \frac{ {y}^{23} }{ {x}^{9} }  \\ ans \: c

8 0
2 years ago
David wants to pack 574 cereal bars into bags of 10 and bags of 8 the number of bags of 8 is 2 more than the bags of 10. How man
BabaBlast [244]

Answer: Number of bags of 8. = 33

Number of bags of 10. = 31

Step-by-step explanation:

Let x = Number of bags of 8.

y = Number of bags of 10.

As per given.

8x+10y= 574   (i)

x=y+2 (ii)

Put value of x from (ii) in (i)

8(y+2)+10y= 574\\\\\Rightarrow\ 8y+16+10y=574\\\\\Rightarrow\ 18y=574-16\\\\\Rightarrow\ 18y=558\\\\\Rightarrow\ y=\dfrac{558}{18}\\\\\Rightarrow\ y=18

Put this in (i), x= 31+2=33

Hence,

Number of bags of 8. = 33

Number of bags of 10. = 31

6 0
2 years ago
The cumulative distribution function f(x) of a discrete random variable x is given by f(0) =. 30, f(1) =. 70, f(2) =. 90 and f(3
Lisa [10]

Correlation between x & y is 0.6125.

In probability theory and statistics, the cumulative distribution function of a real-valued random variable X, or simply the distribution function of X weighted by x, is the probability that X takes a value less than or equal to x.

The cumulative distribution function (CDF) of a random variable X is defined as FX(x)=P(X≤x) for all x∈R. Note that the subscript X indicates that this is the CDF of the random variable X. Also note that the CDF is defined for all x∈R. Let's look at an example.

Learn more about cumulative distribution here: brainly.com/question/24756209

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7 0
1 year ago
You and your friends decide to take a survey to try to convince mall managers to add a new store. You split up to survey parts A
svetlana [45]
33.3 percent or 33 and 1/3
6 0
3 years ago
Read 2 more answers
A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d
cestrela7 [59]

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

7 0
3 years ago
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