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3 years ago

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

#SPJ4

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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

4 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:

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

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

$Pr(5) = 0.267$

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

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,

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

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

$Median = 5$

Person B has a greater median of 5

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