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3 years ago

Which equation is true? A) 4/-9=-4/9 B) 4/-9=2/-3 C) 4/-9=4/9 D) 4/-9=2/3

Mathematics

1 answer:

prisoha [69]3 years ago

6 0

The answer is B, 4/-9 = 2/-3

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I 8. Classify 8 as a perfect square, a perfect cube, both, or neither.

Murljashka [212]

Answer:

It's a perfect cube, but not a perfect square.

Step-by-step explanation:

2^3 = 8, meaning 8 is a perfect cube. But since 2^2 = 4 and 3^2 = 9, and 8 is in between, 8 is not a perfect square.

I hope this helped.

7 0

3 years ago

Can you help me list 5hese from least to greatest

g100num [7]

Ok sweetheart, the easy way to solve these problems is to change them to all the same form for example "decimal" after that you sort them! just don't forget to remember which is which!!! so this is how you solve.

squared 25, 4.26, 3 1/5 2^3,2.58
i converted each so your answer is A.

4 0

3 years ago

WILL GIVE BRAINLIEST! Match the equations that are the same

wolverine [178]

Answer:

1. 1/x = 8

Answer = 8x-1

2. 8x+1=3

Answer ; x=1/4

3. 7= 14/X

Answer ; x = 2

4.1/2x^2 = 2

Answer ;x=2

Step-by-step explanation:

$\frac{1}{x} =8 \\8x = 1\\\\\\8x+1=3\\Collect -like- terms \\8x =3-1\\8x = 2\\Divide -both -sides- by ;8\\\frac{8x}{8} =\frac{2}{8} \\x = 1/4\\\\\\7= \frac{14}{x} \\Cross -multiply\\7x =14\\Divide-both-sides-by-7\\x = 2\\\\\\\frac{1}{2} x^{2} =2\\\frac{x^{2} }{2} =2\\Cross-multiply\\x^{2} =4\\Squre-root -both-sides\\\sqrt{x^2}=\sqrt{4} \\x = 2\\$

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:

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

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

$Pr(5) = 0.267$

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

2 years ago

Choose the infinite geometric series of .15

Karo-lina-s [1.5K]

The answer to your question is option "D"

5 0

3 years ago