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2 years ago

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Hi there i need help in this

1 answer:

MAXImum [283]2 years ago

4 0

Answer:

1.-38

2.36

3.-44

4.44

5.16

6.-24

7.?

8.-2

9.-9

10.18

11.43

12.4

Step-by-step explanation:

lol have a good day!

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HELP <br><br>MMEEE!! i will give brainly

LUCKY_DIMON [66]

Answer:

the answer is 28 in

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2 years ago

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A right triangle has a hypotenuse of 10 feet and a leg of 6 feet. What is the length of the other leg?

Artemon [7]

Answer:

The other length of the leg is <em>8 feet.</em>

Step-by-step explanation:

To find the answer we have to find what<em> 10</em> and <em>6 is squared.</em> <em>10 x 10 = 100</em>.

<em>6 x 6 = 36. </em>Now since we are trying to find the length of the other leg, we will <em>subtract 100 from 36</em> and we get <em>64</em>. Then for the last step, we will figure out what number<em> squared equals 64.</em> Because 8 x 8 = 64, <em>8 is our answer.</em>

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2 years ago

Solve each system using SUBSTITUTION. Remember to check!

Annette [7]

Answer:

$\left \{ {{x=3} \atop {y=-7}} \right.$

Step-by-step explanation:

We can directly substitute y of the first equation to the second equation.

-2x - 1 = 3x - 16

5x = 15

x = 3

Substituting back to any of the two equations, we get y = -2(3)-1 = -7. If you check with the second equation, y = 3(3)-16 = -7 as well.

Therefore $\left \{ {{x=3} \atop {y=-7}} \right.$ .

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2 years ago

Eight less than fout times a number is less than 56. what are the possible values of that number

Arada [10]

Answer:

x<16

Step-by-step explanation:

number<em> n </em>

eight less than four times a number ... 4 x <em>n</em> - 8

is less than 56 ... < 56

4 x <em>n</em> - 8 < 56

4 x <em>n</em> < 56 + 8

4 x<em> n</em> < 64/4

<em> n</em> < 64 / 4

<em>n</em> < 16

3 0

2 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

2 years ago