Ask question

Ask question

All categories

3 years ago

9

I need the answer to 10h+12+8h+4

2 answers:

KengaRu [80]3 years ago

6 0

Answer:

$\huge\boxed{18h+16}$

Step-by-step explanation:

$\text{Combine Like Terms.}\\\\\\10h+12+8h+4\\\\10h+8h+12+4\\\\18h+16\\\\\\\text{The expression can be no longer simplified.}\\\\\\\boxed{\text{The correct answer should be: } 18h+16}\\\\\\\text{ Hope this helps.}$

LiRa [457]3 years ago

4 0

Answer:

18h + 16

Step-by-step explanation:

Since H is an variable, we don’t know what it represents

You might be interested in

The radius of a circle is 4 feet. what is the angle measure of an arc pi feet long?

julsineya [31]

Answer:

Step-by-step explanation:

7 0

2 years ago

valerie has 20 stuff animals. she wants to store them in plastic bags. she estimates she can fit 3 stuffed animals in each bag.

Rudiy27

Answer: it’s not an even number it 6.6 repeating so if round up and say 7 bags

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please help me........

Kobotan [32]

Answer:

dkdkdkdjdjd

Step-by-step explanation:

3 0

3 years ago

Plz help me brainliest will be reward if right and 25 points

slega [8]

Answer:

the answer is f(x) = 6x + 1/6

Step-by-step explanation:

the g(x) ones y intercept is : 0,-6

the -6k(x) ones y intercept is : 0 , -6

the f ( x) ones y inetrcept is ; 0 , 1/6

the h (x) ones y intercept is : 0 , 6

5 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