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3 years ago

Factor x3 + x2 + x + 1 by grouping. What is the resulting expression?

2 answers:

7 0

To factor by grouping, you group up the 1st 2 terms together and the last 2 terms together and then factor those pieces.

(x^3 + x^2) + (x + 1)
x^2(x + 1) + 1(x +1) - now factor out the (x + 1)
(x + 1)(x^2 + 1) - final answer

Lisa [10]3 years ago

3 0

Answer: answer is C.

Step-by-step explanation:

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Write the point-slope form of the line that passes through the origin and is parallel to a line with a slope of 2. include all o

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Point slope form is (y - y1) = m(x - x1), where m is a slope of 2, and x1 = 0 and y1 = 0 are our point at the origin. Therefore, we can plug in these values to get the point slope equation for our line: (y â€“ 0) = 2*(x - 0)

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3 years ago

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In terms of the number of moose, what is the relative frequency of male moose, female moose, adult moose, and baby moose? Write

user100 [1]

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I'm sorry is there a chart/graph, etc. to give us more information?

Without the added information there is no way for me to answer.

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3 years ago

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Pass a brotha some help :(

nikitadnepr [17]

Answer and Step-by-step explanation:

Distribute the 1p to the y and -6 and distribute the 2 to the 3y and 6.

10y - 50 + 6y + 12

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#teamtrees #PAW (Plant and Water)

7 0

3 years ago

Suppose a city with population 800,000 has been growing at a rate of 5% per year. If this rate continues, find the population of

AleksandrR [38]

Answer:

2,340,208.5 or 2,340,209

Step-by-step explanation:

You could answer this by multiplying the current population of 800,000 by 1.05 (1.05 represents the annual growth rate of population) and do that 22 times. But that would take a while.

we got 1.05 because the formula says that y = a( 1 + r ) power t

for a we will give it 1

so y = 1 ( 1 + 0.05) power t

it will give us 1.05

to do this faster, you would first calculate (1.05)22and then multiply this by 800,000.

So, 800,000 x (1.05)22 = 2,340,208.5 or 2,340,209

5 0

2 years ago

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