Ask question

Ask question

All categories

2 years ago

5

What is the domain of this relation?

2 answers:

yan [13]2 years ago

8 0

Answer:

{-3, 3, 5, 9}

Step-by-step explanation:

The domain is simply the inputs (x) used in the function.

defon2 years ago

3 0

Answer:

{-3,3,5,9} this is answer

You might be interested in

Identify the expression equivalent to 16x + 24y

tatuchka [14]

Correct answer is 3: 8 (2x + 3y)

3 0

2 years ago

Read 2 more answers

Which expressions is the same as 1/5 of 7

Harlamova29_29 [7]

The answer is 1/5 < 35/5

8 0

2 years ago

Scott took out a 72 month loan for $35,000 to purchase a new boat. If Scott paid $8,925 in simple interest, what was the interes

diamong [38]

Answer:

Option 4.7% = 3,500 x 4.7% =$164.50 simple annual interest.

82.25 this is what Scott will pay in 6 months at simple interest.

Option 4.2% =3,500 x (1 +0.042/12)^6 =3,500 x 1.0035^6=$3,574.15.

3,500 =$74.15 this is what Scott will pay in 6 months at compounded interest.

The compound option is cheaper by: 74.15 =$8.10.

6 0

3 years ago

Read 2 more answers

1. Evaluate the function f(x)=3x-1 using the domain of -2, 0, 3, and 5. Results are to be shown in a table.

IceJOKER [234]

Answer:

See explanation

Step-by-step explanation:

You are given the function $f(x)=3x-1$

The domain of the function are all possible values of variable x, so you can find f(x) for x=-2, 0, 3, 5:

$f(-2)=3\cdot (-2)-1=-6-1=-7\\ \\f(0)=3\cdot 0-1=0-1=-1\\ \\f(3)=3\cdot 3-1=9-1=8\\ \\f(5)=3\cdot 5-1=15-1=14$

The table is

$\begin{array}{cc}x&f(x)\\-2&-7\\0&-1\\3&8\\5&14\end{array}$

The inverse function $f^{-1}(x)$ has the domain which is the range of the function f(x) (all possible values of f(x)), so the domain of inverse function is {-7,-1,8,14}

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:

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

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

$Pr(5) = 0.267$

<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