All categories

3 years ago

List the domain and range if the relation {(0,2),(4,3),(5,5),(4,7)}

2 answers:

5 0

(x,y)

domain is inputs (normally x or first number)
range is outputs from the given domain (normally y or second number)

domain is just all the first numbers
range is all the second numbers
ignore repeats

domain: {0,4,5}
range: {2,3,5,7}

RoseWind [281]3 years ago

4 0

Hello! The domain is the range of the x-values in a set; the range is the range of the y-values in a set.

Domain: {0,4,5}
Range: {2,3,5,7}

Hope this helps! :)

You might be interested in

Of all the people that attend movies 67% are in the 12-29 age group. At one theater, 300 people attended a showing of a certain

suter [353]

67 multiplied by 3 is 201, so there is 201 people in the age group 12-29. Also, 67% is approximately two thirds, and we can see that reflected in our answer.

6 0

3 years ago

A recipe says to use 4 cups of flour to make 60 cookies. What is the constant of proportionality that relates the number of cook

Sveta_85 [38]

Answer: divide the 60 and the 4 and that's your answer

Step-by-step explanation:

5 0

3 years ago

The table shows the cost of renting ski equipment at a ski lodge. Write a linear function f for the sequence. Then graph the fun

Bingel [31]

Answer:

y=5x+7

or

f(x)=5x+7

Step-by-step explanation:

Slope (m) = 5

Y-intercept (b) = 7

Using the equation y=mx+b, you can find m and b to make y=5x+7.

Steps:

(1, 12) (2, 17) formula: y2-y1 / x2-x1 = m

17-12= 5

2-1 = 1

5/1=5

m=5

12=5(1)+b

12=5+b

7=b

17=5(2)+b

17=10+b

7=b

Answer: y=5x+7

or

f(x)=5x+7

6 0

3 years ago

A five-digit number starts with a number between 4-9 in the first position, with no restrictions on the remaining 4 digits. a

Maurinko [17]

Answer:

(a) $Pr = 0.3024$

(b) $Pr = 0.6976$

Step-by-step explanation:

Given

$Start = \{5,6,7,8\}$ i.e. between 4 and 9

$n(Start) =4$

$Digits = 5$

Solving (a): Probability that each of the 5 digit are different

Since there is no restriction;

The total possible selection is as follows:

$First\ digit = 4$ (i.e. any of the 4 start digits)

$Second\ digit = 10\\$ (i.e. any of the 10 digits 0 - 9)

$Third\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

$Fourth\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

$Fifth\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

So, the total is:

$Total = 4 * 10 * 10 * 10 * 10$

$Total = 40000$

For selection that all digits are different, the selection is:

$First\ digit = 4$ (i.e. any of the 4 start digits)

$Second\ digit = 9$ (i.e. any of the remaining 9)

$Third\ digit = 8$ (i.e. any of the remaining 8)

$Fourth\ digit = 7$ (i.e. any of the remaining 7)

$Fifth\ digit = 6$ (i.e. any of the remaining 6)

So:

$Selection =4 * 9 * 8 * 7 * 6$

$Selection =12096$

So, the probability is:

$Pr = \frac{Selection}{Total}$

$Pr = \frac{12096}{40000}$

$Pr = 0.3024$

Solving (b): At least 1 repeated digit

The probability calculated in (a) is the all digits are different i.e. P(None)

So, using laws of complement

We have:

$P(At\ least\ 1) = 1 - P(None)$

So, we have:

$Pr= 1 - 0.3024$

$Pr = 0.6976$

Solving (c): An expression to model the probability.

Using (a) as a point of reference, we have;

$Pr = \frac{Selection}{Total}$

Where

$Selection =4 * 9 * 8 * 7 * 6$ ---- for selection of 5 i.e. n = 5

$Total = 4 * 10 * 10 * 10 * 10$

$Selection =4 * 9 * 8 * 7 * 6$

This can be rewritten as:

$Selection = 4 * ^9P_4$

4 can be expressed as: 5 - 1

So, we have:

$Selection = (5-1) *^9P_{5-1}$

Substitute n for 5

$Selection = (n-1) *^9P_{n-1}$

$Selection = (n-1)^9P_{n-1}$

$Total = 4 * 10 * 10 * 10 * 10$

This can be rewritten as:

$Total = 4 * 10^4$

$Total = (5-1) * 10^{5-1}$

$Total = (n-1) * 10^{n-1}$

$Total = (n-1) 10^{n-1}$

So, the expression is:

$Pr = \frac{(n-1)^9P_{n-1}}{(n-1)10^{n-1}}$

$Pr = \frac{^9P_{n-1}}{10^{n-1}}$

Where n represents the digit number

5 0

3 years ago

A chemist examines 12 sedimentary samples for bromide concentraction. The mean bromide concentration for the sample date is 0.43

Umnica [9.8K]

Answer:

a) $z = 1.645$

b) Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

Step-by-step explanation:

Population is approximately normal, so we can find the normal confidence interval.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ . This is the critical value, the answer for a).

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645*\frac{0.0325}{\sqrt{12}} = 0.015$

The lower end of the interval is the sample mean subtracted by M. So it is 0.437 - 0.015 = 0.422cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.437 + 0.015 = 0.452 cc/m³.

Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

4 0

3 years ago