All categories

2 years ago

Consider the function represented by the graph. What is the domain of this function

Mathematics

2 answers:

Contact [7]2 years ago

5 0

Answer: The answer is $\{x:x\geq 0\}.$

Step-by-step explanation: We are given a graph that represents a function. We are to find the domain of the graphed function.

Domain and Range: If y=f(x) is a function, then the domain is the set of all first elements of ordered pairs (x-coordinates) and the range is the set of all second elements of ordered pairs (y-coordinates).

In the given graph, 'x' takes the values from 0 to infinity and 'y' takes the values from negative infinity to 9.

Therefore, domain of the function is [0, ∞) and the range is (-∞, 9).

Thus, the domain is given by the set $\{x:x\geq 0\}.$

Neporo4naja [7]2 years ago

3 0

The end of the ray stops the x values from proceeding left at x=0. So your domain is from that point on to infinity. In your solution set x >= 0, since the arrow continues on the right side where x's are positive.

You might be interested in

A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr

erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

$p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034$

The estimated standard error is:

$\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316$

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

$LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096$

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

8 0

2 years ago

P=100a÷t solve for a

pishuonlain [190]

a=1/100pt is the correct answer. First you had to multiply by t from both sides of equation, and it gave us, ⇒ $pt=100a$ . And then you had to flip an equation, and it gave us, ⇒ $100a=pt$ . You can also divide by one-hundred (100) from both sides of equation, and it gave us, $\frac{100a}{100}=\frac{pt}{100}$ . And finally, and it gave us the answer is a=1/100pt is the final answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

8 0

2 years ago

Read 2 more answers

A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100 lb and 5 lb and standard deviation

Bas_tet [7]

Answer:

Option b. None is the correct option.

The Answer is 63%

Step-by-step explanation:

To solve for this question, we would be using the z score formula

The formula for calculating a z-score is given as:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

We have boxes X and Y. So we will be combining both boxes

Mean of X = 100 lb

Mean of Y = 5 lb

Total mean = 100 + 5 = 105lb

Standard deviation for X = 1 lb

Standard deviation for Y = 0.5 lb

Remember Variance = Standard deviation ²

Variance for X = 1lb² = 1

Variance for Y = 0.5² = 0.25

Total variance = 1 + 0.25 = 1.25

Total standard deviation = √Total variance

= √1.25

Solving our question, we were asked to find the percent of filled boxes weighing between 104 lb and 106 lb are to be expected. Hence,

For 104lb

z = (x-μ)/σ,

z = 104 - 105 / √25

z = -0.89443

Using z score table ,

P( x = z)

P ( x = 104) = P( z = -0.89443) = 0.18555

For 1061b

z = (x-μ)/σ,

z = 106 - 105 / √25

z = 0.89443

Using z score table ,

P( x = z)

P ( x = 106) = P( z = 0.89443) = 0.81445

P(104 ≤ Z ≤ 106) = 0.81445 - 0.18555

= 0.6289

Converting to percentage, we have :

0.6289 × 100 = 62.89%

Approximately = 63 %

Therefore, the percent of filled boxes weighing between 104 lb and 106 lb that are to be expected is 63%

Since there is no 63% in the option, the correct answer is Option b. None.

3 0

2 years ago

Reggie heard that as a general rule, he should save at least 10% of his take- home pay. If Reggie's take-home pay is $2340 per m

ivolga24 [154]

We are given to understand as per the problem that one must save 10% of their take home pay.

Reggie's take home pay is $2,340 per month

As per the rule Reggie must save 10% of 2,340 per month = 0.10*2,340 = $234 per month

Since we need to know the savings in a year, we have to multiply the monthly savings by 12 since there are 12 months a year

Minimum amount per year that he should save = $234 *12 = $2,808

Minimum amount per year that he should save = $2,808

4 0

2 years ago

Read 2 more answers

Linear functions are expressed by data in a table and by a graph. Select all that apply.

vovangra [49]

The function expressed in the graph has a steeper slope than the function in the table.

and

The y-intercept is the same for both functions.

I just completed this assignment for edge.

5 0

2 years ago

Read 2 more answers