All categories

3 years ago

What is the domain of the function on the graph

Mathematics

2 answers:

rosijanka [135]3 years ago

6 0

Answer:

The correct option is all real numbers greater than or equal to -3.

Step-by-step explanation:

The concept of domain of a function refers to all the points in the x axis for which the function is defined. In the graph as you can see the function starts in x = -3 (includes that point) and growths towards positive values of x, to +∞. So its domain is [-3, ∞) or {R, x≥ -3} in other words "all real numbers greater than or equal to -3."

Butoxors [25]3 years ago

3 0

All real numbers greater than or equal to -3

You might be interested in

I need help plzzzzz this is really confusing

Anuta_ua [19.1K]

Answer:

B. <u>x</u><u><</u><u> </u>-3

hope it helps.

<h3>stay safe healthy and happy.</h3>

5 0

3 years ago

Read 2 more answers

8. Colby and Cheryl work in different local supermarkets. Colby regularly earns $8.90 per hour,

faust18 [17]

Answer:

colby

hourly rate =$8.90

overtime rate = time and a half = 1.5

the hourly overtime rate is the product of the overtime rate and the hourly rate.

hourly overtime rate= overtime rate × hourly rate

= 1.5 × $8.90

= $13.35

Cheryl

hourly rate = $7.10

overtime rate = double time = 2

the hourly overtime rate is the product of overtime rate and the hourly rate.

hourly overtime rate = overtime rate × hourly rate

= 2 × $7.10

= $14.20

<h3>Comparison</h3>

we note that the hourly overtime rate of cheryl is higher than the hourly overtime rate of colby, which implies that cheryl earns more for one hour of overtime.

we are also interested in the difference of the hourly overtime rate:

hourly overtime rate colby — hourly overtime rate cheryl

= $14.20 – $13.35

= $0.85

thus cheryl earns $0.85 more for one hour of overtime than colby .

<h3>result</h3>

<h2>cheryl,$0.85</h2>

그것이 당신에게 도움이되기를 바랍니다 :)

가장 똑똑한 것으로 표시하십시오 :)

좋은 하루 되세요 :)

5 0

2 years ago

Which of these polygons is not formed by the cross-section created when a plane intersects a cube? Select all that are possible.

lakkis [162]

Answer:

Circle

Octagon

Step-by-step explanation:

There are no curves in a cube to make a circular cross section, and there aren't enough edges to make an octagon (just visualize)

8 0

3 years ago

Read 2 more answers

The standard deviation is used in conjunction with the ▼ to numerically describe distributions that are bell shaped. The ▼ varia

crimeas [40]

Solution: The standard deviation is used in conjunction with the mean to numerically describe distributions that are bell shaped. The mean measures the center of the distribution, while the standard deviation measures the spread of the distribution.

In normal distribution (bell shaped distribution) , the mean and standard deviation are used to describe its distribution. The mean measures the center of the distribution, while the standard deviation takes care of the spread of the distribution.

The mean of the normal distribution is $\mu$ and the standard deviation is $\sigma$

3 0

3 years ago

Below are data on 18 people who fell ill from an incident of food poisoning. The data give the incubation period (the time in ho

Sveta_85 [38]

Answer:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

$Min =15$

$Q_1 = \frac{20+20}{2}=20$

$Median= Q_2= \frac{32+34}{2}=33$

$Q_3 = \frac{46+46}{2}=46$

$Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85

Step-by-step explanation:

For this case we have the following data:

15 16 18 19 20 20 21 28 32 34 36 43 46 46 48 48 72 88

The sample size is n =18

We can calculate the mean with the following formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

The minimum for this case is $Min =15$

Now we can find the 5 number summary.

For the first quartile we work with the first 10 observations: 15 16 18 19 20 20 21 28 32 34. And Q1 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_1 = \frac{20+20}{2}=20$

For the median since we have an even number for the sample size would be the average between the 9th and the 10th position from the dataset ordered and we got:

$Median= Q_2= \frac{32+34}{2}=33$

For the third quartile we work with the last 10 observations: 32 34 36 43 46 46 48 48 72 88. And Q3 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_3 = \frac{46+46}{2}=46$

The maximum is $Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85

3 0

3 years ago