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MAVERICK [17]
3 years ago
11

What is the domain and range of the function? Domain:

Mathematics
2 answers:
ad-work [718]3 years ago
8 0
It depends on the function
Pie3 years ago
8 0

Answer:

Domain: {x∈R}

Range: {f(x)∈R | f(x) ≥ 0}

Step-by-step explanation:

The domain of a function is, in simple terms, all the possible x values of a function (input values), and as the graph shows that all x values are possible (x, 0 and -x) meaning x can be any rational number, represented by the R and the subset symbol (∈).

The range is all the possible f(x) values (outputs/y values) and if you look at the graph you see that it isn't possible for the f(x)/output/y values to be negative so we write:

{f{x}∈R (f(x)/output/y value is rational) | (and) f(x) ≥ 0 (f(x)/output/y value is bigger or equal to 0, not negative)}

Hope I helped :)

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A local food truck specializes in gourmet grilled cheese sandwiches. Each month, the owners of the truck have a constant total o
Citrus2011 [14]
B would be best answer. 5400s is rent

2.25s amount of sandwich cost and 9.00s selling prices. They all need s and has to be less then!
7 0
2 years ago
Find the area of each figure. round to the hundredths place when necessary
notsponge [240]

Answer:

115.48m^{2}

Step-by-step explanation:

This shape can be split into two distinct shapes

Two halves of a semi circle, and a rectangle in between

Circle:

Putting both halves of the semi circle together will give you a full circle. The diameter of the circle is given (7m).

The area of a circle is A = π r^{2}

The radius, r, is half of the diameter, so 7 / 2 = 3.5m

A = π r^{2}

A = π * 3.5^{2}

A = 38.38m^{2}

Rectangle:

The area of a rectangle is A = h b

The height, h, is known at 7m

The base, b, can be found by removing the length from the dot to the end of the semi circles. This length is the radius of the semi circles, 3.5m

Removing the radius from the total length given

18 - 3.5 - 3.5 = 11m

The base is 11m

A = h b

A = 7 * 11 = 77m^{2}

Total Area = Circle area + Rectangle area

Total Area = 38.38 + 77 = 115.48m^{2}

6 0
3 years ago
An airplane started at point K, travelled 560 miles to point L, adjusted its route and travelled another 575 miles to point M. I
S_A_V [24]

Answer:

Step-by-step explanation:

Using the cosine theorem:

c^{2} =a^{2} +b^{2} -abcos\beta

952^{2} =560^{2} +575^{2} -560*575cos\beta

\beta =cos^{-1} (\frac{560^{2}+575^{2}-952^{2}   }{2*560*575} )

The angle is 114.014 degrees if we substract 90 degrees it means that at point L the plane changed its angle by 24 degrees

4 0
3 years ago
Solve pls brainliest
Drupady [299]

91/25, 3.92, 3.928, 2 13/20

im pretty sure this is correct AND IM RLY SRY IF IT ISN’T

7 0
2 years ago
As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a
REY [17]

Answer:

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

Step-by-step explanation:

Before building the confidence interval we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample of 800 teens. 73% said that they use social networking sites.

This means that:

p_T = 0.73, s_T = \sqrt{\frac{0.73*0.27}{800}} = 0.0157

Sample of 2253 adults. 47% said that they use social networking sites.

This means that:

p_A = 0.47,s_A = \sqrt{\frac{0.47*0.53}{2253}} = 0.0105

Distribution of the difference:

p = p_T - p_A = 0.73 - 0.47 = 0.26

s = \sqrt{s_T^2+s_A^2} = \sqrt{0.0157^2+0.0105^2} = 0.019

Confidence interval:

Is given by:

p \pm zs

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Lower bound:

p - 1.96s = 0.26 - 1.96*0.019 = 0.223

Upper bound:

p + 1.96s = 0.26 + 1.96*0.019 = 0.297

The 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social networking sites is (0.223, 0.297). This means that we are 95% sure that the true difference of the proportion is in this interval, between 0.223 and 0.297.

3 0
3 years ago
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