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3 years ago

What are the domain and range of the function f(x)= 2x+1

Mathematics

1 answer:

navik [9.2K]3 years ago

4 0

Answer:

Domain: All the real numbers

Range: All the real numbers

Step-by-step explanation:

The domain of a function is the complete set of possible values of the independent variable 'x'. That is to say, all the values that 'x' can take:

In this case, f(x)= 2x+1, the independent variable has no restrictions. Meaning that 'x' can take all the Real Values. In set notation: x∈ℝ.

The range of a function is the complete set of all possible resulting values of the dependent variable 'y'. In this case, given that the independent variable has no restrictions, the dependent variable 'y' can take any value. So, the range is: y ∈ ( −∞, ∞ ) - All the real numbers.

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Radioactive isotopes are commonly used to estimate artifact ages. Let X be the time in years for a Carbon-14 atom to decay to Ni

Alexeev081 [22]

Answer:

a) 1 / years

b) P ( 5000 < X < 6000) = 0.06222

c) 8267 years

d) T_1/2 = 5730.25 years

Step-by-step explanation:

Given:

- The decay constant λ = 1 / 8267

- X is an exponential random Variable

Find:

a. What are the units of What are the units of λ?

b. What is the probability the decay time is between 5000 and 6000 years?

c. Compute SD(X).

d. Compute the median decay time. The result is called the half-life of Carbon-14?

Solution:

- The decay constant λ means the rate at which the C-14 atoms decays into N-14 atom. The rate is expressed in units of 1 / year.

- The random variable X follows an exponential distribution which has a probability mass function and cumulative density functions as follows:

P ( X = t ) = λ*e^(-λ*t)

P ( X = t ) = e^(-t/8267) / 8267

P ( X < t ) = 1 - e^(-λ*t) = 1 - e^(-t/8267)

- The probability of the decay time between years 5000 and 6000 years is:

P ( 5000 < X < 6000) = 1 - e^(-6000/8267) - 1 + e^(-5000/8267)

P ( 5000 < X < 6000) = 0.06222

- The standard deviation of the exponential distribution is given by:

SD(X) = 1 / λ = 8267 years

- The median decay time for an exponential distribution is given by:

T_1/2 = Ln(2) / λ = Ln(2)*8267

T_1/2 = 5730.25 years

3 0

3 years ago

Read 2 more answers

If you are paid every 3 months, or 4 times a year, what is that called?

ollegr [7]

Thank you! This helped a lot!!! :)

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2 years ago

Solve 40=4y.<br><br> the solution is y=??

Tanzania [10]

Answer:

y = 10

Step-by-step explanation:

40 = 4y

y = 40 : 4

y = 10

6 0

3 years ago

Read 2 more answers

PLEASE HELP

Andrej [43]

Answer: -3 and 0

Step-by-step explanation: f(x) = g(x) where they intersect. Two lines intersect when they have the same x and y values.

5 0

3 years ago

Car A is rated at 32 mi/gal (miles per gallon) for city driving and 29 mi/gal for highway driving. Car B is rated at 27 mi/g

Irina18 [472]

The gallons of gas that each vehicle would consume for the different amounts of driving are as follows:

a) For 300 mi of city driving plus 100 mi of highway driving, car A will consume <u>12.825 gallons</u> of gas, while car B will consume <u>14.05 gallons</u> of gas.

b) For 100 mi of city driving plus 300 mi of highway driving, car A will consume <u>13.475 gallons</u> of gas, while car B will consume <u>12.52 gallons</u> of gas.

<h3>Data and Calculations:</h3><h3>Rating (miles per gallon)</h3>

City Driving Highway Driving

Car A 32 mi/gal 29 mi/gal

Car B 27 mi/gal 34 mi/gal

<h3>Miles of Driving</h3>

City Driving Highway Driving

a) 300 mi 100 mi

b) 100 mi 300 mi

<h3>Car A's Consumption of Gas:</h3>

City Driving Highway Driving Total

a) 9.375 gal.(300/32) 3.45 gal. (100/29) 12.825 gal.

b) 3.125 gal.(100/32) 10.35 gal (300/29) 13.475 gal.

<h3>Car B's Consumption of Gas:</h3>

City Driving Highway Driving Total

a) 11.11 gal. (300/27) 2.94 gal. (100/34) 14.05 gal.

b) 3.70 gal. (100/27) 8.82 gal. (300/34) 12.52 gal.

Thus, different driving ratings (miles per gallon) show the differences in gas consumption by vehicles in city driving and highway driving.

Learn more about the consumption of gas for driving at brainly.com/question/13244789

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2 years ago