Answer:
Step-by-step explanation:
The correct answer is the number of minutes the students studied. The student's grades is dependent on how long they studied. The domain is the independent variable or the amount of time.
Final answer:
The domain in a scatterplot of study time versus test grades represents all the reported study times of the students. It is visualized by plotting this data on a dot plot or analyzing the distribution to see the correlation between study time and grades.
Explanation:
The domain of a function in mathematics represents the set of all possible input values it can accept, typically the x-values on a graph. In the context of a scatterplot of time studying versus test grades created by a math teacher, the domain would consist of the various amounts of time that students reported studying. Analyzing the distribution of these values can showcase how study time correlates to test performance, assuming that this is a function with a one-to-one correspondence between study time and grade received, passing the vertical-line test.
To create a dot plot, as suggested in a collaborative classroom exercise, you would align a number line with the possible amounts of study time and place a dot above the corresponding value for each student's reported study time. Through this process, you could visualize the amount of study time on the x-axis and gain insights into the study habits of the class.
In a given scenario like the one described for Ms. Phan studying for an economics exam, the total benefit and the expected total gain in her score could be plotted to visualize the marginal benefits of additional study hours. This helps to understand the concept of diminishing returns in relation to study time and test score improvement.
What is the quotient: (3x2 + 4x – 15) ÷ (x + 3) ?
is it 3x – 1, r = 1?
i would just like for someone to confirm if it is or isnt ...?
Answer:
No, it is 3x-5 and r=0
Step-by-step explanation:
We can do it by long division method the required quotient is 3x-5 and r=0
not 3x-1 , r=1
multiply the divisor with 3x we will get [tex]3x^2+9x[/tex]to cancel out the first term of dividend
Now after solving we will get [tex]-5x-15[/tex]
Now, multiply the divisor by -5 we will get -5x-15 which will cancel the entire dividend.
A. Line a
B. Line b
C. Line c
D. Line d
A function of the form f(x) = mx + b, where m and b are real numbers, is called a _____ function.
Example: f(x) = 6x - 5
Answer:
Linear
Step-by-step explanation:
Took the test (USA Test prep)
What is the result of adding the system of equations? 2x + y = 4
3x - y = 6
A.x=2
B.x=10
C.5x=10
Hilary wants to go on the latin club trip to italy , it will cost 2,730 for the trip the trip is 30 weeks away and she wants to make equal weekly payments , how much money altogether does hilary need to pay at the end of week 8
Bill had 240 pieces of gum. he gave 1/6 of the piece of gum to his sister. how many pieces of gum did he give to his sister?
What is −20÷45−20÷45 ?
−25−25
−16−16
−116−116
−125
Answer:
the answer is a hope it helps.
Step-by-step explanation:
What is 8-8 to the power of -1 ?
Can someone please help me with this question??!!
An epidemic follows the curve
P = 500 / 1+20,000e^(-0.549t)
; where t is in years. How fast is the epidemic growing after 10 years? (Round your answer to two significant digits.)
The rate at which the epidemic is growing after 10 years is approximately 0.79.
Using the provided formula for the derivative of the population function with respect to time and evaluating it at ( t = 10), we have:
[tex]\[ \frac{dP}{dt} \Bigg|_{t=10} = \frac{-500(20,000)(-0.549)e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \][/tex]
[tex]\[ \approx \frac{-500(20,000)(-0.549)e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \][/tex]
[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(1 + 20,000(0.004088))^2} \][/tex]
[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(1 + 81.76)^2} \][/tex]
[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(82.76)^2} \][/tex]
[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{6856.8976} \][/tex]
[tex]\[ \approx \frac{5431.56}{6856.8976} \][/tex]
[tex]\[ \approx 0.7926 \][/tex]
Rounding to two significant digits, the rate at which the epidemic is growing after 10 years is approximately 0.79.
Answer:
To find the rate of growth of the epidemic after 10 years, we'll first differentiate the epidemic curve equation with respect to time (t) and then plug in t = 10 to find the growth rate.
Therefore, after 10 years, the epidemic is growing at a rate of approximately -0.082 (rounded to two significant digits).
Step-by-step explanation:
To determine the rate of growth of the epidemic after 10 years, we'll first differentiate the given epidemic curve equation with respect to time (t) using the quotient rule and the chain rule of differentiation.
Let [tex]\( P = \frac{500}{1 + 20,000e^{-0.549t}} \)[/tex].
To differentiate P with respect to t, we'll use the quotient rule:
[tex]\[ \frac{dP}{dt} = \frac{d}{dt} \left( \frac{500}{1 + 20,000e^{-0.549t}} \right) \]\[ = \frac{0 - 500 \times \frac{d}{dt}(1 + 20,000e^{-0.549t})}{(1 + 20,000e^{-0.549t})^2} \][/tex]
Now, we'll find [tex]\( \frac{d}{dt}(1 + 20,000e^{-0.549t}) \)[/tex] using the chain rule:
[tex]\[ \frac{d}{dt}(1 + 20,000e^{-0.549t}) = 0 - 20,000 \times (-0.549)e^{-0.549t} \]\[ = 10,980e^{-0.549t} \][/tex]
Substituting this back into the differentiation of P:
[tex]\[ \frac{dP}{dt} = \frac{-500 \times 10,980e^{-0.549t}}{(1 + 20,000e^{-0.549t})^2} \][/tex]
Now, we'll find the growth rate after 10 years by plugging in [tex]\( t = 10 \)[/tex] into [tex]\( \frac{dP}{dt} \)[/tex]:
[tex]\[ \frac{dP}{dt} \bigg|_{t=10} = \frac{-500 \times 10,980e^{-0.549 \times 10}}{(1 + 20,000e^{-0.549 \times 10})^2} \]\[ \approx \frac{-500 \times 10,980 \times e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \]\[ \approx \frac{-500 \times 10,980 \times 0.004056}{(1 + 20,000 \times 0.004056)^2} \]\[ \approx -0.082 \][/tex]
Thus, after 10 years, the epidemic is growing at a rate of approximately -0.082 (rounded to two significant digits).
What is the length of BB'?
By using the distance formula, the length of BB' on the graph is equal to [tex]\sqrt{29}\;units.[/tex]
How to determine the distance between the coordinates of each points?In Mathematics and Geometry, the distance between two (2) end points that are on a coordinate plane can be calculated by using the following mathematical equation:
[tex]Distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]
Where:
x and y represent the data points (coordinates) on a cartesian coordinate.
By substituting the given end points B (0, 2) and B' (5, 4) into the distance formula, we have the following;
[tex]Distance = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\\\Distance \;BB'= \sqrt{(5-0)^2 + (4-2)^2}\\\\Distance \;BB'= \sqrt{(5)^2 + (2)^2}\\\\Distance \;BB'= \sqrt{25 + 4}\\\\Distance \;BB'= \sqrt{29}\;units[/tex]
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How many hours would someone who earns $6.25 per hour have to work to earn $225.65?
Answer: 36.1 hours
Step-by-step explanation:
Given: The amount someone earns for each hour worked = [tex]\$6.25[/tex]
The expected amount to earn by work = [tex]\$225.65[/tex]
Now, to find the number of hours work to earn the expected value , we divide the expected value by the hourly rate, we get
The number of hours work to earn [tex]\$225.65\ =\frac{225.65}{6.25}=36.104\approx36.1[/tex]
Hence, the number of hours work to earn [tex]\$225.65[/tex] about 36.1 hours.
F(x)= kx2, and f(2)=12, then k equals
The value of k in the given function is k =3
From the question, the given function is F(x)= kx2
This can be properly written as
[tex]f(x) =kx^{2}[/tex]
Also, from the question, we have that f(2) = 12
Since [tex]f(x) =kx^{2}[/tex]
∴ [tex]f(2) =k(2)^{2}[/tex]
This becomes
[tex]f(2) = k \times 4[/tex]
[tex]f(2) = 4k[/tex]
Now, to determine the value of k, we will input the value of f(2), that is f(2)=12 in the above equation, that is
[tex]f(2) = 4k[/tex] becomes
[tex]12= 4k[/tex]
Now, divide both sides by 4
[tex]\frac{12}{4} = \frac{4k}{4}[/tex]
[tex]3 = k[/tex]
∴ [tex]k = 3[/tex]
Hence, the value of k in the given function is k =3
Learn more here: https://brainly.com/question/13053668