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Veronika [31]
2 years ago
9

The graph shows the exponential regression model for data representing a rabbit population after x years. Which is true of the r

egression model?
The graph of the regression model is limited to whole-number values for x.

The graph of the regression model is limited to whole-number values for y.

The graph of the regression model cannot be used to approximate the population size for year 1.

The graph of the regression model can be used to predict the population size for any number of years in the future.

Mathematics
2 answers:
cupoosta [38]2 years ago
8 0

Answer:

D is the answer

Step-by-step explanation:

PLZ give me brainliest

tamaranim1 [39]2 years ago
5 0

Answer: The graph of the regression model can be used to predict the population size for any number of years in the future.

Step-by-step explanation:

Since with help of the given graph, we can say that

The graph of the regression model is applied for the every real value of x.

Thus the statement The graph of the regression model is limited to whole-number values for x is false.

Also, the range of an exponential function is always a set of real number.

Therefore, The graph of the regression model is limited to whole-number values for y is false.

Since, for 1 year, the approximate population is given,

Therefore, The graph of the regression model cannot be used to approximate the population size for year 1 is false.

Again according to the given graph we can say that we can get the population for any number of year.

Thus, The graph of the regression model can be used to predict the population size for any number of years in the future is true.


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The ratio of the number of boys to the number of girls at school is 4:5. what fraction of the students are boys?
Arisa [49]
Boys to girls is 4:5.......added = 9

so 4/9 are boys and 5/9 are girls
5 0
3 years ago
Solve the system with elimination 4x+3y=1 -3x-6y=3
Nuetrik [128]

Answer:

x = 1, y = -1

Step-by-step explanation:

If we have the two equations:

4x+3y=1 and -3x - 6y = 3, we can look at which variable will be easiest to eliminate.

y looks like it might be easy to get rid of, we just have to multiply 4x+3y=1  by 2 and y is gone (as -6y + 6y = 0).

So let's multiply the equation 4x+3y=1  by 2.

2(4x + 3y = 1)\\8x + 6y = 2

Now we can add these equations

8x + 6y = 2\\-3x-6y=3\\

------------------------

5x = 5

Dividing both sides by 5, we get x = 1.

Now we can substitute x into an equation to find y.

4(1) + 3y = 1\\4 + 3y = 1\\3y = -3\\y = -1

Hope this helped!

5 0
3 years ago
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
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Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
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ale4655 [162]
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4 0
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Read 2 more answers
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Effectus [21]

Answer:

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Step-by-step explanation:

Hope this helps and have a wonderful day!!!

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