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3 years ago

Suppose that 40 deer are introduced in a protected wilderness area. The population of the herd

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

8 0

The horizontal asymptote of the function is the minimum number of deer in the area.

The equation of horizontal asymptote is: $\mathbf{y = 40}$
The horizontal asymptote means that, the number of deer will never be less than 40

The equation is given as:

$\mathbf{f(x) = \frac{40 + 2x}{1 + 0.05x}}$

Expand the numerator

$\mathbf{f(x) = \frac{40(1 + 0.05x)}{1 + 0.05x}}$

Cancel out the common factor

$\mathbf{f(x) = 40}$

Hence, the equation of horizontal asymptote is:

$\mathbf{y = 40}$

The horizontal asymptote means that, the number of deer will never be less than 40

Read more about horizontal asymptotes at:

brainly.com/question/4084552

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√9 is __________ number *

Lera25 [3.4K]

Answer:

√9 is ____rational______ number *

Step-by-step explanation:

√9 = √(3×3)=3 which is rational

7 0

2 years ago

Find the indicated probability. round to three decimal places. a test consists of 10

Charra [1.4K]

To answer this problem, we use the binomial distribution formula for probability:

P (x) = [n! / (n-x)! x!] p^x q^(n-x)

Where,

n = the total number of test questions = 10

x = the total number of test questions to pass = >6

p = probability of success = 0.5

q = probability of failure = 0.5

Given the formula, let us calculate for the probabilities that the student will get at least 6 correct questions by guessing.

P (6) = [10! / (4)! 6!] (0.5)^6 0.5^(4) = 0.205078

P (7) = [10! / (3)! 7!] (0.5)^7 0.5^(3) = 0.117188

P (8) = [10! / (2)! 8!] (0.5)^8 0.5^(2) = 0.043945

P (9) = [10! / (1)! 9!] (0.5)^9 0.5^(1) = 0.009766

P (10) = [10! / (0)! 10!] (0.5)^10 0.5^(0) = 0.000977

Total Probability = 0.376953 = 0.38 = 38%

There is a 38% chance the student will pass.

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3 years ago

Solve each system of equations; nonlinear algebra 2 x2 + x + y - 26 = 0 x + y = 1

BigorU [14]

Answer:

(5,-4) and (-5,6)

Step-by-step explanation:

Given:

$\left\{\begin{array}{l}x^2+x+y-26=0\\ \\x+y=1\end{array}\right.$

Solve it. First, express y in terms of x from the second equation:

$y=1-x$

Substitute it into the first equation:

$x^2+x+1-x-26=0\\ \\x^2-25=0\\ \\(x-5)(x+5)=0$

Apply zero product property:

$x-5=0\ \text{or}\ x+5=0$

So,

$x=5\ \text{or}\ x=-5$

When $x=5,$ then $y=1-5=-4$

When $x=-5,$ then $y=1-(-5)=6$

We get two solutions: (5,-4) and (-5,6)

8 0

3 years ago

For the school play 40 rows of chairs are set up there are 22 chairs in each row how many chairs are there

liq [111]

If you had one row, you would have 22 chairs.

If you had two rows, you would have 22+22=44 chairs.

If you had three rows, you would have 22+22+22 chairs, or 66 chairs.

See the pattern?

If you had 40 rows, you would add 22 40 times, or 22*40=880 chairs.

Hope this helps!

6 0

3 years ago

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Answer:

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3 years ago