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nevsk [136]
2 years ago
13

A warren of rabbits has a population of 25. The population is increasing at a rate of 20% per year. How can you write an exponen

tial growth function to find the monthly growth rate?
Mathematics
1 answer:
Debora [2.8K]2 years ago
6 0

Answer:

(See explanation)

Step-by-step explanation:

The differential equation for a exponential growth is:

\frac{dn}{dt} = k\cdot n

The formula is rearranged and integrated afterwards:

\frac{dn}{n} = k \cdot dt

\int\limits_{n_{o}}^{n} {\frac{dn}{n} } = k\cdot \int\limits_{0}^{t}\, dt

\ln \frac{n}{n_{o}}  = k\cdot t

\frac{n}{n_{o}} = e^{ k\cdot t}

n = n_{o}\cdot e^{k\cdot t}

The exponential growth function can be used to find the montly growth function as follows:

\ln 1.2 = k\cdot 12

k = 0.015

The coeffcients of the expression are, respectively:

n_{o} = 25

k = 0.015

The exponential growth function with a monthly growth rate is:

n(t) = 25\cdot e^{0.015\cdot t}, where t is measured in months.

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Find the 6th term of the arithmetic sequence -4, 0, 4, 8,
bezimeni [28]

Answer:

a

Step-by-step explanation:

-4,0,4,8,12,16

4 0
2 years ago
Read 2 more answers
Use the data below, showing a summary of highway gas mileage for several observations, to decide if the average highway gas mile
Murrr4er [49]

Answer:

Step-by-step explanation:

Hello!

You need to test at 1% if the average highway gas mileage is the same for three types of vehicles (midsize cars, SUV's and pickup trucks) to compare the average values of the three groups altogether, you have to apply an ANOVA.

                n  |  Mean |  Std. Dev.

Midsize  | 31 |  25.8   |  2.56

SUV’s     | 31 |  22.68 |  3.67

Pickups  | 14 |  21.29  |  2.76

Be the study variables :

X₁: highway gas mileage of a midsize car

X₂: highway gas mileage of an SUV

X₃: highway gas mileage of a pickup truck.

Assuming these variables have a normal distribution and are independent.

The hypotheses are:

H₀: μ₁ = μ₂ = μ₃

H₁: At least one of the population means is different.

α: 0.01

The statistic for this test is:

F= \frac{MS_{Treatment}}{MS_{Error}}~F_{k-1;n-k}

Attached you'll find an ANOVA table with all its components. As you see, to manually calculate the statistic you have to determine the Sum of Squares and the degrees of freedom for the treatments and the errors, next you calculate the means square for both and finally the test statistic.

<u>For the treatments:</u>

The degrees of freedom between treatments are k-1 (k represents the amount of treatments): Df_{Tr}= k - 1= 3 - 1 = 2

<u>The sum of squares is: </u>

SSTr: ∑ni(Ÿi - Ÿ..)²

Ÿi= sample mean of sample i ∀ i= 1,2,3

Ÿ..= grand mean, is the mean that results of all the groups together.

So the Sum of squares pf treatments SStr is the sum of the square of difference between the sample mean of each group and the grand mean.

To calculate the grand mean you can sum the means of each group and dive it by the number of groups:

Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (25.8+22.68+21.29)/3 = 23.256≅ 23.26

SS_{Tr}= (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (25.8-23.26)² + (22.68-23.26)² + (21.29-23.26)²= 10.6689

MS_{Tr}= \frac{SS_{Tr}}{Df_{Tr}}= \frac{10.6689}{2}= 5.33

<u>For the errors:</u>

The degrees of freedom for the errors are: Df_{Errors}= N-k= (31+31+14)-3= 76-3= 73

The Mean square are equal to the estimation of the variance of errors, you can calculate them using the following formula:

MS_{Errors}= S^2_e= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2+(n_3-1)S^2_3}{n_1+n_2+n_3-k}= \frac{(30*2.56^2)+(30*3.67^2)+(13*2.76^2)}{31+31+14-3} = \frac{695.3118}{73}= 9.52

<u>Now you can calculate the test statistic</u>

F_{H_0}= \frac{MS_{Tr}}{MS_{Error}} = \frac{5.33}{9.52}= 0.559= 0.56

The rejection region for this test is <em>always </em>one-tailed to the right, meaning that you'll reject the null hypothesis to big values of the statistic:

F_{k-1;N-k;1-\alpha }= F_{2; 73; 0.99}= 4.07

If F_{H_0} ≥ 4.07, reject the null hypothesis.

If F_{H_0} < 4.07, do not reject the null hypothesis.

Since the calculated value is less than the critical value, the decision is to not reject the null hypothesis.

Then at a 1% significance level you can conclude that the average highway mileage is the same for the three types of vehicles (mid size, SUV and pickup trucks)

I hope this helps!

4 0
2 years ago
Two large numbers of the Fibonacci sequence are F51 = 20,365,011,074 and F52 = 32,951,280,099.What is the approximate value of t
Katen [24]
With the Fibonacci Sequence, we keep adding the 2 previous numbers
1
+1
=2

then
1
+2
=3

2 +
3
=5

And we can continue
<span> <span> <span> 8 </span> <span>
13
</span> <span> 21
</span> <span> 34 </span> <span>
55 </span> <span>
89 </span>
<span> 144 </span>
<span> 233 </span>
<span> 377 </span> </span> </span>

If we take a Fibonacci number and divide it by the PREVIOUS Fibonacci number (For example 377 / 233) we get:
<span> <span> <span> 1.61802575107296 </span> </span> </span>

This is something known as the phi ratio, which equals (1 + sq root(5)) / 2
or <span> <span> <span> 1.6180339887499</span></span></span>
The further we carry out the Fibonacci sequence, the closer the division of (Fibonnaci Number "n") /(Fibonnaci Number "n-1") gets to be
(1 + sq root(5)) / 2
So, I would say that F52 / F51 would  <span><span><span>equal 1.6180339887499 or be very close to it.
Look up the "Golden Ratio" in wkipedia


</span></span></span>
5 0
2 years ago
Read 2 more answers
Glen Davis is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one,
Serjik [45]

Answer:

0.00457 or 0.457%

Step-by-step explanation:

Glen Davis makes a free throw 74% of the time, meaning he misses 26% of the time. Four him to miss four in a row, it would require him to miss his first throw and his second throw and his third throw and his fourth throw. I highlight and because it is a helpful tip to remember that AND situations require multiplication while OR situations require addition.

So in this question we multiply the probabilities of him missing.

P(MMMM) = \frac{26}{100}* \frac{26}{100} *\frac{26}{100}* \frac{26}{100}  = 0.00457

This answer has been rounded off to three significant figures

4 0
2 years ago
Read 2 more answers
Adhesive tape made from fabric has a tensile strength of not less than 20.41 kg/2.54 cm of width. Reduce these quantities to gra
OlgaM077 [116]

Answer:

\frac{20,410 \text{ grams}}{254\text{ mm}}

Step-by-step explanation:

We have been given that adhesive tape made from fabric has a tensile strength of not less than 20.41 kg/2.54 cm of width. We are asked to reduce these quantities to grams and millimeters.

We know 1 kg equals 1000 grams and 1 cm equals 10 mm.

\frac{20.41\text{ kg}}{\text{2.54 cm}}

\frac{20.41\text{ kg}}{\text{2.54 cm}}\times \frac{\text{1 cm}}{\text{10 mm}}

\frac{20.41\text{ kg}}{2.54\times\text{10 mm}}

\frac{20.41\text{ kg}}{254\text{ mm}}

\frac{20.41\text{ kg}}{254\text{ mm}}\times \frac{\text{1000 grams}}{\text{1 kg}}

\frac{20.41\times \text{1000 grams}}{254\text{ mm}}

\frac{20,410 \text{ grams}}{254\text{ mm}}

Therefore, our required measurement would be \frac{20,410 \text{ grams}}{254\text{ mm}}.

8 0
2 years ago
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