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sergeinik [125]
2 years ago
10

Talulah is an ecologist who studies the change in the penguin population of Antarctica over time. She observed that the populati

on decays by a factor of 8/9 every 4 months. The population of penguins can be modeled by a function, P, which depends on the amount of time, t (in months). When Talulah began the study, she observed that there were 27,000 penguins in Antarctica. Write a function that models the population of the penguins t months since the beginning of Talulah's study. P(t) =

Mathematics
2 answers:
Oksana_A [137]2 years ago
8 0

Answer:

P(t) = 27000 * (1/9)^(t/4)

Step-by-step explanation:

This problem can me modelled with an exponencial formula:

P = Po * (1+r)^t

Where P is the final value, Po is the inicial value, r is the rate and t is the amount of time.

In this problem, we have that the inicial population/value is 27000, the rate is -8/9 (negative because the population decays), and the time t is in months, so as the rate is for every 4 months, we use the value (t/4) in the exponencial.

So, our function will be:

P(t) = 27000 * (1-8/9)^(t/4)

P(t) = 27000 * (1/9)^(t/4)

astra-53 [7]2 years ago
5 0

Answer:

P(t) = 27000 * (8/9)^(t/4)

Step-by-step explanation:

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Find the mean, median and mode of the weights of the people shown. 105kg 53kg 76kg 91kg 120kg 61kg 55kg 98kg 61kg
Yuliya22 [10]
First, you need to put them in order
53kg, 55kg, 61kg, 61kg, 76kg, 91kg, 98kg, 105kg, 120kg

For mean, you add them all up and divide by the amount of numbers (9)
720/9 = 80

For median, you find the middle number (76kg)

For mode, you find the number that appears the most (61kg)

Mean: 80
Median: 76
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2 years ago
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The amount of rhubarb in the original recipe is 3 1/2 cups. Using what you know of whole numbers and what you know of fractions,
Alexxx [7]
The easiest way, I think, is to convert the mixed number into an improper fraction, then multiply by 3.
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21/2 = 10 1/2

You could also multiply the whole number by 3 and the fraction by 3, ending up with 9 3/2, but then have to convert the improper fraction into a mixed number
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then add the numbers together
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either way works, whatever is easiest for you.  
8 0
2 years ago
What is 52,634,275,309 in expanded form
DanielleElmas [232]
Fifty-two billion, six hundred and thirty-four million, two hundred and seventy-five thousand, three hundred and nine.
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A men's softball league is experimenting with a yellow baseball that is easier to see during night games. One way to judge the e
sergiy2304 [10]

Answer:

There cannot be equal errors in both and yellow has fewer errors.

Step-by-step explanation:

we can do paired t test for these two colours

H_0: \bar = \bar y\\H_a: \bar <  \bar y\\

(one tailed test)

df = 9

The data can be tabulated as follows:

Yellow white

 

5 7

2 6

6 8

7 5

2 9

5 11

3 8

8 3

4 6

9 10

t-Test: Paired Two Sample for Means  

 

Yellow white

Mean 5.1 7.3

Variance 5.877777778 5.788888889

Observations 10 10

Pearson Correlation -0.139051655  

Hypothesized Mean Difference 0  

df 9  

t Stat -1.908439275  

P(T<=t) one-tail 0.044341411  

t Critical one-tail 1.833112923  

P(T<=t) two-tail 0.088682822  

t Critical two-tail 2.262157158  

Since p value one tailed = 0.0443 and it is <0.05 our significance level, we reject null hypothesis.

There cannot be equal errors in both and yellow has fewer errors.

5 0
2 years ago
Money Spent on Road Repair A politician wishes to compare the variances of the amount of money spent for road repair in two diff
Mkey [24]

Answer:

(A) Null Hypothesis, H_0 : \sigma_1^{2} =\sigma_2^{2}      {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, H_A : \sigma_1^{2} \neq \sigma_2^{2}      {means that there is a significant difference in the variances of the amounts spent in the two counties}

(B) The F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

(C) The value of the F-test statistic is 0.611.

(D) We conclude that there is no significant difference in the variances of the amounts spent in the two counties.

Step-by-step explanation:

We are given that a politician wishes to compare the variances of the amount of money spent on road repair in two different counties.

The data are given here below;

County A              County B

s1 = $11,596         s2 = $14,837

n1 = 15                     n2 = 18

Let \sigma_1^{2} = variances of the amounts spent in County A.

\sigma_2^{2} = variances of the amounts spent in County B.

(A) So, Null Hypothesis, H_0 : \sigma_1^{2} =\sigma_2^{2}      {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, H_A : \sigma_1^{2} \neq \sigma_2^{2}      {means that there is a significant difference in the variances of the amounts spent in the two counties}

The test statistics that would be used here <u>Two-sample F-test statistics </u>distribution;

                             T.S. =  \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ F__n_1_-_1,_ n_2_-_1

where, s_1 = sample standard deviation for County A = $11,596

s_2 = sample standard deviation for County B = $14,837

n_1 = sample size for County A = 15

n_2 =  sample size for County B = 18

(B) Now at 0.05 level of significance, the F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

(C) So, <u><em>the test statistics</em></u>  =  \frac{11,596^{2}  }{14,837^{2} } \times 1  ~  F__1_4,_ 1_7

                                     =  0.611

The value of the F-test statistic is 0.611.

Now, as we can see that our test statistics lie within the range of critical values of F, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

(D) Therefore, we conclude that there is no significant difference in the variances of the amounts spent in the two counties.

7 0
2 years ago
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