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rodikova [14]
3 years ago
14

There are 20 parrots at the animal sanctuary. Their population is increasing at a rate of 15% per year. There are also 24 snakes

at the sanctuary. Each year 4 more snakes are born. Part A: Write functions to represent the numbers of parrots and snakes at the sanctuary throughout the years. (4 points) Part B: How many parrots are at the sanctuary after 10 years? How many snakes are at the sanctuary after the same number of years? Assume there are no deaths to the animals during this time. (2 points) Part C: After approximately how many years is the number of parrots and snakes the same? Justify your answer mathematically. (4 points)
Mathematics
1 answer:
Mila [183]3 years ago
6 0
A.
y = 20(1.15)^x.....parrots
y = 4x + 24.....snakes

B. after 10 yrs
y = 20(1.15^10)
y = 20(4.05)
y = 81 parrots

y = 4(10) + 24
y = 40 + 24
y = 64 snakes

C.
20(1.15)^x = 4x + 24
x = 6.63 years they will be the same
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Using number sets, it is found that the correct classification of the number is:

C. rational number, integer, whole number, natural number , real number.

Numbers are classified as:

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In this problem, the number given is:

\frac{100}{5} = 20

  • Which is a whole(natural) number, thus it is also integer, rational and real, and the correct option is C.

A similar problem is given at brainly.com/question/10814303

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Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist
satela [25.4K]

Answer:

t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

p_v =2*P(t_{(9)}>1.539)=0.158  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

Step-by-step explanation:

Data given and notation  

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And the sample deviation given by:

s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And we got:

\bar X=7.55 represent the sample mean

s=0.103 represent the sample standard deviation

n=10 sample size  

\mu_o =7.5 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 7.5  

Alternative hypothesis:\mu \neq 7.5  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=10-1=9  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(9)}>1.539)=0.158  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

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