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Troyanec [42]
2 years ago
15

The number of rainbow smelt in Lake Michigan had an average rate of change of −19.76 per year between 1990 and 2000. The bloater

fish population had an average rate of change of −92.57 per year during the same time. If the initial population of rainbow smelt was 227 and the initial population of bloater fish was 1,052, after how many years were the two populations equal? The linear function that models the population of rainbow smelt is y1 = −19.76x + 227, where x = the years since 1990 and y1 = the number of rainbow smelt. The linear function that models the population of bloater fish is y2 = . The linear equation that determines when the two populations were equal is . The solution is x =
Mathematics
2 answers:
mojhsa [17]2 years ago
5 0
Let x =  number of years between 1990 and  2000.

Rainbow smelt:
Initial population = 227
Average yearly rate of change = -19.76
The linear function that models the population is
y₁ = 227 - 19.76x

Bloater fish
Initial population = 1052
Average yearly rate of change = -92.57
The linear function that models the population is
y₂ = 1052 - 92.57x

When the two populations are equal, then y₁ = y₂.
That is
227 - 19.76x = 1052 - 92.57x
(92.57 - 19.76)x = 1052 - 227
72.81x = 825
x = 11.33 years
   = 11 years, 0.33*12 months
   = 11 years, 4 months
Add x to 1990 to obtain the year April 2011.

Answer:
x = 11.33 years.
This occurs in April 2001.
Oduvanchick [21]2 years ago
4 0
X= 11.33

That's hopefully gonna help you a bit
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Answer:

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Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. 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)  

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We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

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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{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

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

df=n-1=1700-1=1699  

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

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

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Answer:

option C. Angle BTZ Is-congruent-to Angle BUZ

Step-by-step explanation:

Point Z is equidistant from the vertices of triangle T U V

So, ZT = ZU = ZV

When ZT = ZU  ∴ ΔZTU is an isosceles triangle ⇒ ∠TUZ=∠UTZ

When ZT = ZV  ∴  ΔZTV is an isosceles triangle ⇒ ∠ZTV=∠ZVT

When ZU = ZV  ∴ ΔZUV is an isosceles triangle ⇒ ∠ZUV=∠ZVU

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Answer: 999 games

Step-by-step explanation:

There are many ways to illustrate the rooted tree model to calculate the number of games that must be played until only one player is left who has not lost.

We could go about this manually. Though this would be somewhat tedious, I have done it and attached it to this answer. Note that when the number of players is odd, an extra game has to be played to ensure that all entrants at that round of the tournament have played at least one game at that round. Note that there is no limit on the number of games a player can play; the only condition is that a player is eliminated once the player loses.

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We could also use the formula for rooted trees to calculate the number of games that would be played.

i=\frac{l - 1}{m - 1}

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Answer:

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Step-by-step explanation:

(This exercise has been presented in Spanish and for that reason explanation will be held in Spanish)

La cantidad de botellas de 0,25 litros que se llenan con 49 litros de yogurt es:

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La cantidad remanente de yogurt es:

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Ahora, la cantidad de botellas de 0,5 litros que se llenan con el volumen remanente es:

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