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Stels [109]
3 years ago
10

A biologist studied the populations of common guppies and Endler's guppies over a 6-year period. the biologist modeled the popul

ations, in tens of thousands, with the following polynomials where x is time, in years.
common guppies: 3.1x^2+6x+0.3

Endler's guppies: 4.2x^2-5.2x+1

what polynomial models the total number of common and Endler's guppies
Mathematics
1 answer:
Mice21 [21]3 years ago
3 0
We can say that total number of guppies would simply be the sum of two polynomials.
To get our answers we need to add those two polynomials. We add polynomials by adding their like terms. Like terms are those that have the same exponent.
3.1x^2+6x+0.3+4.2x^2-5.2x+1=\\=(3.1x^2+4.2x^2)+(6x-5.2x)+(0.3+1)=\\
=7.3x^2+0.8x+1.3
So the total amount of guppies would be modeled using this polynomial:
7.3x^2+0.8x+1.3
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Quadrilateral ABCD ​ is inscribed in this circle.
Sedbober [7]
The answer to this is 65. 

Remember that the angles opposite of each other equal to 180. So a+c=180 and b+d=180. 

First, we solve for x. Since we know that b and d equal to 180, we subtract 148 from 180. 
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Then we plug that in the expression for angle a. 
2x+1 becomes 2(32)+1
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3 years ago
A road bike has a wheel diameter of 622 mm. What is the circumference of the wheel? Use 3.14 for π?
Mkey [24]
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3 0
3 years ago
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Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,
Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0
3 years ago
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Shtirlitz [24]
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6 0
2 years ago
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how much would $300 invested at 7% interest be worth after 4 years? Round your answer to the nearest cent
Sav [38]
It would be 840cents

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