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blagie [28]
2 years ago
6

The weights of Sarus crane chicks and hooded crane chicks H (both in grams) during the 10 days following hatching can be modeled

by the functions S = –0.122t3 + 3.49t2 – 14.6t + 136 H = –0.115t3 + 3.71t2 – 20.6t + 124 where t is the number of days after hatching. According to the models, what is the difference in weight between 5-day-old Sarus crane chicks and hooded crane chicks? Round your answer to the nearest thousandth, if necessary.
Mathematics
1 answer:
Katarina [22]2 years ago
6 0

Answer:

  • 35.875 grams

Step-by-step explanation:

Given

<u>Functions:</u>

  • S(t) = –0.122t³ + 3.49t² – 14.6t + 136
  • H(t) = –0.115t³ + 3.71t² – 20.6t + 124  

<u>Finding the difference between S(5) and H(5)</u>

  • S(5) - H(5) =
  • –0.122t³ + 3.49t² – 14.6t + 136  - (–0.115t³ + 3.71t² – 20.6t + 124) =
  • (-0.122 + 0.115)*5³ + (3.49 -3.7)*5² - (14.6 -20.6)*5 + 136 - 124 = 35.875

Answer is 35.875 grams

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Elizabeth and her sister earned money shoveling snow over winter break. Altogether, they shoveled for a
Phoenix [80]

Answer:

Elizabeth spent 17.5 hours and her sister spent 12.5 hours.

Step-by-step explanation:

Let x be the number of hours that Elizabeth shoveled and y be the number of hours that her sister shoveled.

Number of hours they shoveled = 30

So, x + y = 30 or

y = 30 - x

Jenna charged $12 per sidewalk.

So, her total earnings = 12x

Her sister charged $11 per sidewalk.

So, her total earnings = 11y.

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Substitute y = 30 - x,

12x + 11(30 - x) = 347.5

12x + 330 - 11x = 347.5

x + 330 = 347.5

x = 347.5 - 330

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4 0
3 years ago
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A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point est
liq [111]

Answer:

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And for the deviation we have:

s= \sqrt{0.00132}=0.0363

And that value represent the best estimator for the population deviation since:

E(s) =\sigma

Step-by-step explanation:

For this case we have the following data:

1.48,1.45,1.54,1.52,1.52

The first step for this cae is find the sample mean with the following formula:

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

And replacing we got:

\bar X= \frac{1.48+1.45+1.54+1.52+1.52}{5} = 1.502

And now we can calculate the sample variance with the following formula:

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

And replacing we got:

s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132

And for the deviation we have:

s= \sqrt{0.00132}=0.0363

And that value represent the best estimator for the population deviation since:

E(s) =\sigma

4 0
3 years ago
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Add 12 everytime

4 0
3 years ago
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