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sineoko [7]
3 years ago
15

Estimate square root of 90 to the nearest hundredths place

Mathematics
2 answers:
Vladimir [108]3 years ago
5 0
The method for doing this will depend on the required accuracy of your estimate.

Using linear approximation between 9^2 = 81 and 10^2 = 100, the estimate is
.. 9 +(90-81)/(100-81) = 9 +9/19 ≈ 9.47

Using one round of "Babylonian Method" on the above estimate,
.. (9.47 +90/9.47)/2 ≈ 9.486 ≈ 9.49 . . . . . . actually accurate to the hundredths place

Using derivatives and linear approximation,
.. √90 ≈ √81 +1/(2√81)*(90-81) = 9.50
or
.. √90 ≈ √100 +1/(2√100)*(90 -100) = 9.50

netineya [11]3 years ago
3 0
Assume no calculators, no paper, no pen, all in the head.

90=9*10, so 9.5 is a good approximation.
in fact, 9.5=9*10+0.562=90.25
We will use this initial estimate x0=9.5 to improve to accurate to the hundredth (and more).

Newton's method gives a better estimate by the following formula (all done in the head)
x1=x0-(x0^2-90)/(2*x0)
=9.5-(90.25-90)/(2*9.5)
=9.5-0.25/19
=9.5-0.013125   [in the head, 0.25/19=0.25/20+5%=0.125+0.0625=0.013125]
=9.486875
=9.49  (to the hundredth)

Exact value = 9.48683298... 
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c and e because statistical questions can be answered by collecting data that can have variability.

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3 years ago
A group of physical education majors was discussing the heights of female runners and whether female runners tended to be tall,
Anna11 [10]

Answer:

The 90% confidence interval for the mean µ of the population of female runners.

( 65.0328 , 66.5672)

Step-by-step explanation:

<u>Step(i)</u>

Given  A sample of 12 runners showed a sample mean height of 65.80 inches and a sample standard deviation of 1.95 inches.

Given sample size is n = 12 <30 so small sample

Given sample mean (x⁻) = 65.80 inches

sample standard deviation (S) = 1.95 inches.

<u>Step(ii)</u>

Assume the population is approximately normal.

The 90% confidence interval for the mean µ of the population of female runners.

(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )

substitute  all above interval

(65.80 - t_{\alpha } \frac{1.95}{\sqrt{12} } ,65.80 +t_{\alpha } \frac{1.95}{\sqrt{12} } )

The degrees of freedom γ=n-1 = 12-1=11

From t- table = 1.363 at 90 % 0r 0.10 level of significance

(65.80 -1.363 \frac{1.95}{\sqrt{12} } ,65.80 +1.363 \frac{1.95}{\sqrt{12} } )

on calculation , we get

(65.80 -0.7672 ,65.80 + 0.7672)

( 65.0328 , 66.5672)

8 0
3 years ago
Solve the simultaneous equations by substitution <br>5x+6y=−11 <br> 5x=2y−3
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Answer:

Step-by-step explanation:

we have

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we will multiply first equation with -1

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5x-2y=-3

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we go to our original equation and replace y=-1

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3 0
3 years ago
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Answer:

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Step-by-step explanation:

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For a limit to exist, it must be the same regardless of the direction of approach.

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Answer:

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Step-by-step explanation:

Given:

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Now, again on day 3, each of the 25 bacteria got split into additional 5 bacteria.

So, number of of bacteria on day 3 = 25\times 5 =125

Therefore, the population of bacteria on day 3 is 125.

8 0
3 years ago
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