Compare using >, <, or =. 75 minutes_7 hours
2 answers:
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Answer:
459
Step-by-step explanation:
The "long division method" algorithm for square root makes use of the relation described by the square of a binomial.
(a +b)² = a² +2ab +b² = a² +b(2a +b)
<h3>Steps</h3>The value for which the root is desired is written with digits marked off in pairs either side of the decimal point.
The initial digit of the root is the integer part of the square root of the most-significant pair. Here that is floor(√21) = 4. This is shown in the "quotient" spot above the leftmost pair. The square of this value is subtracted, and the next pair brought down for consideration. Here, that means the next "dividend" is 506.
The next "divisor" will be 2 times the "quotient" so far, with space left for a least-significant digit. Here, that means 506 will be divided by 80 + some digits. As in regular long division, determining the missing digit involves a certain amount of "guess and check." We find that the greatest value 'b' that will give b(80+b) ≤ 506 is b=5. This is the next "quotient" digit and is placed above the "dividend" pair 06. The product 5(85) = 425 is subtracted from 506, and the next "dividend" pair is appended to the result. This makes the next "dividend" equal to 8181.
As in the previous step, the next "divisor is 2 times the quotient so far: 2×45 = 90, with space left for the least significant digit. 8181 will be divided by 900-something with a "quotient" of 9. So, we subtract the product 9(909) = 8181 from the "dividend" 8181 to get the next "dividend." That result is zero, so we're finished.
The root found here is 459.
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<em>Additional comment</em>
In practice, roots are often computed using iterative methods, with some function providing a "starter value" for the iteration. Some iterative methods can nearly double the number of good significant digits in the root at each iteration.
Using this "long division method," each "iteration" adds a single significant digit to the root. Its advantage is that it always works, and is generally suitable for finding roots by hand. Once the number of root digits begins to get large, the "divisor" starts to be unwieldy.
Answer:
Step-by-step explanation:
Hello!
The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.
Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.
Be:
X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)
n₁= 12
X[bar]₁= 20.50 hs
S₁²= 199.76hs² (S₁= 14.13hs)
X₂: Battery duration between charges when the primary usage of the phone is internet applications.
n₂= 10
X[bar]₂= 8.50
S₂²= 33.29hs² (S₂= 5.77hs)
Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)
The parameters of interest are σ₁² and σ₂²
a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1
The hypotheses are:
H₀: σ₁²/σ₂² ≤ 1
H₁: σ₁²/σ₂² > 1
There is no level of significance listed so I've chosen α: 0.05
b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:
S²= *[∑X²-(∑X)²/n]
For the first sample:
n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36
S₁²= *[7240.36-(246)²/12]= 199.76hs²
S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs
For the second sample:
n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12
S₂²= *[1022.12-(85)²/10]= 33.2911hs²
S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs
c)
For this hypothesis test, the statistic to use is a Snedecors F:
This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:
The rejection region is then F ≥ 3.10
p-value: 0.006
Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.
Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.
I hope this helps!
