I think it's 230 inches
Cheers :3
<span>Assuming that the particle is the 3rd
particle, we know that it’s location must be beyond q2; it cannot be between q1
and q2 since both fields point the similar way in the between region (due to
attraction). Choosing an arbitrary value of 1 for L, we get </span>
<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>
Rearranging to calculate for d:
<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207 </span>
<span>
We pick the value that is > q2 hence,</span>
d = 2.72075922005613*L
<span>d = 2.72*L</span>
Recognize that both 0.96 and 0.144 are divisible by 12:
(0.96/12) / (0.144/12) = 0.08 / 0.012. This reduces to 0.02 / 0.003, or
20/3 or approx. 6.666.
You could also begin by eliminating the decimal fractions. Mult. 0.96 and 0.144 each by 1000 results in 960/144.
Since both 960 and 144 can be divided evenly by 24, we get 40 and 6.
40/6 = 20/3, or approx. 6.666, as before.
Answer:C. (77.29, 85.71)
Step-by-step explanation:
We want to determine a 95% confidence interval for the mean test score of randomly selected students.
Number of sample, n = 25
Mean, u = 81.5
Standard deviation, s = 10.2
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean +/- z ×standard deviation/√n
It becomes
81.5 +/- 1.96 × 10.2/√25
= 81.5 +/- 1.96 × 2.04
= 81.5 +/- 3.9984
The lower end of the confidence interval is 81.5 - 3.9984 =77.5016
The upper end of the confidence interval is 81.5 + 3.9984 =85.4984
Therefore, the correct option is
C. (77.29, 85.71)
you need to reword that it doesnt even make senseStep-by-step explanation:
