Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
I think it’s a carrot I just need points
Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:
In which:
- is the sample proportion.
In this problem, we have a 95% confidence level, hence, z is the value of Z that has a p-value of , so the critical value is z = 1.96.
The sample size and the estimate are given by:
Hence:
The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).
More can be learned about the z-distribution at brainly.com/question/25890103