Step-by-step explanation:
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Every week (w) Sandra will save $10. Which algebraic expression could Sandra use to calculate how much money she has saved after
1 answer:
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Answer:
So the value of height that separates the bottom 85% of data from the top 15% is 38.32.
C) 38.32
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
2) Solution to the problem
Let X the random variable that represent the waiting time in minutes of a population, and for this case we know the distribution for X is given by:
Where and
And the best way to solve this problem is using the normal standard distribution and the z score given by:
We want to find a value a, such that we satisfy this condition:
(a)
(b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.04. On this case P(Z<1.04)=0.85 and P(z>1.04)=0.15
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
So the value of height that separates the bottom 85% of data from the top 15% is 38.32.
C) 38.32
Answer:
34.43
Step-by-step explanation:
A differential of length in terms of t will be ...
dL(t) = √(x'(t)^2 +y'(t)^2)
where ...
x'(t) = 4cos(4t)
y'(t) = 7cos(7t)
The length of c(t) will be the integral of this differential on the interval [0, 2π].
Dividing that interval into 10 equal pieces means each one has a width of (2π)/10 = π/5. The midpoint of pieces numbered 1 to 10 will be ...
(π/5)(n -1/2), so the area of the piece will be ...
sub-interval area ≈ (π/5)·dL((π/5)(n -1/2))
It is convenient to let a spreadsheet or graphing calculator do the function evaluation and summing of areas.
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The attachment shows the curve c(t) whose length we are estimating (red), and the differential length function (blue) we are integrating. We use the function p(n) to compute the midpoint of the sub-interval. The sum of sub-interval areas is shown as 34.43.
The length of the curve is estimated to be 34.43.
