Compare the following two numbers. Which statement is true?
1 answer:
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Answer:
The proportion of scores reported as 1600 is 0.0032
Step-by-step explanation:
Let X be the score for 1 random person in SAT combining maths and reading. X has distribution approximately N(μ = 1011,σ = 216).
In order to make computations, we standarize X to obtain a random variable W with distribution approximately N(0,1)
The values of the cummulative distribution function of the standard Normal random variable, lets denote it are tabulated, you can find those values in the attached file. Now, we are ready to compute the probability of X being bigger than 1600
Hence, the proportion of scores reported as 1600 is 0.0032.
33) and
, so
34) By the distance formula,
Also, EC = 8. So, .
35)
36)
Answer:
- -108.26
- -108.13
- -108.052
- -108.026
- -108
Step-by-step explanation:
A graphing calculator or spreadsheet is useful for making the repeated function evaluations required.
The average velocity on the interval [a,b] will be ...
v avg = (y(b) - y(a))/(b-a)
Here, all the intervals start at a=3, so the average velocity for the given values of t will be ...
v avg = (y(3+t) -y(3))/((3+t) -3) = (y(3+t) -y(3))/t
This can be computed for each of the t-values given. The results are shown in the attached table.
__
We note that the fractional part of the velocity gets smaller in proportion to t getting smaller. We expect it to go to 0 when t goes to 0.
The estimated instantaneous velocity is -108 ft/s.
_____
We can simplify the average velocity equation to ...
v avg = ((48(3+t) -26(t+3)^2) -(48(t+3) -26(3)^2)) / t
= (48t -26(t^2 +6t))/t
= 48 -26t -156
<em> v avg = -108 -26t</em>
Then the average velocity at t=0 is -108.
