A satellite circles the earth in such a manner that it is y miles from the equator (north or south, height from the surface no
t considered) t minutes after its launch, where y is given by the equation below.
(picture)
At what times t in the interval , the first 4 hr, is the satellite mi north of the equator?
(picture)
At what times t in the interval , the first 4 hr, is the satellite mi north of the equator?
1 answer:
9514 1404 393
Answer:
t ∈ {19, 85, 109, 175, 199}
or
t ∈ {40, 64, 130, 154, 220}
Step-by-step explanation:
A graphing calculator relieves the tedium of solving for t. The attachments show the times the satellite is 4000 miles from the equator. Since y may be either north or south, the satellite may be 4000 north of the equator for y = 4000 or for y = -4000.
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If y is miles <em>north</em> of the equator, then ...
4000 = 6000cos(z) . . . . . where z=(π/45)(t-7)
z = arccos(4000/6000) ≈ 0.84107 radians
There is another solution at 2π -0.84107 radians, about 5.44212 radians
The corresponding time values are ...
t = 7 +(45/π)(z) ≈ 19.047 and 84.953 . . . . minutes
The period of the function is 90 minutes, the times will be the above times and at 90-minute intervals after each.
The satellite is 4000 miles north of the equator at ...
t ≈ 19, 85, 109, 175, 199 . . . minutes after launch
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The solution is similar if y represents miles <em>south</em> of the equator. Then we have ...
-4000 = 6000cos(z)
z = arccos(-4000/6000) ≈ 2.30052 radians (and 3.98266 radians)
The corresponding time values are ...
t = 7 +(45/π)(z) ≈ 39.952 and 64.047 . . . . minutes
Again, considering the period, the satellite is 4000 miles north of the equator at ...
t ≈ 40, 64, 130, 154, 220 . . . minutes after launch
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<em>Additional comment</em>
We might usually think of y as being positive in the northerly direction. However, in this age of political correctness and bias sensitivity, we have to consider that "y is north or south" means exactly that. So, we have also shown the solutions when y is positive south of the equator.
Usually, we prefer that the variables are defined unambiguously.
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Complete question :
Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.
What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.
Answer:
0.56 ; 0.60
Step-by-step explanation:
From The attached Venn diagram :
C = attend college ; J = has a job
P(C) = (35+45)/300 = 80/300 = 8/30
P(J) = (30+45)/300 = 75/300 = 0.25
P(C n J) = 45 /300 = 0.15
1.)
P(J | C) = P(C n J) / P(C)
P(J | C) = 0.15 / (8/30)
P(J | C) = 0.5625 = 0.56
2.)
P(C | J) = P(C n J) / P(J)
P(C | J) = 0.15 / (0.25)
P(C | J) = 0.6 = 0.60
