Answers:
- Domain is (-4, 3]
- Range is (-5, 5]
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Explanation:
The domain is the set of allowed x input values, aka the set of all allowed x coordinates of the points. We see that 
. It might help to draw vertical lines through the endpoints until you reach the x axis. Note the open hole at x = -4 to indicate we do not include this as part of the domain (hence the lack of "or equal to" for the first inequality sign).
The interval then can be condensed into the shorthand form (-4, 3] which is the domain in interval notation.
It says: x is between -4 and 3. It can't equal -4 but it can equal 3.
So the use of parenthesis versus square brackets tells the reader which endpoint is included or not.
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The range describes all possible y outputs. We see that y = 5 is the largest it gets and y = -5 is the lower bound. It might help to draw horizontal lines through the endpoints until you reach the y axis. The open hole means -5 is not part of the range.
The range as a compound inequality is 
. This condenses into the shorthand of (-5, 5] which is the range in interval notation.
Verbally, the range is the set of y values such that y is between -5 and 5. It can't equal -5 but it can equal 5.
150 grams because 5 oz is equal to 141 grams
Answer:
dependent events since P(A and B) is not equal to P(A) * P(B)
Step-by-step explanation:
According to the Question,
- Given, The probability that Jane will go to a ballgame (event A) on a Monday is 0.73, and the probability that Kate will go to a ballgame (event B) the same day is 0.61. The probability that Kate and Jane both go to the ballgame on Monday is 0.52.
Thus, The events A, B and A∩B are:
A - Jane will go to a ballgame on Monday;
B - Kate will go to a ballgame on Monday;
A∩B - Kate and Jane both go to the ballgame on Monday.
- P(A)=0.73, P(B)=0.61, P(A∩B)=0.52.
- Pr(A)⋅Pr(B) = 0.73⋅0.61 = 0.4453 ≠ 0.52
So, events A and B are dependent events since P(A and B) is not equal to P(A) * P(B)
Answer:
a. see attached
b. H(t) = 12 -10cos(πt/10)
c. H(16) ≈ 8.91 m
Step-by-step explanation:
<h3>a.</h3>
The cosine function has its extreme (positive) value when its argument is 0, so we like to use that function for circular motion problems that have an extreme value at t=0. The midline of the function needs to be adjusted upward from 0 to a value that is 2 m more than the 10 m radius. The amplitude of the function will be the 10 m radius. The period of the function is 20 seconds, so the cosine function will be scaled so that one full period is completed at t=20. That is, the argument of the cosine will be 2π(t/20) = πt/10.
The function describing the height will be ...
H(t) = 12 -10cos(πt/10)
The graph of it is attached.
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<h3>b. </h3>
See part a.
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<h3>c.</h3>
The wheel will reach the top of its travel after 1/2 of its period, or t=10. Then 6 seconds later is t=16.
H(16) = 12 -10cos(π(16/10) = 12 -10cos(1.6π) ≈ 12 -10(0.309017) ≈ 8.90983
The height of the rider 6 seconds after passing the top will be about 8.91 m.
