Answer:
4cotα=tanα
4(1/tanα)=tanα
(4/tanα)=tanα
cross multiply
=> 4=tan²α
√4=√tan²α
±2=tanα
α=arc( tan) |2|
α=63.4° ( in first quadrant)
and
α=180+63.4=243.4 in the third quadrant
since we also found a negative answer( i.e –2) then α also lies in quadrants where it gives a negative value(i.e second and fourth quadrants)
α=180–63.4=116.6° in the second quadrant
α=360–63.4=296.6 in the fourth quadrant
therefore theta( in my case, alpha) lies in all four quadrants and is equal to:
α=63.4°,243.4°,116.6°and 296.6°
Answer:
The equation is 1.96- 0.75
The answer is 1.21
Step-by-step explanation:
Hope this helps :)
OK SURE WHAT IS IT? SO I CAN ANSWER YOUR QUEST.?
Answer:
Step-by-step explanation:
Hello!
The objective is to estimate the average time a student studies per week.
A sample of 8 students was taken and the time they spent studying in one week was recorded.
4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4
n= 8
X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94
S= 1.39
Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:
X[bar] ±
* (S/√n)

6.74 ± 2.365 * (1.36/√8)
[5.6;7.88]
Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]
I hope this helps!
I don’t know I can’t help you