Bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,
![\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2} \\\\\\ sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1} \\\\\\ sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)](https://tex.z-dn.net/?f=%5Cbf%202sin%5E2%282x%29%3D2%5Cimplies%20sin%5E2%282x%29%3D%5Ccfrac%7B2%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%5E2%282x%29%3D1%5Cimplies%20%5Bsin%282x%29%5D%5E2%3D1%5Cimplies%20sin%282x%29%3D%5Cpm%5Csqrt%7B1%7D%0A%5C%5C%5C%5C%5C%5C%0Asin%282x%29%3D%5Cpm%201%5Cimplies%20sin%5E%7B-1%7D%5Bsin%282x%29%5D%3Dsin%5E%7B-1%7D%28%5Cpm%201%29)
Answer:
bro try by your self
Step-by-step explanation:
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Answer:
2000
Step-by-step explanation:
S.I=P×rate / 100×time
30000×8 / 100×5/6
=2000
Answer:
The null and alternative hypotheses are:


Under the null hypothesis, the test statistic is:

Where:
is the sample mean
is the sample standard deviation
is the sample size


Now, we can find the right tailed t critical value at 0.01 significance level for df = n-1 = 10 - 1 = 9 using the t distribution table. The t critical value is given below:
Since the test statistic is less than the t critical value, we therefore, fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim that the people do better with the new edition.