Verify that [tan(theta) + cot(theta)]^2 = sec^2(theta) + csc^2(theta)
1 answer:
(tanθ + cotθ)² = sec²θ + csc²θ
<u>Expand left side</u>: tan²θ + 2tanθcotθ + cot²θ
<u>Evaluate middle term</u>: 2tanθcotθ = = 2
⇒ tan²θ + 2+ cot²θ
= tan²θ + 1 + 1 + cot²θ
<u>Apply trig identity:</u> tan²θ + 1 = sec²θ
⇒ sec²θ + 1 + cot²θ
<u>Apply trig identity:</u> 1 + cot²θ = csc²θ
⇒ sec²θ + csc²θ
Left side equals Right side so equation is verified
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A
using the Cosine rule in ΔSTU
let t = SU, s = TU and u = ST, then
t² = u² + s² - (2us cos T )
substitute the appropriate values into the formula
t² = 5² + 9² - (2 × 5 × 9 × cos68° )
= 25 + 81 - 90cos68°
= 106 - 33.71 = 72.29
⇒ t = ≈ 8.5 in → A
Answer:
z=-3.25
Step-by-step explanation:
1) Data given and notation
represent the mean production for the sample
represent the sample standard deviation for the sample
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
Part a: State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean production is different from 5.5 tons, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the info given we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
Since this is a two side test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the production its significant different compared to the desired percentage of SiO2 of 5.5 at 5% of signficance.