If sin y° = 9 divided by c. and tan y° = 9 divided by d. ,what is the value of cos y°?
2 answers:
Answer: The fraction d/c
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Explanation:
Check out the attached image. I've labeled the three sides of the triangle based on the reference angle y. The opposite side is 9, the adjacent is d, and the hypotenuse is c.
How did I get those values? Well
sine = opposite/hypotenuse = 9/c
tangent = opposite/adjacent = 9/d
matching terms we see that opposite = 9, adjacent = d, hypotenuse = c
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Now we compute the ratio for cosine
cosine = adjacent/hypotenuse
cos(y) = d/c
We don't have enough info to know the numeric value of d/c
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The probability plot of this distribution shows that it is approximately normally distributed..
Check explanation for the reasons.
Step-by-step explanation:
The complete question is attached to this solution provided.
From the cumulative probability plot for this question, we can see that the plot is almost linear with no points outside the band (the fat pencil test).
The cumulative probability plot for a normal distribution isn't normally linear. It's usually fairly S shaped. But, when the probability plot satisfies the fat pencil test, we can conclude that the distribution is approximately linear. This is the first proof that this distribution is approximately normal.
Also, the p-value for the plot was obtained to be 0.541.
For this question, we are trying to check the notmality of the distribution, hence, the null hypothesis would be that the distribution is normal and the alternative hypothesis would be that the distribution isn't normal.
The interpretation of p-valies is that
When the p-value is greater than the significance level, we fail to reject the null hypothesis (normal hypothesis) and but if the p-value is less than the significance level, we reject the null hypothesis (normal hypothesis).
For this distribution,
p-value = 0.541
Significance level = 0.05 (Evident from the plot)
Hence,
p-value > significance level
So, we fail to reject the null or normality hypothesis. Hence, we can conclude that this distribution is approximately normal.
Hope this Helps!!!
