Answer:
See the attached figure which represents the problem.
As shown, AA₁ and BB₁ are the altitudes in acute △ABC.
△AA₁C is a right triangle at A₁
So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)
△BB₁C is a right triangle at B₁
So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)
From (1) and (2)
∴ A₁C/AC = B₁C/BC
using scissors method
∴ A₁C · BC = B₁C · AC
Answer:
6.18% of the class has an exam score of A- or higher.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
What percentage of the class has an exam score of A- or higher (defined as at least 90)?
This is 1 subtracted by the pvalue of Z when X = 90. So
has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% of the class has an exam score of A- or higher.