Question 3 <
B0/2 pts
A friend has a 81% average before the final exam for a course. That score includes everything but
the final, which counts for 20% of the course grade.
What is the best course grade your friend can earn?
What is the minimum score would your friend would need on the final to earn a 75% for the course?
Give answers accurate to at least one decimal place.
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The best course grade your friend can earn is 0.848 = 84.8%.

The minimum score would your friend would need on the final to earn a 75% for the course is of 0.51 = 51%.

Step-by-step explanation:

This is a weighed average problem, in which we multiply each grade by its weight.

We have that:

In 80% of the course, the friend has a grade of 81%.

In the other 20%, he will have x.

What is the best course grade your friend can earn?

This will happen if he earns 100% = 1 on the final test. So

The best course grade your friend can earn is 0.848 = 84.8%.

What is the minimum score would your friend would need on the final to earn a 75% for the course?

This is x, when the grade is 0.75. So

The minimum score would your friend would need on the final to earn a 75% for the course is of 0.51 = 51%.

i pretty sure it is

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.