An educational technology company advertises English language fluency after one year of using their online language courses. Researchers test their curriculum by randomly selecting 100 new students and following their project after one year of using the curriculum. They test the students using Cambridge English Test and assign each student a score between 1 to 25. They consider scores below 15 to indicate a lack of fluency: H0:μ=15 HA:μ≥15 Based on the data they conclude that the evidence is not strong enough to conclude that the students are fluent in English. What type of error is possible in this situation? Type I Type II

In this case, as the null hypothesis is not rejected, the only error possible is Type II error. That means that the researchers conclude that the students are not fluent when they really are fluent in English.

Step-by-step explanation:

They test an hypothesis to test the claim that students are strong in english.

The null and alternative hypothesis are:

If they reject the null hypothesis, that means that the studens are fluent in English.

With the data they have, they conclude that the evidence is not strong enough to conclude that the students are fluent in English.

That means they failed to reject the null hypothesis.

The type of errors are:

Type I error: reject a true null hypothesis.

Type II error: failed to reject a false null hypothesis.

In this case, as the null hypothesis is not rejected, the only error possible is Type II error. That means that the researchers conclude that the students are not fluent when they really are fluent in English.

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18