Amath teacher claimed that the average grade of the students in her algebra 2 classes this year would be equal to the average grade of the same students in algebra 1 classes last year. the average grade of last year’s algebra 1 students was an 87%. in a random sample of 30 current algebra 2 students, the average grade was 86%, with a standard deviation of 3%. is there enough evidence to reject the teacher's claim? enter your answer as a z-value and either yes or no, separated by a comma, like this: -4.25, yes

These questions should be asked more clearly.

If 3% is the standard deviation of the average then a 1% decrease in score is a z=-1/3, which is small in absolute value, so we can't reject the claim. In this interpretation we'd -0.33, no.

But if we'll assume 3% is the standard deviation of the individual samples, it's a different story. We use percent as our unit. The standard deviation of the average is

.

So we get a z (really a t) here of

That's kinda borderline, a one sided t test with 29 dfs will give about the same probability as the normal distribution for a tail z=1.8, p=.036 aka p=3.6%. Typically we'd choose a 1% or 5% rejection threshold before we started; this one is in between.

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

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step-by-step explanation: