a. The standard error is equal to the standard deviation divided by the square root of the sample size:
SE = (23 ppm)/√18 ≈ 5.42 ppm
b. The t statistic is given by the difference between the true and sample means, divided by the standard error:
t = (192 ppm - 180 ppm)/SE ≈ 2.21
c. The p-value is approximately 0.0204.
d. Since p < 0.05, the difference is significantly different, so we reject the null hypothesis.
e. A type I error might have occurred, since it's possible that the null hypothesis was true, but we ended up rejecting it.
Answer:
=
Step-by-step explanation:
both numbers are the same siooooooooooo
A right parenthesis ) is missing from this problem statement. I'm assuming that you meant
<span>3 1/2+( 2+4 1/2).
There are at least two ways in which you could do this problem. One would be to simply drop the parentheses entirely:
</span><span>3 1/2+ 2+4 1/2 It'd make sense to add 3 1/2 and 4 1/2 together FIRST, and then to add 2 to the result.
Or, do the work inside the parentheses FIRST: </span><span>3 1/2+( 2+4 1/2).
</span>
This simplifies the problem to <span>3 1/2+( 6 1/2).
You MUST get the same answer either way.</span>
Answer:
8⁵ = 32768
Step-by-step explanation:
Answer:
Step-by-step explanation:
The max and min values exist where the derivative of the function is equal to 0. So we find the derivative:
Setting this equal to 0 and solving for x gives you the 2 values
x = .352 and -3.464
Now we need to find where the function is increasing and decreasing. I teach ,my students to make a table. The interval "starts" at negative infinity and goes up to positive infinity. So the intervals are
-∞ < x < -3.464 -3.464 < x < .352 .352 < x < ∞
Now choose any value within the interval and evaluate the derivative at that value. I chose -5 for the first test number, 0 for the second, and 1 for the third. Evaluating the derivative at -5 gives you a positive number, so the function is increasing from negative infinity to -3.464. Evaluating the derivative at 0 gives you a negative number, so the function is decreasing from -3.464 to .352. Evaluating the derivative at 1 gives you a positive number, so the function is increasing from .352 to positive infinity. That means that there is a min at the x value of .352. I guess we could round that to the tenths place and use .4 as our x value. Plug .4 into the function to get the y value at the min point.
f(.4) = -48.0
So the relative min of the function is located at (.4, -48.0)