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When graphing, its how much you rise and run
thats the formula when u have a graph and the question is asking for the slope.
you chose a point, and go down or up however many times it goes to the next point, and you run to the left however many times it gets you to get to that point.
if the line goes downward the slope is therefore negative, and if the line goes upward, the slope is therefore, positive.
m in math is meant to be slope, so
thats the formula when u have a graph and the question is asking for the slope.
you chose a point, and go down or up however many times it goes to the next point, and you run to the left however many times it gets you to get to that point.
if the line goes downward the slope is therefore negative, and if the line goes upward, the slope is therefore, positive.
m in math is meant to be slope, so
<h2>Answer : (2.67, 3.53)</h2><h2></h2>
Step to step explanation:
Confidence interval for mean, when population standard deviation is unknown:
, where = sample mean
n= sample size
s= sample standard deviation
= Critical t-value for n-1 degrees of freedom
We assume the family size is normal distributed.
Given, n= 31 , , s= 1.42 ,
Critical t value for and degree of 30 freedom
= 1.697 [By t-table]
The required confidence interval:
Hence, the 90% confidence interval for the estimate = (2.67, 3.53)
Please take the time to type in the question by yourself. I assume you use some kind of an apps to transcribe problems. It only confuses and discourages helpers. The fact that the question was reported means it is not clear.
I ASSUME the question is as follows:
Need to find P-G.
We see clearly that each expression P, G contains 3 terms, namely w^4, w^2z^2 and z^4. We will have to add/subtract like terms to get the answer.
(distribute negative sign into expression G)
(group like terms)
(add/subtract like terms)
I ASSUME the question is as follows:
Need to find P-G.
We see clearly that each expression P, G contains 3 terms, namely w^4, w^2z^2 and z^4. We will have to add/subtract like terms to get the answer.