The probability is found from a suitable table or probability calculator.
p(z < 1.3) ≈ 0.9031995154
Answer:
V=pi r² h/3
Step-by-step explanation:
V=volume
R=radius
H=height
Either B or D I’m not really sure but I know it’s one of this answers
Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
Answer:
no. see below
Step-by-step explanation:
This much of Josh's working is correct:
x^2-6x=7
x^2-6x+9=7+9
(x-3)^2=16
At this point Josh apparently overlooked the fact that he needed to take the square root of both sides of the equation. Had he done that, he would have ...
x -3 = ±4
x = 3+4 . . . or . . . 3 -4
x = 7 or -1
_____
Josh reported values of x that would match ...
x -3 = ±16
He <em>violated the equal sign</em> by taking the square root on the left, and multiplying by ±1 on the right. Doing different operations on the two sides of the equation will mean the value of x is changed to something other than what you're looking for.
