You want to isolate the x-term from the constant term, so you can subtract x/3 and add 10. This gives you
... 4/9x -10 -x/3 +10 > x/3 -12 -x/3 +10
... 1/9x > -2 . . . . . . collect terms
Now, you can multiply by 9 to see the condition on x.
... 9(1/9x) > -2(9)
... x > -18
On the x-y plane, the graph of this will be a dashed line at x=-18, and the half-plane to the right of that line will be shaded.
On a number line, there will be an open circle at x=-18, and the number line to the right of that circle will be marked (bold, colored, shaded, whatever).
Answer:
1850
Step-by-step explanation:
The problem is asking you how many kids (predicted by the superintendent) will buy lunches for 1-2 days. Since you know the percentage predicted and the student amount, just multiply as a decimal (convert by putting the number over 100 as a fraction.)
5000*0.37=1850
Answer:
3 is the answer
Step-by-step explanation:
5.........4............3..............2...............1
so 3 follows 4
Your anwser is quit school ):73
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z=
where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=
=-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.
