Answer:
A = 50x^2+3
B = 75
C = 55
Step-by-step explanation:
A. f(g(x))
B. g(g(3))
C. g(f(-2))
The conclusion is that since the value of 0.028 is less than 0.05 so we reject the null hypothesis at 5% level of significance and conclude that the mean of the difference in SAT writing scores for all students who take SAT prep class is equal to 0.
<h3>What is a null hypothesis?</h3>
A null hypothesis is a hypothesis that there is no significant difference between specified populations, any observed difference being due to sampling error.
Here, the effect of a one-day SAT prep class at a 5% level of significance was assessed. Since the value of 0.028 is less than 0.05 so we reject the null hypothesis at 5% level of significance and conclude that the mean of the difference in SAT writing scores for all students who take SAT prep class is equal to 0.
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Answer:
hundred
Step-by-step explanation:
