Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
0.92 , 0.479 , 0.387 , 0.2
[See the last first digit after the decimal. I chose 0.92 since 9 is the greater than 2,3 and 4]
Answer:
Step One: Identify two points on the line.
Step Two: Select one to be (x1, y1) and the other to be (x2, y2).
Step Three: Use the slope equation to calculate slope.
Step-by-step explanation:
Answer:
At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll
Step-by-step explanation:
Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.
Let five years ago be group I X and as of now be group II Y

(Two tailed test at 5% level of significance)
Group I Group II combined p
n 270 300 570
favor 120 140 260
p 0.4444 0.4667 0.4561
Std error for differene = 
p difference = -0.0223
Z statistic = p diff/std error = -1.066
p value =0.2864
Since p value >0.05, we accept null hypothesis.
At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll
Answer:
0.5 cups or 1/2 a cup per person
Step-by-step explanation:
becasue there are 2.5 SERVED to 5 people served is a key word btw. and it would make sence for people to get less than the total