Answer: 81%
Step-by-step explanation:
From the question, we are informed that a student received the following test scores: 71%, 89%, 72%,
84% and 83% in 5 tests and the student wants to maintain an average of 80%.
The lowest score/grade they can receive on the next test to maintain at least an 80% average first thus:
First, to make it easy we can remove the percent sign. Then we multiply 80 by 6 since we're calculating for 6 tests scores. This will be:
= 80 × 6
= 480
We then add all the 5 test scores. This will be:
= 71 + 89 + 72 + 84 + 83
= 399
We then subtract the values gotten. This will be:
= 480 - 399
= 81
This means the student must get at least 81%
13. 1,6
18. -15,12
19. -7, 1/2
=(3a-4a-3a)+(2b+6b-2b)+(-7c+9c-7c)
=-4a+6b-5c #
In a throw of 2 fair dice, there are 6*6=36 equiprobability outcomes.
To get a sum of 5, there are 4 ways, (1,4),(2,3),(3,2),(4,1) with probability of 4/36=1/9
To get at least one 5, there are 6+6-1=11 outcomes (note (5,5) has been counted in both, so subtracted from sum). The probability is 11/36
Since the two events are mutually exclusive (once we have a five, the sum can no longer be 5), we can add the probabilities to get the probability of one event or the other.
P(sum of 5 OR at least one 5)=1/9+11/36=4/36+11/36=15/36=5/9
Answer:
The point estimate for this problem is 0.48.
Step-by-step explanation:
We are given that a University wanted to find out the percentage of students who felt comfortable reporting cheating by their fellow students.
A survey of 2,800 students was conducted and the students were asked if they felt comfortable reporting cheating by their fellow students. The results were 1,344 answered "Yes" and 1,456 answered "no".
<em>Let </em>
<em> = proportion of students who felt comfortable reporting cheating by their fellow students</em>
<u></u>
<u>Now, point estimate (</u><u>) is calculated as;</u>
where, X = number of students who answered yes = 1,344
n = number of students surveyed = 2,800
So, Point estimate () = 
= <u>0.48 or 48%</u>
