To answer this
problem, we use the binomial distribution formula for probability:
P (x) = [n!
/ (n-x)! x!] p^x q^(n-x)
Where,
n = the
total number of test questions = 10
<span>x = the
total number of test questions to pass = >6</span>
p =
probability of success = 0.5
q =
probability of failure = 0.5
Given the
formula, let us calculate for the probabilities that the student will get at
least 6 correct questions by guessing.
P (6) = [10!
/ (4)! 6!] (0.5)^6 0.5^(4) = 0.205078
P (7) = [10!
/ (3)! 7!] (0.5)^7 0.5^(3) = 0.117188
P (8) = [10!
/ (2)! 8!] (0.5)^8 0.5^(2) = 0.043945
P (9) = [10!
/ (1)! 9!] (0.5)^9 0.5^(1) = 0.009766
P (10) = [10!
/ (0)! 10!] (0.5)^10 0.5^(0) = 0.000977
Total
Probability = 0.376953 = 0.38 = 38%
<span>There is a
38% chance the student will pass.</span>
Answer: cot B
Step-by-step explanation: if i am right mark me as brainliest
Answer:
The Correct Answer is 7.071
Step-by-step explanation:
Trust me, if its wrong report me, if its right mark me brinaliest pls
PART1:
First Combination:
Pizza ($7) + Chicken Strips ($6) + Biscuits ($3) + Grapes ($4) = $20
Second Combination:
Dog Food ($13) + Bread ($3) + Crackers ($2) + Broccoli ($2) = $20
Third Combination:
Shampoo ($4) + Tissues ($3) + Pizza ($7) + Eggs ($3) + Biscuits ($3) = $20
PART 2:
First Combination:
$7.20 + $5.70 + $2.90 + $3.70 = $19.60
No, I wouldn’t have gone over the limit
Second Combination:
$13.40 + $3.50 + $2.00 + $1.90 = $20.80
Yes, I would have gone over the limit
Third Combination:
$3.50 + $2.60 + $7.20 + $2.50 + $2.90 = $18.70
No, I wouldn’t have gone over the limit
Hope this helps!!
