2(m - 5) + 6PEMDASu multiply 2 into (m-5)2*m and 2*-5that would be 2m-10 + 6now subtract common variables-10 + 6 = -42m - 4
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:
Infinitely many solutions.
Step-by-step explanation:
Let's begin by carrying out the indicated multiplications, which must be done before any addition or subtraction:
2(8r+5)-3=4(4r-1)+11 becomes 16r + 10 - 3 = 16r - 4 + 11.
Subtracting 16r from both sides, we get 10 - 3 = - 4 + 11, or 7 = 7
This is always true, so we can conclude that this equation has infinitely many solutions.
Answer:
Step-by-step explanation:
Given that :
the null and the alternative hypothesis are computed as :
This is a two tailed test
This is because of the ≠ sign in the alternative hypothesis which signifies that the rejection region in the alternative hypothesis are at the both sides of the hypothesized mean difference .
Decision Rule: at the level of significance ∝ = 0 . 10
The decision rule is to reject the null hypothesis if z < - 1 . 64 and z > 1 . 64
NOTE: DURING THE MOMENT OF TYPING THIS ANSWER THERE IS A TECHNICAL ISSUE WHICH MAKES ME TO BE UNABLE TO SUBMIT THE FULL ANSWER BUT I'VE MADE SCREENSHOTS OF THEM AND THEY CAN BE FOUND IN THE ATTACHED FILE BELOW
