Answer:
9 servings
There should be an answer to that. Answer: There are 9 servings. step- by- step Explanation: We would first divide 22.5/2.5. This will give us the amount of time 2.5 will go into 22.5, providing us with the amount of servings
Step-by-step explanation:
Complete Question
Assume that random guesses are made for 7 multiple-choice questions on a test with 5
choices for each question, so that there are n=7 trials, each with probability of success (correct) given by p=0.20.
Find the probability of no correct answers.
Answer:
The probability is ![P(X = 0 ) = 0.210](https://tex.z-dn.net/?f=P%28X%20%3D%200%20%29%20%3D%200.210)
Step-by-step explanation:
From the question we are told that
The number of trial is n = 7
The probability of success is p = 0.20
Generally the probability of failure is
![q = 1- 0.20](https://tex.z-dn.net/?f=q%20%3D%20%201-%200.20)
![q = 0.80](https://tex.z-dn.net/?f=q%20%3D%200.80)
Given that this choices follow a binomial distribution as there is only two probabilities i.e success or failure
Then the probability is mathematically represented as
![P(X = 0 ) = \left 7} \atop {}} \right. C_0 * (0.2)^{0} * (0.8)^{7- 0}](https://tex.z-dn.net/?f=P%28X%20%3D%200%20%29%20%3D%20%5Cleft%207%7D%20%5Catop%20%7B%7D%7D%20%5Cright.%20C_0%20%2A%20%20%280.2%29%5E%7B0%7D%20%2A%20%20%280.8%29%5E%7B7-%200%7D)
Here ![\left 7} \atop {}} \right. C_0 = 1](https://tex.z-dn.net/?f=%5Cleft%207%7D%20%5Catop%20%7B%7D%7D%20%5Cright.%20C_0%20%3D%20%201)
=> ![P(X = 0 ) = 1 * 1* (0.8)^{7- 0}](https://tex.z-dn.net/?f=P%28X%20%3D%200%20%29%20%3D%201%20%2A%20%201%2A%20%20%280.8%29%5E%7B7-%200%7D)
=> ![P(X = 0 ) = 0.210](https://tex.z-dn.net/?f=P%28X%20%3D%200%20%29%20%3D%200.210)
Answer:
Step-by-step explanation:
AREA: 42 cm^2
h = 14 cm
area of a triangle formula:
![\frac{height * base}{2} \\\frac{14 * base}{2} = 42](https://tex.z-dn.net/?f=%5Cfrac%7Bheight%20%2A%20base%7D%7B2%7D%20%5C%5C%5Cfrac%7B14%20%2A%20base%7D%7B2%7D%20%3D%2042)
(14)(base) = 42 * 2
(14)(base) /14 = 84 /14
base = 6 cm
Answer:
<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>
By De morgan's law
![(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq 1-P(A)+1-P(B)\\\\-P(A\cap B)\leq 1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1](https://tex.z-dn.net/?f=%28A%5Ccap%20B%29%5E%7Bc%7D%3DA%5E%7Bc%7D%5Ccup%20B%5E%7Bc%7D%5C%5C%5C%5CP%28%28A%5Ccap%20B%29%5E%7Bc%7D%29%3DP%28A%5E%7Bc%7D%5Ccup%20B%5E%7Bc%7D%29%5Cleq%20P%28A%5E%7Bc%7D%29%2BP%28B%5E%7Bc%7D%29%20%5C%5C%5C%5C1-P%28A%5Ccap%20B%29%5Cleq%20%20P%28A%5E%7Bc%7D%29%2BP%28B%5E%7Bc%7D%29%20%5C%5C%5C%5C1-P%28A%5Ccap%20B%29%5Cleq%20%201-P%28A%29%2B1-P%28B%29%5C%5C%5C%5C-P%28A%5Ccap%20B%29%5Cleq%20%201-P%28A%29-P%28B%29%5C%5C%5C%5CP%28A%5Ccap%20B%29%5Cgeq%20P%28A%29%2BP%28B%29-1)
which is Bonferroni’s inequality
<h3>Result 1: P (Ac) = 1 − P(A)</h3>
Proof
If S is universal set then
![A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)](https://tex.z-dn.net/?f=A%5Ccup%20A%5E%7Bc%7D%3DS%5C%5C%5C%5CP%28A%5Ccup%20A%5E%7Bc%7D%29%3DP%28S%29%5C%5C%5C%5CP%28A%29%2BP%28A%5E%7Bc%7D%29%3D1%5C%5C%5C%5CP%28A%5E%7Bc%7D%29%3D1-P%28A%29)
<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>
Proof:
If S is a universal set then:
![A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)](https://tex.z-dn.net/?f=A%5Ccup%28B%5Ccap%20A%5E%7Bc%7D%29%3D%28A%5Ccup%20B%29%20%5Ccap%20%28A%5Ccup%20A%5E%7Bc%7D%29%5C%5C%3D%28A%5Ccup%20B%29%20%5Ccap%20S%5C%5CA%5Ccup%28B%5Ccap%20A%5E%7Bc%7D%29%3D%28A%5Ccup%20B%29)
Which show A∪B can be expressed as union of two disjoint sets.
If A and (B∩Ac) are two disjoint sets then
B can be expressed as:
![B=B\cap(A\cup A^{c})\\](https://tex.z-dn.net/?f=B%3DB%5Ccap%28A%5Ccup%20A%5E%7Bc%7D%29%5C%5C)
If B is intersection of two disjoint sets then
![P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)](https://tex.z-dn.net/?f=P%28B%29%3DP%28B%5Ccap%28A%29%2BP%28B%5Ccup%20A%5E%7Bc%7D%29%5C%5CP%28B%5Ccup%20A%5E%7Bc%7D%3DP%28B%29-P%28B%5Ccap%20A%29)
Then (1) becomes
![P(A\cup B) =P(A) +P(B)-P(A\cap B)\\](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%20%3DP%28A%29%20%2BP%28B%29-P%28A%5Ccap%20B%29%5C%5C)
<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>
Proof:
If A and B are two disjoint sets then
![A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\](https://tex.z-dn.net/?f=A%3DA%5Ccap%28B%5Ccup%20B%5E%7Bc%7D%29%5C%5CA%3D%28A%5Ccap%20B%29%20%5Ccup%20%28A%5Ccap%20B%5E%7Bc%7D%29%5C%5CP%28A%29%3DP%28A%5Ccap%20B%29%20%2B%20P%28A%5Ccap%20B%5E%7Bc%7D%29%5C%5C)
<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>
Proof:
If B is subset of A then all elements of B lie in A so A ∩ B =B
where A and A ∩ Bc are disjoint.
![P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})](https://tex.z-dn.net/?f=P%28A%29%3DP%28B%5Ccup%20%28%20A%5Ccap%20B%5E%7Bc%7D%29%29%5C%5C%5C%5CP%28A%29%3DP%28B%29%2BP%28%20A%5Ccap%20B%5E%7Bc%7D%29)
From axiom P(E)≥0
![P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)](https://tex.z-dn.net/?f=P%28%20A%5Ccap%20B%5E%7Bc%7D%29%5Cgeq%200%5C%5C%5C%5CP%28A%29-P%28B%29%3DP%28%20A%5Ccap%20B%5E%7Bc%7D%29%5C%5CP%28A%29%3DP%28B%29%2BP%28A%5Ccap%20B%5E%7Bc%7D%29%5Cgeq%20P%28B%29)
Therefore,
P(A)≥P(B)