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ryzh [129]
3 years ago
6

Someone help me with question 6 please!

Mathematics
1 answer:
Alika [10]3 years ago
3 0
1h=60min\\1min=60s\\therefore\ 1h=3600s\to1s=\dfrac{1}{3600}h\\\\57.1s=\dfrac{57.1}{3600}=\dfrac{571}{36000}\\\\\dfrac{1mi}{57.1s}=\dfrac{1mi}{\frac{571}{36000}h}=\dfrac{36000}{571}mph\approx63mph
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Find the equivalent expression.<br><br> 3+5x−4y+2y−8x
Luba_88 [7]

Answer:

-3x - 2y + 3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Terms/Coefficients

Step-by-step explanation:

<u>Step 1: Define</u>

3 + 5x - 4y + 2y - 8x

<u>Step 2: Simplify</u>

  1. Combine like terms (x):                    -3x - 4y + 2y + 3
  2. Combine like terms (y):                    -3x - 2y + 3
8 0
3 years ago
Read 2 more answers
For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.
nordsb [41]

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

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)

<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)

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\

B can be  expressed as:

B=B\cap(A\cup A^{c})\\

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)

Then (1) becomes

P(A\cup B) =P(A) +P(B)-P(A\cap B)\\

<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})\\

<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

A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})

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})

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)

Therefore,

P(A)≥P(B)

8 0
3 years ago
Bored anyone wanna talk? or do a zo om?​
Serga [27]

Answer:

Step-by-step explanation:

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⣿⣿⣿⣿⣿⡏⠉⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⣿

⣿⣿⣿⣿⣿⣿⠀⠀⠀⠈⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⠛⠉⠁⠀⣿

⣿⣿⣿⣿⣿⣿⣧⡀⠀⠀⠀⠀⠙⠿⠿⠿⠻⠿⠿⠟⠿⠛⠉⠀⠀⠀⠀⠀⣸⣿

⣿⣿⣿⣿⣿⣿⣿⣷⣄⠀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠠⣴⣿⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⡟⠀⠀⢰⣹⡆⠀⠀⠀⠀⠀⠀⣭⣷⠀⠀⠀⠸⣿⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠈⠉⠀⠀⠤⠄⠀⠀⠀⠉⠁⠀⠀⠀⠀⢿⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⢾⣿⣷⠀⠀⠀⠀⡠⠤⢄⠀⠀⠀⠠⣿⣿⣷⠀⢸⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⡀⠉⠀⠀⠀⠀⠀⢄⠀⢀⠀⠀⠀⠀⠉⠉⠁⠀⠀⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠈⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿

LOL

3 0
3 years ago
Read 2 more answers
Mr. Jones waffle recipe uses 3/4 cups of milk to make 8 waffles how many cups of milk should be used to make 24 waffles?
ella [17]

Answer:

9/4

Step-by-step explanation:

If you need 3/4 cups of milk to make 8 waffles, you would need 9/4 cups to make 24 waffles because you are making 3 times the amount of waffles, so you need 3 times the amount of milk.

5 0
3 years ago
1. - x + 5y =-23 4x - y = 16​
Contact [7]
6y=5x=39
Have a good day
8 0
3 years ago
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