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Mila [183]
3 years ago
12

YALL HELP!!!!!!!!!!!!!!

Mathematics
1 answer:
QveST [7]3 years ago
8 0

Answer:

The second choice

Step-by-step explanation:

In most cases, the dependent variable is the y value in the graph.

The dependent variable is the thing that is being measured or changing.

In this example it can't be the time because time is always moving at a constant rate and there is nothing that can be done to change it.

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PLEASE SOMEONE DO THIS IM RUSHING SO BAD AND IM CRYING - GIVING BRAINLIEST!!!!!
LUCKY_DIMON [66]

Answer:

1.) $1.07

2.) $2.25

3.) $14.20

Step-by-step explanation:

1.) Toonie = 2 CAD

CAD = USD $1.57

$1.57 - $0.50 = $1.07 (USD)

2.) $2.75 - $5 = $2.25 (USD)

3.) $20 x 4 = 80

$80 - $65.80 = $14.20 (USD)

Hope this helped!

8 0
3 years ago
Solve for y<br> 4(4y-7) +3y=24
Dima020 [189]

Answer:

y= 52/19 or 2 14/19

Step-by-step explanation:

first we open the brackets

16y-28+3y=24

then we shift the terms

16y+3y= 24+28

19y= 52

y= 52/19 or 2 14/19

hope it helps,pls mark me as brainliest

7 0
3 years ago
Which of the following statements contain a variable? Check all that apply. A. There are 60 seconds in a minute. B. He is 15 yea
sdas [7]

The answer is the option C and the option D, which are:

C. Double the number of pages in the book.

D. The age of my friend.

The explanation for this problem is shown below:

By definition, a variable is an unkown quantity. In the Option A the quantity is given: 60 seconds in a minute. In the option B you know the quantity too, which is 15 years old. But the option C and the option D does not give any quantity, you don't know the number of pages in the book or the age of the friend yet. Therefore, the answer is the options C and D.

8 0
3 years ago
What is the value of 6x - 3y if x = 5 and y= -1
nadezda [96]

Answer:

33

Step-by-step explanation:

6x - 3y

=> x = 5, y = -1 Substitute in the above equation,

=> 6(5) - 3(-1)

=> 6(5) + 3

=> 30 + 3

=> 33

5 0
2 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
2 years ago
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