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Kipish [7]
3 years ago
7

HELP ASAP PLZ &THANK YOU :)

Mathematics
1 answer:
Andre45 [30]3 years ago
8 0

The standard form:

Ax+By=C

The slope-intercept form:

y=mx+b

m - slope

m=\dfrac{y_2-y_1}{x_2-x_1}

b - y-intercept (0, b).

We have the point (0, 43) → b= 43.

Next point (2, 55). Substitute to the slope formula:

m=\dfrac{55-43}{2-0}=\dfrac{12}{2}=6

Therefore we have:

y=6x+43      <em>subtract 6x from both sides</em>

\boxed{-6x+y=43}


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Which pair of numbers are not opposites?
mrs_skeptik [129]

Answer:

The answer is C, 17 and | -17|

Step-by-step explanation:

We can solve this using the process of elimination.

Looking at the first one, we know that those | | around the 7 mean to find the absolute value. The absolute value is the number of units the number is from zero, and it is always positive. This means that it is saying 7 and -7. Those are opposites, so we know that this isn't right.

17 and -17 are opposites, since the opposite of a positive is a negative.

The last one is the same. 71 and -71 are opposites. That leaves us with C.

Remember that those | | mean to find the absolute value of. -17 is 17 unites from zero. So is 17 the oppostie of 17? No, so these are not opposites, and therefore are our answer.

8 0
3 years ago
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The formula y-y1=m(x-x1)
bagirrra123 [75]

what's the question? you only put a formula

4 0
3 years ago
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)

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svetoff [14.1K]

The length is 3 times longer than the width (7)

3 x 7 = 21

the length is 21.

To find perimeter, add all sides together:

21 + 21 + 7 + 7 = 42 + 14 = 56

56 ft is your answer

hope this helps

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