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1 year ago

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pls help i hate math. three forces act on an object. one force pulls right with 60n. the second pulls left with 30n. the third p

ulls left 20n. what is the net force?

1 answer:

inn [45]1 year ago

5 0

Answer:

20 is.the net force act in object

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Plz just give the eqution.

Nina [5.8K]

Answer:

16 = <u>3x</u>

Step-by-step explanation:

It is an equilateral triangle. The formula for the perimeter of an equilateral triangle is P = 3a.

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2 years ago

Write the polynomial in factored form<br><br> x^3-2x^2-35x=

slamgirl [31]

Answer:

<span>x=6</span>, <span>x=−5</span> or <span>x=9</span>

Explanation:

<span><span>f<span>(x)</span></span>=<span>(x−6)</span><span>(x+5)</span><span>(x−9)</span></span>

If all of the linear factors are non-zero, then so is their product <span>f<span>(x)</span></span>.

If any of the linear factors is zero, then so is their product <span>f<span>(x)</span></span>.

So the zeros of <span>f<span>(x)</span></span> are precisely the values of x which make at least one of the linear factors 0, namely: 6, <span>−5</span> or 9.

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3 years ago

12. Elly's credit card record for the last 7 months is below. Based on the information from the table, what will be her new bala

blagie [28]

Answer:

52145.55

Step-by-step explanation:

4 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

teds age is now four less than twice the age of his brother. The sum of their age is 50. what will be their ages three years fro

AnnyKZ [126]

Answer:

Sorrry i don't know about it

4 0

3 years ago