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3 years ago

Commutative property of addition 9+8=17

Mathematics

2 answers:

Andrei [34K]3 years ago

5 0

Commutative property 8+9 = 17

luda_lava [24]3 years ago

5 0

The commutative property is when you move the numbers around.

So the commutative property for this is 8 + 9 =17.

I hope this helped!!

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If RY=7x+4, YC=10x-2, then what is the value of RC?

irina1246 [14]

If RY + YC = RC
7x+4=10x-2
3x=6
x=2
18+18=36=RC
The value of RC is 36

5 0

3 years ago

AWARDING DOUBLE POINTS question in pic, will report anything other than the answer

mash [69]

Answer:

m∠R is 72°

Step-by-step explanation:

In the given figure

∵ ΔPQR ≅ ΔUVW

→ From congruency

∵ m∠P = m∠U

∵ m∠Q = m∠V

∴ m∠R = m∠W

∵ m∠R = (10x - 18)°

∵ m∠W = 8x°

∵ m∠R = m∠W

→ Equate their measures

∴ 10x - 18 = 8x

→ Add 18 to both sides

∵ 10x - 18 + 18 = 8x + 18

∴ 10x = 8x + 18

→ Subtract 8x from both sides

∴ 10x - 8x = 8x - 8x + 18

∴ 2x = 18

→ Divide both sides by 2 to find x

∴ x = 9

→ Substitute the value of x in the m∠R

∵ m∠R = 10(9) - 18

∴ m∠R = 90 - 18

∴ m∠R = 72°

∴ m∠R is 72°

7 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)

8 0

3 years ago

Solve for x. <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B2%7D%20%20%3D%20%20-%207" id="TexFormula1" title=" \frac{

weeeeeb [17]

Answer:

x = -14

Step-by-step explanation:

$\frac{x}{2}$ = -7

multiply 2 on both sides

the 2's cancel out on the fraction side, leaving x = -7(2)

multiply -7 and 2 to get -14

x = -14

4 0

3 years ago

Read 2 more answers

Number 31 please guys!!!!!

monitta

the answer is c
this is because, first you use the distributive property. Multiply 'y' by the number 3 and you get 3y
then multiply 3 by the number 4.
you get 12
subtract 7 from 12 and you get 5

So, the expression becomes 3y+5

Hope this helped!!

3 0

4 years ago