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3 years ago

Please help me!!! I suck at math :((

Mathematics

1 answer:

stiks02 [169]3 years ago

7 0

If we know that line $AB$ is equal to line $AD$ , then this means that ∠ $ADB$ has to be equal to ∠ $ABD$ in order to form a closed triangle. All angle measures of a triangle add up to $180$ °, so use the equation: $2(35) + x = 180$ .

70 + x = 180

Subtract 70 from both sides to isolate x:

180 - 70 = 110

110 = x

Hope this helped.

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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

How do you change 24 2\5 into decimal?

seropon [69]

24 and 2/5=24+2/5
24=24

0.1=1/10
2/5=x/10
times both sides by 10
20/5=x
4=x
2/5=4/10=0.4

24 and 2/5=24+0.4=24.4

5 0

3 years ago

Read 2 more answers

4.8/1 7/9 =(1.2b):6 2/3

Hunter-Best [27]

Answer:

b = 15

Step-by-step explanation:

Solve for b:

$\dfrac{4.8}{1\frac{7}{9}}=\dfrac{1.2b}{6\frac{2}{3}}$

Cross multiply:

$6\dfrac{2}{3}\cdot 4.8=1.2b\cdot 1\dfrac{7}{9}\\ \\\dfrac{6\cdot 3+2}{3}\cdot 4.8=1.2b\cdot \dfrac{9+7}{9}\\ \\\dfrac{20}{3}\cdot \dfrac{48}{10}=\dfrac{12b}{10}\cdot \dfrac{16}{9}\\ \\20\cdot 4=b\cdot \dfrac{16}{3}\\ \\5=b\cdot \dfrac{1}{3}\\ \\b=5\cdot 3\\ \\b=15$

8 0

3 years ago

MissTica

The graph of the inequality will be B. on a coordinate plane, a dashed straight line has a positive slope and goes through (0, negative 4) and (3, negative 2). Everything to the right of the line is shaded.

<h3>How to get the graph?</h3>

It can be deduced that the coordinates satisfy the equation of the line.

3y = 2x - 12

Divide through by 3.

y = 2/3x - 4.

The slope is 2/3.

In conclusion, the correct option is B.

Learn more about inequalities on:

brainly.com/question/14074138

#SPJ1

5 0

2 years ago

The minute hand of a clock in a clock tower has a length of 36 inches.

Paul [167]

45 i guessStep-by-step explanation:

4 0

3 years ago