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2 years ago

Solve for x. Round to the nearest tenth

Mathematics

2 answers:

aksik [14]2 years ago

8 0

Answer:

Answers are in corresponding pictures.

marshall27 [118]2 years ago

6 0

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Answer: 2) 33.3

4) 31.9

6) 11

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

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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

2 years ago

The two amino acids animated above

andrew-mc [135]

Answer:

its peptide bond and water

Step-by-step explanation:

Two amino acids animated above are joined together by a PEPTIDE BOND and a form of a molecule of WATER in the process. Hope this helps Mate :)

3 0

3 years ago

Read 2 more answers

I need help on this loll

Serhud [2]

Answer: The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

Step-by-step explanation:

Let us revise the translation

If the point (x , y) translated horizontally to the right by h units then its image is (x + h , y)

If the point (x , y) translated horizontally to the left by h units then its image is (x - h , y)

If the point (x , y) translated vertically up by k units then its image is (x , y + k)

If the point (x , y) translated vertically down by k units then its image is (x , y - k)

(x , y) → (x ± h , y ± k) the right arrow symbol used to show the

translation from a point to its image

Example:

∵ P (0 , 0) → P' (1 , 2)

∴ The rule is (x , y) → (x + 1 , y + 2)

Let us find the translation rule that maps point D ( 7 , − 3 ) onto

point D' (2 , 5)

∵ Point (x , y) = (7 , -3)

∵ Its image after translation (x + h , y + k) = (2 , 5)

∴ x + h = 2

∵ x = 7

∴ 7 + h = 2

- Subtract 7 from both sides

∴ h = -5

∵ y + k = 5

∵ y = -3

∴ -3 + k = 5

- Add 3 to both sides

∴ k = 8

∴ The rule of translation is (x , y) → (x - 5 , y + 8)

The translation rule that maps point D ( 7 , − 3 ) onto point D ' ( 2 , 5 )

is (x , y) → (x - 5 , y + 8)

3 0

2 years ago

Gina decided to order some clothes from a catalog she ordered three pairs of jeans at $39 each for t-shirts at $15 each and 2 st

Blizzard [7]

39 times 3 and 15 times 3 then 27 times 2 then add together to get 216

4 0

3 years ago

A construction company is hired to build a wheelchair ramp for a restaurant. The ramp connects the handicap parking space (point

makvit [3.9K]

Answer:

1. diagram in the attachment

2. AB = 32.3 ft

Step-by-step explanation:

2. from the diagram:

$sin \theta = \frac{opposite}{hypotenuse}\\sin (8) = \frac{4.5}{AB} \\AB = \frac{4.5}{sin(8)} \\AB = \frac{4.5}{0.1392} \\AB = 32.33ft$

4 0

3 years ago