All categories

3 years ago

It’s not A I failed and I stuck at math can someone plz help me thxxx

Mathematics

1 answer:

12345 [234]3 years ago

8 0

Answer:

b. 12,960ft

Step-by-step explanation:

half of 24 gives you 12 therefore the answer must start from 12.

You might be interested in

Mamie saved $161.25. this is 25% of the amount she needs to save how mutch money does she need to save

kipiarov [429]

Answer:

483.75 is needed

Step-by-step explanation:

161.25 * 4 = 645.00

645.00 - 161.25 =

4 0

3 years ago

A submarine dives as shown in the diagram.

Marianna [84]

Answer:

Tan x = 125/500= 0.25

so x = arctan (0.25)

=14 degree

8 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

Convert the expression to simplified radical form.<br> 32^(1/2)x^(5/2)y^(3/2)

Citrus2011 [14]

Hi, the attached image has your answer.

3 0

3 years ago

Solve −4(x + 10) − 6 = −3(x − 2). (1 point)

AURORKA [14]

Hey there!!

Given equation :

... -4 ( x + 10 ) - 6 = -3 ( x - 2 )

Using the distributive property :

... -4x - 40 - 6 = -3x + 6

Combining like terms :

... -4x - 46 = -3x + 6

Adding 46 on both sides :

... -4x = -3x + 52

Adding 3x on both sides :

... -x = 52

... x = -52

Hope helps!

8 0

3 years ago