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How to do this in math

igomit [66]

Multiply both sides by 5

3 0

3 years ago

Find the exact value of x.

il63 [147K]

Answer:

here is the exact value of x=16

5 0

2 years ago

Sarah Sandoval is a coffee farmer trying to decide how many tons of coffee to produce. She can sell each ton of coffee for $2,40

Alexandra [31]

Answer:

Sarah should produce four tons of coffee and total cost for four ton is $6,000.

Step-by-step explanation:

According to the scenario, calculation of the given data are as follows,

Selling price per ton = $2,400

Cost of production is increasing by $600 for each ton.

So, Cost of production first ton = $600

Cost of production for second ton = $600 + $600 = $1,200

Similarly, Cost of production for third ton = $1,200 + $600 = 1,800

Cost of production of fourth ton = $1,800 + $600 = $2,400

Here, Sarah can produce coffee till selling price = cost of production.

As, cost of production of four ton = selling price, then Sarah can produce only four tons of coffee.

Total cost for four ton coffee = $600 + $1,200 + $1,800 + $2,400

= $6,000

8 0

3 years ago

Need some help please?

n200080 [17]

The value of the unknown length is 24

<h3>How to determine the unknown length?</h3>

Represent the unknown length with x.

So, we have the following equivalent ratio:

30 : 30 + x = 25 : 45

Express as fraction

30/30 + x = 25/45

Simplify the fraction

30/30 + x = 5/9

Cross multiply

150 + 5x = 270

Evaluate the like terms

5x = 120

Divide by 5

x = 24

Hence, the value of the unknown length is 24

Read more about similar shapes at:

brainly.com/question/24214480

#SPJ1

5 0

2 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