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vova2212 [387]
3 years ago
11

The Smurfs family goes grocery shopping once a week they spent $250 in a month for the groceries how much money would you spend

per day assume the month has 30 days how much money would be spent per year how much money would you spend per week
Mathematics
2 answers:
aalyn [17]3 years ago
7 0
$250/30 days= money spent per day
$8.33 = money spent per day

$250 x 12 months = money spent per year
$3,000.00 = money spent per year

$250/4 weeks= money spent per week
$62.50 = money spent per week

hope this helps :)






Vilka [71]3 years ago
4 0
The answer is 8.333333333
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Its B

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The reliance of an event on the outcine of another event gives rise to a dependent event. For any two or more event, whereby the outcome of event A has no effect on the possible outcome of event B, then such scenario is termed as being Independent. In the scenario described above, the outcome oof Sam's pick does not depend on the outcine of Steve's pick. This is because, once Steve makes his selection, the marble is returned into the box before Sam makes his own selection.Therefore, no matter the color of marble picked by Steven, it will have no bearing on the outcome of Sam's pick.

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2 years ago
The first Ferris wheel was a hit at the 1893 Chicago World’s Fair. Each of its 36 cars carried 40 riders. How many riders filled
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800

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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
7m + -10 =-10 what is m?
tigry1 [53]

Answer:

m = 0

Step-by-step explanation:

Given

7m + -10 = -10

Positive negative gives negative.

So the expression can be rewritten as

7m - 10 = -10

Add 10 to both sides of the equation

7m - 10 + 10 = -10 + 10

7m = 0

Divide both sides by 7

7m /7 = 0/7

m = 0

8 0
3 years ago
Read 2 more answers
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