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Marysya12 [62]
3 years ago
10

Can you please help me with this question :)

Mathematics
1 answer:
timurjin [86]3 years ago
4 0

Answer:

B

Step-by-step explanation:

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A basketball that sold for $55 is discounted 15%. what is the sale price?​
Mamont248 [21]

Answer: $46.75

Step-by-step explanation:

10% of $55=$5.50 plus

5% of $55=$2.75

Total

15% of $55 =$8.25

$55.00 less $8.25

=$46.75

5 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
2 years ago
Brainliest Brainliest Brainliest Brainliest Brainliest Brainliest Brainliest Brainliest Brainliest
djverab [1.8K]

Answer:

<u>n</u>   =  <u>9000</u>

180    180

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
40 points and brainly math problem easy!
Nat2105 [25]

Answer: the third one

Step-by-step explanation:

8 0
3 years ago
For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les
kaheart [24]

Answer:

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation  

n=1000 represent the random sample taken    

\hat p=0.52 estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

\alpha=0.05 represent the significance level (no given, but is assumed)    

Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

5 0
2 years ago
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