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salantis [7]
2 years ago
14

Adrienne has several receipts from recent transactions that she entered in her records. The receipts include an ATM receipt for

a $60.00 withdrawal (plus a $2.00 transaction fee), a grocery store receipt for $32.50, and a $1,200 paycheck deposit slip. When she finishes entering her transactions, Adrienne realizes that her balance is incorrect. Assuming that Adrienne’s beginning account balance was $320.00, why is her balance incorrect?
a. Adrienne forgot to include the $2.00 ATM transaction fee.
b. Adrienne did not use $320.00 as her starting balance.
c. Adrienne deducted $23.50 from her balance instead of $32.50. d. Adrienne did not enter her ATM withdrawal correctly.
Mathematics
2 answers:
Rina8888 [55]2 years ago
6 0

Answer:

D.

Step-by-step explanation:

Adrienne did not enter her ATM withdrawal correctly.

svp [43]2 years ago
3 0

Answer:D

Step-by-step explanation:    :)

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b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

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Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

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n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

With y2 ∈ {0,1}

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P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

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P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

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