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yawa3891 [41]
3 years ago
15

Jayne’s checkbook balance is $8,075.10. Her bank statement indicated an ending balance of $7,617.17. After comparing her records

with the bank statement, she found a deposit from November 1 in the amount of $1,043.73 was not credited to her account. A check written for $626.32 had not been presented for payment. The bank paid her interest in the amount of $53.23. There was a charge on her statement in the amount of $37.90 for printed checks. Jayne found a check written for $55.85 that was not recorded in her check register. What was Jayne’s reconciled balance?
Mathematics
1 answer:
seropon [69]3 years ago
4 0
I’m guessing that you add these
$1043.73 + $626.32 + $53.23 +
$37.90 + $55.85 = $1817.03

and then you subtract them from

ANSWER 1 :

$8,075.10 - $1,817.03 = $6,258.07

or you subtract it from

ANSWER 2 :

$7,617.17 - $1,817.03 = $5,800.15

I’m not sure which answer is it, but it’s one of the answers, (for me I think it’s answer 1) :)

Jaydon
2 years ago
those ain't even options
Jaydon
2 years ago
the answer options are $8,034.85 $8,043.58 $8,034.58 $8,304.58
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2

Step-by-step explanation:

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3 years ago
The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it
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Answer:

(a) 11.25 and 1.68  

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....

where,  n = number of trials (samples) taken = 15

             r = number of success

             p = probability of success which in our question is % of cases in

                  which customer problems are resolved on the same day, i.e.;75%

So, here X ~ Binom(n=15,p=0.75)

(a) Expected number of problems to be resolved today = E(X)

            E(X) = \mu = n * p = 15 * 0.75 = 11.25

    Standard deviation = \sigma = \sqrt{n*p*(1-p)} = \sqrt{15*0.75*(1-0.75)} = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

     P(X = 10) = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}

                    = 3003*0.75^{10} *0.25^{5} = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

    = \binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}

    = 3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4} = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

    given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)  

= \binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}

= 1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}

= 0.6865

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^^^^

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2(8w + 1)

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2 appears on both sides of the plus sign.

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