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ehidna [41]
3 years ago
7

There are 24 sixth graders and 32 seventh graders attending an after school event. For this event, sixth graders and 7th graders

need to be grouped together on teams. If each team needs the same number of sixth and seventh graders, what is the greatest amount of teams that could be formed? How many sixth graders will be on each team?
Mathematics
1 answer:
madam [21]3 years ago
8 0

Answer:

8

Step-by-step explanation:

if there are 24 in 6 th grade the highest will be 8 because 8 it the only large number that can go in 24

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The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it
brilliants [131]

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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Answer:

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