I need help with #2 help if u can

PLSSS HELP IF YOU TURLY KNOW THISS

xxMikexx [17]

Answer:

2 1/2

Step-by-step explanation:

Turn both fractions into improper fractions, multiply the whole number by the denominator and add the numerator.

4 3/4

4 * 4 + 3 = 19

19/4

2 1/4

2 * 4 + 1 = 9

9/4

Now since she drank 2 1/4 gallons of juice, subtract 9/4 from 19/4.

19/4 - 9/4 = 10/4

Now turn back into a mixed fraction, do this by putting 4 into ten twice and taking the remainder and putting it as numerator:

4 * 2 = 8

10 - 8 = 2

2 1/2

8 0

2 years ago

6. The first quartile (rounded to the nearest tenth if needed) of the following data set is: 15. 16. 17. 23. 11. 19. 20. 15, 18,

Rainbow [258]

Answer:

Step-by-step explanation:

first of all arrange this data from ascending to descending order.

11,15,15,15,16,17,18,19,20,22,23

use formula

N+1/4 (N means no of observations)

11+1/4

12/4

3 rd term

therefore Q1=15

6 0

3 years ago

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)

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