Help plz

A city stadium has 7,128 seats. The city wants to add 1,694 seats to the stadium. The stadium will now hold blank seats.

Naya [18.7K]

Answer: idk i wanna know the answer too

Step-by-step explanation:

3 0

3 years ago

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

4 years ago

Read 2 more answers

What’s the answer Hint: it’s not y = x + 2

Lyrx [107]

Answer:

y = 2x

the rise is the 2, and the the run is positive 1, because with graphing its always rise/run, and if there isn't anything below it, it is 2x/1

4 0

3 years ago

Lim ln(tan x) as tends to pi/2 from the left

ioda

Let $x=\arctan y$ . Then $\tan x=y$ , and so as $x\to\dfrac\pi2^-$ , you have $y\to+\infty$ . The limit is then equivalent to

$\displaystyle\lim_{x\to\frac\pi2^-}\ln(\tan x)=\lim_{y\to\infty}\ln y=\infty$

3 0

3 years ago

Me ayuden por favor es para Hoy ✌✌

DIA [1.3K]

ninguna de las preguntas se puede dividir, pero puede simplificar la última.

Lo siento pero, Espero que esto sea de ayuda!

8 0

3 years ago