Find the (real-valued) general solution to the differential equation. z′′+5z′=0

1 answer:

7 0

Try this solution:1. to make up the characteristic equation:α²+5α=0; $\left[\begin{array}{ccc} \alpha=0 \\ \alpha =-5\\\end{array}$ 2. to make up the final answer:y=C₁+C₂e⁻⁵ˣ

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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

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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

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