Factor completely. 81x^2+180x+100

A basic cellular package costs $20 per month for 60 minutes of calling, with an additional charge of $0.20/min beyond that time.

Anika [276]

Answer: you can use 260 min, 60 min at a fixed price ($20) and extra 200.

Step-by-step explanation:

C(x) = 20 + 0.20(x − 60)

C(x) ≤ 60

20 + 0.20(x − 60) ≤ 60

0.20(x − 60) ≤ 60 - 20

0.20x - 12 ≤ 40

0.20x ≤ 40 + 12

0.20x ≤ 52

x ≤ 52/0.2

x ≤ 260

This way, you can use 260 min, 60 min at a fixed price ($20) and extra 200.

7 0

2 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

2 years ago

Read 2 more answers

In the figure below, what is the value of angle Z?

Hoochie [10]

Answer:

=2.83 the second option

Step-by-step explanation:

Using the trigonometric ratios we can find the sides of the triangle with the acute angles.

In the triangle provided we will use COSINE

Cos ∅=adjacent/hypotenuse

Let us substitute with the values in the question into the formula.