Y=3(1.9)x in Exponential Function

On a typical day, flights at a local airport arrive at a rate of 10 every 15 minutes. At this rate, how many flights would you e

Hunter-Best [27]

40 because 15×4 is 60 and there is 60 minutes in an hour

3 0

3 years ago

If a football player passes a football from 4 feet off the ground with an initial velocity of 36 feet per second, how long will

tiny-mole [99]

0 = -16t² + 36t + 4
0 = -8t² + 18t + 2 ← Complete the square to solve
-8(t² - 9t/2) = -2
8(t² - 9t/4 + 81/64) = 2 + 81/8
8(t - 9/8)² = 97/8
(t - 9/8)² = 97/64
t - (t = 0)
(h = 0) at:
<span>t = (9 + √97)/8 ≅ 2.356 seconds
</span>

3 0

3 years ago

Read 2 more answers

The ratio of black pencils to colored pencils in Lauren's backpack is 1:9

Softa [21]

Answer:

I don't really understand the question but...

2:18

3:27

4:36

5:45

6: 54

7:63

8:72

8 0

3 years ago

Which of the following is an irrational number

erastova [34]

Answer:

c. π

Step-by-step explanation:

Intgers, fractions, and repeating or finite-length decimals are all rational number.

Pi (π) is an irrational number.

8 0

4 years ago

There are currently 16 teams in the National Football Conference (NFC). Of those teams, 11 have won at least one super bowl and

svp [43]

You are given 16 teams, 11 have won at least one super bowl and 5 have not.

A. The probability that both selected teams have won at least 1 super bowl is

$Pr(A)=\dfrac{C_{11}^2}{C_{16}^2}=\dfrac{\dfrac{11!}{2!(11-2)!}}{\dfrac{16!}{2!(16-2)!}}=\dfrac{\dfrac{10\cdot 11}{2}}{\dfrac{15\cdot 16}{2}}=\dfrac{55}{120}=\dfrac{11}{24}.$

B. The probability that neither selected team has won at least 1 super bowl is

$Pr(B)=\dfrac{C_{5}^2}{C_{16}^2}=\dfrac{\dfrac{5!}{2!(5-2)!}}{\dfrac{16!}{2!(16-2)!}}=\dfrac{\dfrac{4\cdot 5}{2}}{\dfrac{15\cdot 16}{2}}=\dfrac{10}{120}=\dfrac{1}{12}.$

C. The probability that at least one selected team has won at least 1 super bowl is

$Pr(C)=1-Pr(B)=1-\dfrac{1}{12}=\dfrac{11}{12}.$

D. to find the probability that the second team selected has won at least 1 super bowl given that the first team selected has not won a super bowl, consider such events:

P - the second team selected has won at least 1 super bowl;

Q - the first team selected has not won a super bowl.

Then

$Pr(P|Q)=\dfrac{Pr(P\cap Q)}{Pr(Q)}=\dfrac{\dfrac{5\cdot 11}{C_{16}^2}}{\dfrac{C_5^1\cdot C_{15}^1}{C_{16}^2}}=\dfrac{55}{75}=\dfrac{11}{15}.$

E. To find the probability that the second team selected has won at least 1 super bowl given that the first team selected has won at least 1 super bowl, consider events:

M - the second team selected has won at least 1 super bowl;

N - the first team selected has won at least 1 super bowl.

Then

$Pr(M|N)=\dfrac{Pr(M\cap N)}{Pr(N)}=\dfrac{\dfrac{11\cdot 10}{C_{16}^2}}{\dfrac{C_{11}^1\cdot C_{15}^1}{C_{16}^2}}=\dfrac{110}{165}=\dfrac{2}{3}.$

3 0

3 years ago