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3 years ago

Will someone help me

Mathematics

2 answers:

faust18 [17]3 years ago

8 0

Answer:

1.one solution

2.no solution

3.ifinitly many

Step-by-step explanation:

jolli1 [7]3 years ago

7 0

Answer:

1, one solution

2, no solution

3, infinitely many solution

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Find the nth term 4,9,14,19 WHOEVER ANSWERS FIST CORRECTLY WILL GET GIVEN BRAINEST

irina1246 [14]

Answer:

$4,9,14,19$

$a=4$

$d=9-4=5$

$a^n=a+(9-1)d$

$=4+(9-1)5$

$=4+5n-5$

$=5n-1$

Step-by-step explanation:

Thanks!!!!

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2 years ago

Rotundo to the nearest 10 of 569

Brums [2.3K]

569 rounded to the nearest 10 would be 570.
The 6 is your 10's place and anything after it (aka the 9) that is equal to or greater than 5 makes your next number go up one

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3 years ago

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zheka24 [161]

Answer: The answer is 10

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3 years ago

Unit 4

Dennis_Churaev [7]

He did this then that plus to this and that =334.2

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3 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$

Hence, probability of the stronger team winning overall is:

$=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103$

#Define Y as the random variable for winning 2/3 games.:

$P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480$

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

8 0

3 years ago