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1 year ago

Sketch over the interval [0, 2π] the graph of: y = sin(x + π)

Mathematics

1 answer:

LuckyWell [14K]1 year ago

7 0

Answer:

solve the inequality /2x-1/≥3

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{x + 3y =- 5<br> {9x + 3y = 3

LiRa [457]

Answer:

1. y-intercept(s):

(0,−53)

x-intercept(s):

(−5,0)

2. x-intercept(s):

(13,0)

y-intercept(s):

(0,1)

Step-by-step explanation:

7 0

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

2 years ago

My teacher never explained how to do this so im kinda lost so can i get help and get the answers on these please(ASAP)

ivolga24 [154]

Answer: This shat def a virus

Step-by-step explanation:

7 0

2 years ago

Can someone help me please

Taya2010 [7]

Answer:

n = -1

Step-by-step explanation:

n^2 + 5n + 1 = 3n

N^2 +2n +1 = 0 (subtract 3n from both sides)

(n+1)(n+1) = 0 (factor

n+1 =0; n= -1

n+1 = 0 ; n=-1

4 0

2 years ago

I'm so confused! what answer am i missing???

Marina86 [1]

The point P could be located in quadrant I, quadrant IV and the x-axis

<h3>How to determine the quadrant?</h3>

The coordinate of point P is given as:

P = (a, b)

Where a is positive

When a is positive, then the likely locations of P are:

On the x-axis, when b = 0
In quadrant I, when b is positive
In quadrant IV, when b is negative

Hence, point P could be located in quadrant I, quadrant IV and the x-axis

Sketch over the interval [0, 2π] the graph of: y = sin(x + π)​

Sketch over the interval [0, 2π] the graph of: y = sin(x + π)