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an airport offers two shuttles that run on different schedules.If both shuttles leave the airport at 4:00 p.m.,at what time will

otez555 [7]

What are both of their schedules? If they are completely different, then the next time they will leave together is 4:00 pm the next day

3 0

3 years ago

In the picture above which represents a linear relationship? A. B. C. D.

MissTica

Possibly d, because it’s a reflection

8 0

3 years ago

Read 2 more answers

Please help me with these questions.

Debora [2.8K]

4. k less than 3.5, so a number k is being taken away from 3.5:
The answer is 3.5-k

5. the sum of g and h, reduced by 11, so you need a set of parentheses to write this one: (g+h)-11

6. 5 less than y, plus g, so do a combination of the strategies used in the previous two problems and you will get: (y-5)+g

7. 5 less than the sum of y and g, you are taking away 5 from the sum of two numbers, y and g, so: (y+g)-5

Hope this helps :-)

5 0

3 years ago

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

koban [17]

Answer:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) P=0.42

c) P=0.08

d)

X | P(X)

------------------

0 | 0.42

1 | 0.50

2 | 0.08

e) E(x)=0.66

s.d.=0.62

Step-by-step explanation:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) The probability of losing the first game is

$P(G_1=L)=1-P(G_1=W)=1-0.4=0.6$

The probability of losing the second game, given that the first game was lost is:

$P(G_2=L|G_1=L)=1-P(G_2=W|G_1=L)=1-0.3=0.7$

So the probability of losing both games is:

$P(G_2=L\&G_1=L)=P(G_1=L)*P(G_2=L|G_1=L)=0.6*0.7=0.42$

c) The probability of winning both games is:

$P(G_2=W\&G_1=W)=P(G_1=W)*P(G_2=W|G_1=W)=0.4*0.2=0.08$

d) The variable X can take values 0, 1 and 2.

X=0 is when both games are lost. This happens with probability P=0.42.

X=2 is when both games are won. This happens with probability P=0.08.

X=1 is when one game is won and the other is lost. This happens with probability P=1-0.42-0.08=0.50.

Then the table of probabilities become:

X | P(X)

------------------

0 | 0.42

1 | 0.50

2 | 0.08

e) The expected value is:

$E(x)=\sum p_ix_i=0.42*0+0.50*1+0.08*2=0.00+0.50+0.16=0.66$

The variance and standard deviation of x are:

$V(x)=\sum p_i(x_i-E(x))^2\\\\V(x)=0.42(0-0.66)^2+0.50(1-0.66)^2+0.08(2-0.66)^2\\\\V(x)=0.42*0.4356+0.50*0.1156+0.08*1.7956\\\\V(x)=0.183+0.058+0.144=0.385\\\\\\\sigma=\sqrt{V(x)}=\sqrt{0.385}=0.62$

The standard deviation can

5 0

3 years ago

Out of 28 cookies, are chocolate chip. How many cookies are chocolate chip? Plssss

vredina [299]

I believe you left out a picture

4 0

3 years ago