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Lilit [14]
3 years ago
6

13. Which is Not independent 14 . 2 coins tossed , probability

Mathematics
2 answers:
morpeh [17]3 years ago
4 0

Answer:

13. Option 1

14. Option 2 : 1/4

Step-by-step explanation:

13. First option is correct.

If two marbles are drawn from a bag without replacement then the second event will depend upon the first event i.e. the occurrence of first event will affect the occurrence of second event.

14. If two coins are tossed, the outcomes are

S = {HH, HT, TH, TT}

There is only one outcome that favors the given scenario i.e. first coin showing heads and second showing tails.

So, probability of showing heads first and then second showing tails = 1/4

So second option is correct..

Luden [163]3 years ago
3 0

Answer:

13. B) You pull a green tile from a bag of tiles, return it, and then pull a yellow tile.

14. A) 1/2.

How this helps you! (:

-Hamilton1757

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Consider the expressions 7y + 5 − 3 and 7y + 2. Which statement is true?
romanna [79]
The two expressions are equivalent because they have the same value regardless of the number of substituted in for y
4 0
2 years ago
A baseball is thrown into the air from the top of a 224-foot tall building. The baseball's approximate height over time can be r
Studentka2010 [4]

9514 1404 393

Answer:

  A.  7 seconds

Step-by-step explanation:

We assume your factored equation is something like ...

  h(t) = -16t(t -7)(t +2)

The time it takes the ball to reach the ground is the positive value of t that makes a factor zero:

  t -7 = 0   ⇒   t = 7

The ball will land on the ground in 7 seconds.

7 0
3 years ago
The daily revenues of a cafe near the university are approximately normally distributed. The owner recently collected a random s
lbvjy [14]

Answer:

The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

MOE=z_{crit}\times\dfrac{\sigma}{\sqrt{n}}

Rearranging this equation in terms of n gives

n=\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2

Now the Margin of Error is reduced by 2 so the new M_2 is given as M/2 so the value of n_2 is calculated as

n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n

As n is given as 40 so the new sample size is given as

n_2=4n\\n_2=4*40\\n_2=160

So the sample size to obtain the desired margin of error is 160.

4 0
3 years ago
6 sweets cost 24p all together how much do 5 sweets cost
Vlada [557]

Cost of 6 sweets = 24p

so, cost of 1 sweet = 24p/6 = 4p

Now, cost of 5 sweets will be =4p*5 = 20p

3 0
3 years ago
1) -10m(-3 +3m) – 5(8m – 6)
Juliette [100K]

- 30m^2 - 10m + 30

distribute your negative 10, you will first get 30m, then - 30m^2.

next distribute the negative 5, you will get - 40m, then 30.

combine like terms, and make sure its in standard form, - 30m^2 - 10m + 30

7 0
3 years ago
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