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Agata [3.3K]
3 years ago
8

A jar has 6 red marbles and 4 blue marbles. If 2 marbles are chosen, one at a time without replacement, what is the probability

that both marbles will be red
Mathematics
1 answer:
Juliette [100K]3 years ago
7 0

1/3 Probability both are red

red picked first:

6/10

red picked second:

5/9 (9 as first red marble not replaced so 5 red marbles remaining and 4 blue marbles still)

6/10 * 5/10 = 1/3

Hope that helps!!

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Write 2 as a complex number.
balandron [24]

Answer:

2 + 0i

Step-by-step explanation:

Our real number in this scenario would be 2. And since we don't need an imaginary part because we are just converting 2 by itself, we plug in 0 for the imaginary number.

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Anybody please help me :( I don’t know the answer ??
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Which is not a statistical question?
Helga [31]

Answer:

d.

Step-by-step explanation:

A, B, and C ask questions that will be answered with different sets of data.

<u>example a:</u>

purple- 11 students

blue- 5 students

orange- 7 students

<u>example b:</u>

math- 27

history- 30

art- 25

<u>example c:</u>

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If it were to ask how many windows are there in each of the classrooms then it would be statistical.

you get the gist lol.

3 0
3 years ago
The number of cars running a red light in a day, at a given intersection, possesses a distribution with a mean of 2.4 cars and a
liberstina [14]

Answer:

The sampling distribution of the sample mean is:

\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

Let <em>X</em> = number of cars running a red light in a day, at a given intersection.

The information provided is:

E(X)=\mu=2.4\\SD(X)=\sigma=4\\n=100

The sample selected is quite large, i.e. <em>n</em> = 100 > 30.

The Central limit theorem can be used to approximate the sampling distribution of the sample mean number of cars running a red light in a day, by the Normal distribution.

The mean of the sampling distribution of the sample mean is:

\mu_{\bar x}=\mu=2.4

The standard deviation of the sampling distribution of the sample mean is:

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{100}}=0.40

The sampling distribution of the sample mean is:

\bar X\sim N(\mu_{\bar x}=2.4,\ \sigma_{\bar x}=0.40)

8 0
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Answer:

A solution to a system of equations means the point must work in both equations in the system. So, we test the point in both equations. It must be a solution for both to be a solution to the system.

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