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Gennadij [26K]
4 years ago
14

Suppose I pick a jelly bean at random from a box containing two red and ten blue ones. I record the color and put the jelly bean

back in the box. What is the probability of getting a red jelly bean both times if I do this twice? (Round your answer to three decimal places.)
Mathematics
1 answer:
Lubov Fominskaja [6]4 years ago
6 0

Answer:

The probability of getting a red jelly bean both of the times is 0.028.

Step-by-step explanation:

First we have that in the box are 12 jelly beans having 2 red and 10 blue. The probability of getting one of the red jelly beans are the sum of the probability of the 2 red jelly beans, that is:

ProbabilityRed=\frac{1}{12} +\frac{1}{12}=\frac{2}{12}

Now we can calculate the probability of getting the red jelly bean twice, this will be the product of the probability of getting each one, is the same probability for the second chance because we are getting the jelly bean back in the box, this is:

ProbabilityBothRed=\frac{2}{12} *\frac{2}{12}=\frac{4}{144}  =\frac{1}{36} =0.028

Now we have that the probability of getting a red jelly bean both of the times is 0.028.

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Tom opens a bank account and makes an initial deposit of $500. The banker tells Tom that he is going to revive an annual rate of
daser333 [38]

So for this you will be using an exponential equation, which is y=ab^x with a=initial value, b=growth/decay, y = total balance, and x = # of years


In this case, a = 500, and since with this problem the initial value is growing, you will add 100% to 6% to get 106%, or 1.06. The equation will be formed as such: y=500(1.06)^x


With this problem, just plug in 15 into x and solve for y: y=500(1.06)^{15}\\ y=1198.28


In short, after 15 years Tom will have $1198.28 in his account.

5 0
3 years ago
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1)the difference between the upper and the lower quartile is the
nirvana33 [79]
A) interquartile range

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3 0
3 years ago
Let me know the answer plz
Gala2k [10]

<u>Answer:</u>

The correct answer option is H. 4374.

<u>Step-by-step explanation:</u>

We are given the following geometric sequence and we are to find its 8th term:

\frac{2}{9}, \frac{2}{3}, 2,...

Here a_1=\frac{2}{9} and common ratio (r) = \frac{\frac{2}{3} }{\frac{2}{9} } =3.

The formula we will use to find the 10th term is:

nth term = a_1 \times r^{(n-1)}

Substituting the values in the formula to get:

10th term = \frac{2}{9} \times 3^{(10-1)}

10th term = 4374

7 0
3 years ago
A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence
Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

n = 30 -- sample size

\bar x = 50 -- sample mean

\sigma = 10 --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

\sigma_x = \frac{\sigma}{\sqrt n}

\sigma_x = \frac{10}{\sqrt {30}}

\sigma_x = \frac{10}{5.478}

\sigma_x = 1.825

The 95% confidence interval for the z value is:

z = 1.960

Calculate margin of error (E)

E = z * \sigma_x

E = 1.960 * 1.825

E = 3.577

The confidence bound is:

Lower = \bar x - E

Lower = 50 - 3.577

Lower = 46.423

Lower = 46.4 --- approximated

Upper = \bar x + E

Upper = 50 + 3.577

Upper = 53.577

Upper = 53.6 --- approximated

<em>So, the 95% confidence interval is (46.4, 53.6)</em>

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

\sigma_x = \frac{\sigma}{\sqrt n}

\sigma_x = \frac{10}{\sqrt {90}}

\sigma_x = \frac{10}{9.49}

\sigma_x = 1.054

The 95% confidence interval for the z value is:

z = 1.960

Calculate margin of error (E)

E = z * \sigma_x

E = 1.960 * 1.054

E = 2.06584

The confidence bound is:

Lower = \bar x - E

Lower = 50 - 2.06584

Lower = 47.93416

Lower = 47.9 --- approximated

Upper = \bar x + E

Upper = 50 + 2.06584

Upper = 52.06584

Upper = 52.1 --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

n = 30     E = 3.577

In (b), we have:

n = 90    E = 2.06584

<em>Notice that the margin of error decreases when the sample size increases.</em>

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