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katen-ka-za [31]
2 years ago
14

Which of the following is a valid probability distribution? A 2-column table labeled Probability Distribution A has 4 rows. The

first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.42, 0.38, 0.13, 0.07. A 2-column table labeled Probability Distribution B has 4 rows. The first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.27, 0.28, 0.26, 0.27. A 2-column table labeled Probability Distribution C has 4 rows. The first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.16, 0.39, 0.18, 0.17. A 2-column table labeled Probability Distribution D has 4 rows. The first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.63, 0.12, 0.14, 0.13.
Mathematics
1 answer:
dsp732 years ago
8 0

Answer:

I believe it is A

Step-by-step explanation:

all variables should add up to 1.00 and all in the first answer add up to 1.00

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Legalization of marijuana, Part I. The 2010 General Social Survey asked 1,578 US res- idents: \Do you think the use of marijuana
mixer [17]

Answer:

Step-by-step explanation:

a) Sample statistics are used to estimate population value. Since 48% is a sample proportion, therefore, it is a sample statistic.

b) For 95% confidence level, z* = 1.96.

\hat{p}\pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}= 0.61\pm 0.61\sqrt{\frac{0.61(1-0.61)}{1578}}=0.61\pm 0.024 \ or (0.586, 0.634).

We are 95% confident that the true proportion of US residents who think marijuana should be made legal lies between 58.6% and 63.4%.

c)

\\np=1578(0.61)=962.58

\\n(1-p)=1578(1-0.61)=615.42

Since both np and n(1-p), are at least 10, the normal model is a good approximation for these data.

d) As the lower limit of confidence interval is less than 0.5, less than 50% population is also a plausible value of true proportion. This means the statement "Majority of Americans think marijuana should be legalized" is not justified.

4 0
2 years ago
If f Superscript negative 1 Baseline (x) = negative one-fifth x, what is f Superscript negative 1 Baseline (x) = one-fifth x?
Artyom0805 [142]

Answer:

  a different inverse function

Step-by-step explanation:

If f^{-1}(x)=-\frac{1}{5}x then f^{-1}(x)=\frac{1}{5}x is a different inverse function.

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The second (inverse) function is the first reflected over the y-axis.

5 0
2 years ago
Read 2 more answers
Each vertical cross-section of the triangular prism shown below is an isosceles triangle.
vova2212 [387]

Answer:

Height is 3

Step-by-step explanation:

4.24 x 4.24 x 6

Right triangle base = a + b + c

                         a^2 + 3^2  =  4.24^2

= a^2 + 9                             = 17.98

We cross out b to subtract.

                                   a^2  = 17.98 - 9

                                    a^2 = 8.98

We then square         √a^2 = √8.98

                                    a = 2.996

We round up               a = 3                                

We have found the height is 3

5 0
2 years ago
Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1
astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

x^{2} - 24 = 1

Given that resulting expression is a second order polynomial of the form x^{2} - a^{2}, there are two real and distinct solutions. Roots of the expression are:

x_{1} = -5 and x_{2} = 5.

Now, it is also required to determine which part of the interval (x_{1}, x_{2}) is equal to a number greater than zero (positive). That is:

x^{2} - 24 > 0

x^{2} > 24

x < -4.899 and x > 4.899.

Therefore, exists two sub-intervals: [-5, -4.899] and \left[4.899,5\right]. Besides, x^{2} - 24 > y = 1 in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx

A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx

A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}

A = 2.525 -2.474+9.798 + 2.525 - 2.474

A = 9.9\,units^{2}

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

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Mrs. Blackwell gives each of her students two pencils. How many pencils did she hand out?
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<span>no está en mi equipo lo siento</span>
6 0
1 year ago
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