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suter [353]
3 years ago
6

Let Î be a Bernoulli random variable that indicates which one of two hypotheses is true, and let P(Î = 1) = p. Under the hypothe

sis Î = 0, the random variable X is uniformly distributed over the interval [0,1]. Under the alternative hypothesis Î = 1, the PDF of X is given by fX| Î(x|1) = 2x if 0<=x<=1 and 0 otherwise. Consider the MAP rule for deciding between the two hypotheses, given that X=x. Find the probability of error associated with the MAP rule as a function of p.
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
5 0
I am so confused on this can u help me
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What number must be placed in the box in the equation below to produce an equation that has more than one solution: 5u+3(4u-6)=-
Sergeeva-Olga [200]

Answer: the number must be placed in the box is 9

suppose: the number must be placed in the box is x

have:

5u + 3(4u - 6) = -3u -27 + 20u + x

⇔ 5u + 12u - 18 = -3u - 27 + 20u +x

⇔ 17u + 3u - 20u = -27 + 18 + x

⇔ 0 = x - 9

⇔ x = 9

Step-by-step explanation:

7 0
3 years ago
Find the vertex and the axis of symmetry of the graph of the function?
Orlov [11]
3) Vertex: (-7, 0) Axis of Symmetry: x=-7
6) Vertex: (-2, -1) Axis of Symmetry: x=-2
5 0
4 years ago
30 POINTS please help me!
Stels [109]
To prove a similarity of a triangle, we use angles or sides.

In this case we use angles to prove

∠ACB = ∠AED (Corresponding ∠s)
∠AED = ∠FDE (Alternate ∠s)

∠ABC = ∠ADE (Corresponding ∠s)
∠ADE = ∠FED (Alternate ∠s)

∠BAC = ∠EFD (sum of ∠s in a triangle)

Now we know the similarity in the triangles.

But it is necessary to write the similar triangle according to how the question ask.

The question asks " ∆ABC is similar to ∆____. " So we find ∠ABC in the prove.

∠ABC corressponds to ∠FED as stated above.
∴ ∆ABC is similar to ∆FED

Similarly, if the question asks " ∆ACB is similar to ∆____. "
We answer as ∆ACB is similar to ∆FDE.

Answer is ∆ABC is similar to ∆FED.
7 0
4 years ago
Solve square root <br> (2x+3)^2=36
REY [17]
X=3/2 btw its a fraction
6 0
4 years ago
Read 2 more answers
The data to the right represent the weights​ (in grams) of a random sample of 50 candies. Complete parts​ (a) through​ (f). 0.92
Inessa05 [86]

Answer:

0.069

Step-by-step explanation:

The given data set is

0.92, 0.87, 0.88, 0.82, 0.82, 0.87, 0.97, 0.86, 0.89, 0.84, 0.81, 0.88, 0.77, 0.86, 0.93, 0.84, 0.72, 0.82, 0.74, 0.83, 0.93, 0.75, 0.79, 0.91, 0.84, 0.91, 0.88, 0.88, 0.83, 0.78, 0.99, 0.81, 0.78, 0.75, 0.82, 0.76, 0.82, 0.87, 0.91, 0.77, 0.72, 0.94, 0.71, 0.73, 0.81, 0.81, 0.86, 0.93, 0.93, 0.82.

Formula for mean:

Mean=\frac{\sum x}{n}

Sum of all terms = 41.98

Mean of the data set is

Mean=\frac{41.98}{50}

Mean=0.8396

Formula for standard deviation for population:

\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}

Formula for standard deviation for sample:

\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n-1}}

\sigma=\sqrt{\frac{0.231792}{50-1}}

\sigma=\sqrt{0.004730449}

\sigma=0.06877826

\sigma\approx 0.069

Therefore, the standard deviation of the data set is 0.069.

5 0
3 years ago
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