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3 years ago

A computer maker receives parts from three suppliers, S1, S2, and S3. Fifty percent

1 answer:

3 0

The Bayes' theorem is being used to compute the posterior probabilities. This is a way to measure probability distribution. It's being used to estimate the posterior probabilities using the prior probabilities. Following are the solution to the given question:

By the law of total probability, every probability area is divided into $\bold{S1,S2,S3}$ .

$\to \bold{P(D) = P(D|S1)P(S1) + P(D|S2)P(S2) + P(D|S3)P(S3)}$

Using the Bayes' rule:

$\to \bold{P(S2|D) = P(D|S2) \frac{P(S2)}{P(D)}}$

Therefore, the final answer is " $\bold{ P(D|S1)P(S1) + P(D|S2)P(S2) + P(D|S3)P(S3)}$ and $\bold{P(D|S2) \frac{P(S2)}{P(D)}}$ "

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Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in s

Elodia [21]

2+5i is complex, and MUST be accompanied by its conjugate, 2-5i.

Thus, the factors of this 4th degree poly are (x-4)(x+8)(x-2-5i)(x-2+5i).

Multiply this out to obtain the poly in question. Then arrange the powers of x in descending order, beginning with x^4.

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3 years ago

catriona earned 16 more stickers than Peter. she earned 35 stickers. how many stickers did Peter earn?

expeople1 [14]

Answer:

Step-by-step explanation:

If Catriona has 16 more stickers than Peter than the only thing you have to do is subtract and find the difference

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4 years ago

Use the following vectors to answer parts (a) and (b). v1equals=[Start 3 By 1 Matrix 1st Row 1st Column 1 2nd Row 1st Column n

erik [133]

Answer:

(1) No matter what's the value of $h$ , $\vec{v}_3$ is never in the span of $\vec{v}_1$ and $\vec{v}_2$ .

(2) The three vectors $\vec{v}_1$ , $\vec{v}_2$ , and $\vec{v}_3$ are always linearly dependent for all real $h$ .

Step-by-step explanation:

If $\vec{v}_3$ is in the span of $\vec{v}_1$ and $\vec{v}_2$ , there need to exist real $a$ and $b$ such that

$a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3$ .

Assume that such $a$ and $b$ do exist.

In other words,

$\displaystyle a \left[\begin{array}{c}{1 \\ -4\\2} \end{array}\right] + b\left[\begin{array}{c}{-4 \\ 16\\-8}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right]$ .

$\displaystyle \left[\begin{array}{c}{a \\ -4a\\2a} \end{array}\right] + \left[\begin{array}{c}{-4b \\ 16b\\-8b}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right]$ .

$\left\{\begin{array}{rrcr} a & - 4b &=& 5\\ -4a & + 16b &= &7\\2a & -8b & =&h\end{aligned}\right.$ .

Rewrite as an augmented matrix and row-reduce:

$\displaystyle \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ -4 & 16 & 7 \\ 2 & -8 & h\end{array}\right]$ .

(Add four times row one to row two and $-2$ times row one to row three.)

$\displaystyle \sim \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ 0 & 0 & 27 \\ 0 & 0 & h - 10\end{array}\right]$ .

Note that in row two,

Left-hand side: $0$ ;
Right-hand side: $27\neq 0$ .

In other words, this system is inconsistent. There's no real $a$ and $b$ that would satisfy the condition

$a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3$ .

Hence $\forall h \in \mathbb{R}, \quad \vec{v}_3\not \in \text{Span}\{\vec{v}_1, \vec{v}_2\}$ .

There's no real $h$ that allows $h$ , $\vec{v}_3$ to be part of the span of $\vec{v}_1$ and $\vec{v}_2$ .

If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.

Note that

$\vec{v}_2 = (-4)\vec{v}_1 + 0 \; \vec{v}_3$ . In other words, $\vec{v}_2$ can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of $\vec{v}_3$ is zero, neither the exact value of $\vec{v}_3$ nor the value of $h$ will make a difference. Therefore, for all $h \in \mathbb{R}$ , the three vectors $\vec{v}_1$ , $\vec{v}_2$ , and $\vec{v}_3$ are linearly dependent.