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guapka [62]
3 years ago
15

A line segment is 20 units long. The line segment is translated up 6 units and left 8 units. The length of the resulting line se

gment is
either 4 or 20 or 34 units because translations
decrease or increase or preserve length.

Is the resulting lin segment 4 or 20 or 34? And do the translations decrease or increase or preserve? That is my question please help me.
Mathematics
1 answer:
lapo4ka [179]3 years ago
3 0

Answer:

The length of the resulting line segment is 20 units and the length of a line segment is preserved on translation.

Step-by-step explanation:

The given length of the line segment is 20 units.

The line segment is translated up 6 units and left 8 units.

Here, the line segment is translated, so the whole line segment is at a new position after translation, there is no change in the length of the line.

So, the length of the line segment is preserved on translation.

After translation, the length of the line remains unchanged.

So, the length of the resulting line segment is 20 units.

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A= 47 sq. in.<br><br> Find the value of x. If necessary, round to the nearest tenth.
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Answer:

x = 9.7

Step-by-step explanation:

The area of a triangle is A = b * h / 2 where A, b, and h are the area, base and height respectively. We know that A, b, and h equal 47, x, and x respectively so we can write:

47 = x * x / 2

47 = x² / 2

94 = x²

x = ±√94

In context, x can't be negative so the answer is x = √94 = 9.7.

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The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(
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Answer:

(a) A_1 and A_2 are indeed mutually-exclusive.

(b) \displaystyle P(A_1\; \cap \; B) = \frac{1}{20}, whereas \displaystyle P(A_2\; \cap \; B) = \frac{1}{25}.

(c) \displaystyle P(B) = \frac{9}{100}.

(d) \displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}, whereas P(A_1 \; |\; B) = \displaystyle \frac{4}{9}

Step-by-step explanation:

<h3>(a)</h3>

P(A_1 \; \cap \; A_2) = 0 means that it is impossible for events A_1 and A_2 to happen at the same time. Therefore, event A_1 and A_2 are mutually-exclusive.

<h3>(b)</h3>

By the definition of conditional probability:

\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}.

Rearrange to obtain:

\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot  P(A_1) = 0.25 \times 0.20 = \frac{1}{20}.

Similarly:

\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot  P(A_2) = 0.80 \times 0.05 = \frac{1}{25}.

<h3>(c)</h3>

Note that:

\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}.

In other words, A_1 and A_2 are collectively-exhaustive. Since A_1 and A_2 are collectively-exhaustive and mutually-exclusive at the same time:

\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}.

<h3>(d)</h3>

By Bayes' Theorem:

\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}.

Similarly:

\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}.

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