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3 years ago

Find the area of a rhombus with diagonals 9 ft. and 12 ft

Mathematics

1 answer:

aleksklad [387]3 years ago

3 0

The answer would be 54, because A= PQ over 2 = 9 x 12 over 2 = 54

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Monica walks 2 miles each day Monday through Friday. She walks 3 miles on Saturday. She does not walk on Sunday. Monica calculat

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She is wrong. She would have walked 337 miles after 29 weeks.

Step-by-step explanation:

13x29=337

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2 years ago

What is 1 1/3 times 1 4/9 in fraction form

nikitadnepr [17]

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52 / 27

Step-by-step explanation:

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2 years ago

Evaluate the range of the function f(x) = 3 -6x, when the domain is x=0

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3 years ago

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I need help with this question please answer asap

saveliy_v [14]

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9 months

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2 years ago

urn I contains 1 red chip and 2 white chips; urn II contains 2 red chipsand 1 white chip. One chip is drawn at random from urn I

vitfil [10]

Answer:

Following are the answer to this question:

Step-by-step explanation:

In the question first calls the W if the transmitted chip was white so, the W' transmitted the chip is red or R if the red chip is picked by the urn II.

whenever a red chip is chosen from urn II, then the probability to transmitters the chip in white is:

$P(\frac{w}{R}) = \frac{P(W\cap R)}{P(R)} \ \ \ \ \ _{Where}\\\\P(R) = P(W\cap R) + P(W'\cap R) \\$

The probability that only the transmitted chip is white is therefore $P(W) = \frac{2}{3}\\$ , since urn, I comprise 3 chips and 2 chips are white.

But if the chip is white so, it is possible that urn II has 4 chips and 2 of them will be red since urn II and 2 are now visible, and it is possible to be: $P(\frac{R}{W}) = \frac{2}{3}$

$P(W\cap R) = P(W) \times P(\frac{R}{W}) \\$

$= \frac{2}{3}\times \frac{2}{4} \\\\= \frac{2}{3}\times \frac{1}{2} \\\\= \frac{2}{3}\times \frac{1}{1} \\\\=\frac{1}{3}\\\\= 0.333$

Likewise, the chip transmitted is presumably red $(P(W')= \frac{1}{3})$ and the chip transferred is a red chip of urn II $(P(\frac{R}{W'})= \frac{3}{4}$ , and a red chip is likely to be red $(\frac{R}{W'})$ .

Finally, $P(W'\cap R) = P(W') \times P(\frac{R}{W'})\\$

$= \frac{1}{3} \times \frac{3}{4} \\\\ = \frac{1}{1} \times \frac{1}{4} \\\\=\frac{1}{4}\\\\= 0.25$

The estimation of $P(R)$ and $P(\frac{W }{R})$ as:

$P(R) = 0.3333 + 0.25\\\\ \ \ \ \ \ \ \ \ \ = 0.5833 \\\\ P(\frac{W}{R}) = \frac{0.3333}{0.5833} \\\\\ \ \ \ \ \ \ = 0.5714$

4 0

2 years ago