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2 years ago

Round 3.751 to the nearest tenth.

Mathematics

2 answers:

Lady bird [3.3K]2 years ago

8 0

Answer:

3.8

Step-by-step explanation:

If it is 5 or above, round up, 4 or down, round down.

0.751 round to tenth=0.8 since hundredths is 5

ss7ja [257]2 years ago

5 0

Answer: It's 3.8

Step-by-step explanation:

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25 x (10+3) in word form

FinnZ [79.3K]

Answer:

Step-by-step explanation:

25 times the sum of 10 and 3

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2 years ago

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Z1 and Z2 are vertical angles. Z2 has measure of 93° what is the measure of Z1? enter your answer in the box.

Soloha48 [4]

Answer:

m∠Z1 = 93°

Step-by-step explanation:

Vertical Angles are the angles opposite each other when two lines cross.

They are congruent (the same as each other).

Therefore, m∠Z1 = m∠Z2, so if m∠Z2 = 93° then m∠Z1 = 93°

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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:

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

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