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3 years ago

This should be easy

Mathematics

2 answers:

Ivanshal [37]3 years ago

8 0

Answer:

Step-by-step explanation:

it's the holy number jesus christ can confirm

dsp733 years ago

6 0

Answer:4

Step-by-step explanation:

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Help plz:)))I’ll mark u Brainliest

DENIUS [597]

9514 1404 393

Answer:

x = 7

Step-by-step explanation:

We can write a proportion relating the short segment to the full length:

(x+1)/18 = 4/(4+5)

x +1 = 8 . . . . . . multiply by 18 and simplify

x = 7 . . . . . . . . subtract 1

3 0

3 years ago

If half of a number is greater than 16, then the number must be

Evgesh-ka [11]

Answer:

Step-by-step explanation:

let's call the number 2x

then half of it will be x

if x > 16

smallest this number can be is 33

6 0

3 years ago

Please lok at the screen shot and answer like number 5. is...... and number 6. is...... with label and equation. Please and than

Andrews [41]

The answer to number 5 is 16 because if 1/8 goes per bag then every one pound is stretched into eight bags and you multiply by two for the two pounds and get 16 bags for number six you just divide 1/4 by 3 and get 0.0833 pounds per child

7 0

3 years ago

Mr. Hartman purchased 2-pounds of beef that cost

elena-14-01-66 [18.8K]

Answer:

$5.60

Step-by-step explanation:

multiply $2.80 by 2

7 0

3 years ago

Read 2 more answers

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

3 years ago