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3 years ago

Help me with this simple question please!! answer #7 for me !

Mathematics

2 answers:

Elena L [17]3 years ago

8 0

Answer:

They are proportional

AleksAgata [21]3 years ago

3 0

Answer:

Yes

Step-by-step explanation:

The rate is $3 every 25 texts

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The angle of elevation from a viewer to the top of a flagpole is 50°. The viewer is 80 ft away and the viewer's eyes are 5.5 ft

Sunny_sXe [5.5K]

Easy 67 I don’t really know

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3 years ago

Whats the answer to log <img src="https://tex.z-dn.net/?f=log_%7Bx%2B2%7D%20%289x%29%20%3D2" id="TexFormula1" title="log_{x+2} (

lyudmila [28]

Answer:

x= 1, 4

Step-by-step explanation:

$log_{x+2} (9x) =2\\\\9x = (x+2)^2 \\\\(x+2)^2 - 9x = 0\\\\x^2 +4x + 4-9x = 0\\\\x^2 - 5x +4= 0\\\\x^2 - 4x-x+4=0\\\\x(x-4)-1(x-4)=0\\\\(x-4)(x-1)=0\\\\x-4 =0\: or\: x - 1=0\\\\x = 4\: or\: x = 1\\\\x = 1,\: 4$

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3 years ago

Find the minimum value of the parabola y = x2 + 10.

natulia [17]

Answer:

(0,10)

Step-by-step explanation:

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3 years ago

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An object travels 4/5 miles in one-half hour. What is its speed?

Ilia_Sergeevich [38]

Answer:

8/5 miles or 1.6 mile per hour

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3 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

3 years ago