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2 years ago

I need help with these questions fast hury

Mathematics

1 answer:

PilotLPTM [1.2K]2 years ago

4 0

LOLLLLLOLLOLLLOLLL !

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11 ounces is to one cup as<br> ounces is to 10 cups.

Margaret [11]

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Step-by-step explanation:

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2 years ago

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Find a numerical value of one trigonometric function of x for cscx = sinx tanx + cosx

Alex787 [66]

Answer:

Step-by-step explanation:

we have

$cscx = sinx tanx + cosx\\\\\frac{1}{sinx}=sinx.\frac{sinx}{cosx} +cosx\\\\\frac{1}{sinx}.sinx.cosx= sinx.\frac{sinx}{cosx}.sinx.cosx+cosx.sinx.cosx\\ cosx= (sinx)^{3} +sinx.(cosx)^{2} \\cosx= sinx[(sinx)^{2}+(cosx)^{2})\\ cosx=sinx.1\\cosx=sinx\\tanx=1\\x=\frac{\pi}{4} +k\pi,k-integer.$

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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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2 years ago

3/5n+15=10+2/5n what is n

Nutka1998 [239]

$\textsf{Hey there!}$

$\mathsf{\dfrac{3}{5}n +15=10+\dfrac{2}{5}n}$

$\mathsf{\bf{SUBTRACT}\ \mathsf{by\ \dfrac{2}{5}n\ from\ both\ of\ your\ sides}}$

$\mathsf{Like: \dfrac{3}{5}n+15-\dfrac{2}{5}n=\dfrac{2}{5}n+10-\dfrac{2}{5}n}$

$\mathsf{\bf{CANCEL: \dfrac{3}{5}+\dfrac{2}{5}}}\textsf{out\ because that gives you\ 1}$

$\textsf{Solve\ the set we have left over}$

$\mathsf{Our\ new\ equation\ becomes: \dfrac{1}{5}n+15=10}$

$\mathsf{\bf{SUBTRACT}}\textsf{ by 15 on both of your sides}$

$\mathsf{\dfrac{1}{5}n+15-15=10-15}$

$\mathsf{Cancel\ out: 15-15\ because\ that\gives\ you\ 0 }$

$\mathsf{Keep: 10-15\ because\ it\ helps\ us\ solve\ for\ your\ n }$

$\mathsf{10-15=-5}$

$\mathsf{New\ equation: \dfrac{1}{5}n=-5}$

$\mathsf{\bf{MULTIPLY}\mathsf{\ both\ sides\ by\ 5}}$

$\mathsf{\dfrac{1}{5}n\times5=5\times-5}$

$\mathsf{Cancel\ out\ \dfrac{1}{5}\times5 \ because \ it \ gives \ us \ 1}$ $\mathsf{Keep: 5\times -5\ because\ it\gives\ us\ the\ value\ of\ your\ n}$