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Marysya12 [62]
2 years ago
12

Consider the following formula used to find the volume of a cone.

Mathematics
2 answers:
zhuklara [117]2 years ago
6 0

Answer:

\hookrightarrow \: { \rm{V = \pi {r}^{2}  \frac{h}{3} }} \\

• simplify making h the subject:

{ \rm{3V = \pi {r}^{2} h}} \\  \\ { \boxed{ \boxed{ \rm{h =  \frac{3V}{\pi {r}^{2} } \:  }}}}

frutty [35]2 years ago
4 0

\\ \sf\longmapsto V=\pi r^2\dfrac{h}{3}

\\ \sf\longmapsto \dfrac{h}{3}=\dfrac{V}{\pi r^2}

\\ \sf\longmapsto h=\dfrac{3V}{\pi r^2}

Option B

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There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit
DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = nCr * p^{r} * q^{n-r}

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = 15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}

p(x=3) = 455 * 0.4^{3} * 0.6^{12}

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday  = 1 - 0.0633 = 0.9337

3 0
3 years ago
__ are the symbols 0,1,2,3,4,5,6,7,8,and9.
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The answer is probably one of the followig:
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5 0
3 years ago
Write each word form in standard form. <br> Four thousand
oksano4ka [1.4K]
4,000 +000 +00 + 0     is stander form for four thousand 

3 0
3 years ago
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Your friend gave you a gift card to a new restaurant. after you spend $36 on dinner, there is a balance of $44 left on the gift
Sergio [31]
36+44=80 when you add them
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3 years ago
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ccording to the U.S. National Weather Service, at any given moment of any day, approximately 1000 thunderstorms are occurring wo
viva [34]

Answer:

The critical value of <em>z</em> for 99% confidence interval is 2.5760.

The 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

Step-by-step explanation:

Let <em>X</em> = number of lightning strikes on each day.

A random sample of <em>n</em> = 23 days is selected to observe the number of lightning strikes on each day.

The random variable <em>X</em> has a sample mean of, \bar x=8.1\ mn and the population standard deviation, \sigma=0.51\ mn.

The (1 - <em>α</em>)% confidence interval for population mean <em>μ</em> is:

CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Compute the critical value of <em>z</em> for 99% confidence interval is:

z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5760

*Use a <em>z</em>-table.

The critical value of <em>z</em> for 99% confidence interval is 2.5760.

Compute the 99% confidence interval for population mean number of lightning strike as follows:

CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\=8.1\pm2.5760\times\frac{0.51}{\sqrt{23}}\\=8.1\pm0.2738\\=(7.8262, 8.3738)\\\approx(7.83, 8.37)

Thus, the 99% confidence interval for population mean number of lightning strike is (7.83 mn, 8.37 mn).

The 99% confidence interval for population mean number of lightning strike  implies that the true mean number of lightning strikes lies in the interval (7.83 mn, 8.37 mn) with 0.99 probability.

3 0
3 years ago
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