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Anastasy [175]
3 years ago
14

The cone has a height of 9 in and a diameter of 6 in. Find the volume of the popcorn container. Use 3.14 for π

Mathematics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

The volume of the popcorn container is V=84.78\ in^3

Step-by-step explanation:

we know that

The volume of the cone is equal to

V=\frac{1}{3}\pi r^{2}h

where

r is the radius of the base of cone

h is the height of the cone

In this problem we have

r=6/2=3\ in ----> the radius is half the diameter

h=9\ in

\pi =3.14

substitute the values

V=\frac{1}{3}(3.14)(3)^{2}(9)

V=84.78\ in^3

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Step-by-step explanation:

Consider the quadratic equation, we have

ax^2+bx+c=0

Dividing the equation by a, we get

x^2+\frac{b}{a}x+\frac{c}{a}=0

Put \frac{c}{a} on the other side,

x^2+\frac{b}{a}x=\frac{-c}{a}

Add (\frac{b}{2a})^2 on both the sides,

x^2+\frac{b}{a}x+(\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2

completing the square, we have

(x+\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2

Now, solving for x,

x+\frac{b}{2a}={\pm}\sqrt{\frac{-c}{a}+(\frac{b}{2a})^2}

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Step-by-step explanation:

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3 years ago
A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select
Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; \mu = 57,800  and  \sigma = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( \frac{X-\mu}{\sigma} <= \frac{58000-57800}{750} ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( \frac{X-\mu}{\sigma} < \frac{57000-57800}{750} ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = \frac{\sigma}{\sqrt{n} } = \frac{750}{\sqrt{100} } = 75

The z probability is given by;

           Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{58000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{57000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

7 0
3 years ago
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