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yan [13]
3 years ago
14

The equation above can be used to calculate the volume V, of a cone with a height h and radius r. If the volume of a cone is 243

?ft3, and the height is equal to the radius of the base, what is the height of the cone in feet?
Mathematics
1 answer:
WARRIOR [948]3 years ago
5 0

Answer:

The height of the cone is h=9\ ft

Step-by-step explanation:

we know that

The volume of a cone is equal to

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

In this problem we have

V=243\pi\ ft^{3}

r=h\ ft

substitute

243\pi=\frac{1}{3}\pi (h^{2})h

simplify

729=(h^{3})

h=\sqrt[3]{729}\ ft

h=9\ ft

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Find the value of x in the below complementary angle 3x(5x+2)​
Natali [406]

Answer:

The value of x is 11

Step-by-step explanation:

<em>Two angles are complementary if the sum of their measures is 90°</em>

Let us use this rule to solve our question

∵ The angle of measure (3x)° and the angle of measure (5x + 2)°

   are complementary angles

→ That means their sum equals 90°

∴ 3x + 5x + 2 = 90

→ Add the like terms in the left side

∵ (3x + 5x) + 2 = 90

∴ 8x + 2 = 90

→ Subtract 2 from both sides to move 2 from the left side to the right side

∵ 8x + 2 - 2 = 90 - 2

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7 0
3 years ago
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen
devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = \frac{\sum X}{n} = 441

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 14.34

            n = sample size = 16

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.131 < t_1_5 < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.131) = 0.95

P( -2.131 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.131 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X -2.131 \times {\frac{s}{\sqrt{n} } } , \bar X +2.131 \times {\frac{s}{\sqrt{n} } } ]

                                      = [ 441-2.131 \times {\frac{14.34}{\sqrt{16} } } , 441+2.131 \times {\frac{14.34}{\sqrt{16} } } ]

                                      = [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

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