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Find the total volume. Please show your work.

elena-s [515]

Answer:

Step-by-step explanation:

You have a half sphere and a cylinder.

Trick question! Note that the diameter of the cylinder is in meters, while the height is given in centimeters.

For cylinder:

radius r = 5 m

height h = 6 cm = 0.06 m

volume = πr²h = π(5²)(0.06) = 1.5π m³

For sphere:

radius r = 5 m

volume of sphere = (4/3)πr³

volume of half sphere = (2/3)πr³ = (2/3)π5³ = (250/3)π m³

Total volume = 1.5π + 250π/3 ≅ 266.5 m³

4 0

2 years ago

Joe has 4/5 as much money as bob. Tim has $50 more than joe. Rita has $150, which is $35 more than bob. How much money does tim

lesantik [10]

Rita = $150

Bob = $150 - $35 = $115 Bob has $115

Joe = $115 divided by 5 = $23 $23 x 4 = $92 Joe Has $92

Tim = Has $50 more then joe, $92 + $50 = $142 Tim has $142

3 0

3 years ago

What year is 7 years after 2003

mezya [45]

Answer:

2010

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago

Answer my question please help

Alchen [17]

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I ᕼOᑭᗴ IT ᕼᗴᒪᑭՏ

ǝɹnsʇouɯᴉ#

6 0

3 years ago