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Ugo [173]
2 years ago
15

You’ve bought a cylindrical thermos 8 in. high with a base 6 in. across. How much coffee can you put in the thermos?

Mathematics
1 answer:
lana66690 [7]2 years ago
4 0

Answer:

3.706 liters

Step-by-step explanation:

Hello

assuming that these measures correspond to the internal measures of the thermos,

we can calculate the volume as, the volume of a cylinder, which is given by

V=\pi r^{2} h

where r is the radius of the base, and h the height

also

Diameter=2r

r=Diameter/2

Step 1

let

h=8 in

base = diameter = 6 inch

r=diameter/2

r=6/2 inch,r=3 inch

Step 2

Put the values into the equation

V=\pi r^{2} h\\V=\pi (3in)^{2} (8 in)\\V=\pi *72 \ in^{3} \\\\V=72\pi\ in^{3} \\\\

Step 3

Now, the unit (cubic inches) is not an usual unit for measuring coffee, let´s convert cubic inches into liters by knowing

1 inch = 2.54 cm

1 litro =  1000 cubic centimeters

if\ 1\ inch\ = 2.54\ cm\\(1\ inch)^{3} =(2.54\ cm)^{3}\\\\1\ inch^{3}=16.3870\ cm^{3}

1 cubic inch = 16.3870 cubic centimeters

1\ cubic\ inch\ = 16.3870\ cubic\ centimeters\\\\\\72\pi =x?\ cubic\ centimeters\\\\x=\frac{72\pi \cubic \ inch\ \*16.3870 }{1\ cubic\ inch} \\\\x=3706.65\ cubic\ centimeters

Step 4

Finally, convert cubic centimeters into liters1000\ cubic\ centimeters =1\ liter \\\\\\3706.65\ cubic\ centimeters = x?\ liters\ centimeters\\\\x=\frac{3706.65 \cubic \ centimeters\ \*1 liter}{1000 cubic centimeters} \\\\x=3706.65\ cubic\ centimeters\\x=3.706 Liters

I can put 3.706 liters in the thermos

Have a good day

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Step-by-step explanation:

When we have a random variable X that is normally distributed with mean \large\bf \mu and standard deviation \large\bf \sigma, then  

The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean \large\bf \mu and standard deviation \large\bf \sigma to the left of a.

The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean \large\bf \mu and standard deviation \large\bf \sigma to the right of b.

The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤  b) = P(a < X ≤  b)= P(a ≤ X < b) is the area under the normal curve with mean \large\bf \mu and standard deviation \large\bf \sigma between a and b.

In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml

(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?

As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).

You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.

<em>If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula: </em>

<em>1- NORMDIST(62,63,5.4,1) in Excel </em>

<em>1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc </em>

<em>and we will get a value of 0.5735 or 57.35% </em>

(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?

Now we want the area to the left of 90

<em>NORMDIST(90,63,5.4,1) in Excel </em>

<em>NORMDIST(90;63;5.4;1) in OpenOffice Calc </em>

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(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?

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<em>In Excel </em>

<em>NORMDIST(68.4,63,5.4,1) - NORMDIST(57.6,63,5.4,1)  </em>

<em>In OpenOffice Calc  </em>

<em>NORMDIST(68.4;63;5.4;1) - NORMDIST(57.6;63;5.4;1)  </em>

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