Find the volume. Please

An icicle with a diameter of 15.5 centimeters at the top, tapers down in the shape of a cone with a length of

Helga [31]

Answer:

Step-by-step explanation:

Note: I will leave the answers as fraction and in terms of pi unless the question states rounding conditions to ensure maximum precision.

From the question, we can tell it is a inversed-cone (upside down)

Volume of Cone = $\pi r^{2} \frac{h}{3}$

a) Given Diameter , d = 15.5cm and Length , h = 350cm,

we first find the radius.

$r = \frac{d}{2} \\=\frac{15.5}{2} \\=7.75cm$

We will now find the volume of the cone.

Volume of cone $\pi (7.75)^{2} \frac{350}{3} \\= \frac{168175\pi }{24}$

We know the density of ice is 0.93 grams per $1cm^{3}$

$1cm^{3} =0.93g\\\frac{168175\pi }{24} cm^{3} =0.93(\frac{168175\pi }{24} )\\= 20473 g$ (Nearest Gram)

b) After 1 hour, we know that the new radius = 7.75cm - 0.35cm = 7.4cm

and the new length, h = 350cm - 15cm = 335cm

Now we will find the volume of this newly-shaped cone.

Volume of cone = $\pi (7.4)^{2} \frac{335}{3} \\= \frac{91723\pi }{15} cm^{3}$

Volume of cone being melted = New Volume - Original volume

= $\frac{168175\pi }{24} -\frac{91723\pi }{15} \\= \frac{35697\pi }{40} cm^{3}$

c) Lets take the bucket as a round cylinder.

Given radius of bucket, r = 12.5cm (Half of Diameter) and h , height = 30cm.

Volume of cylinder = $\pi r^{2} h\\=\pi (12.5)^{2} (30)\\=\frac{9375\pi }{2} cm^{3}$

To overflow the bucket, the volume of ice melted must be more than the bucket volume.

Volume of ice melted after 5 hours = $5(\frac{35697\pi }{40} )\\=\frac{35697\pi }{8} cm^{3}$

See, from here of course you are unable to tell whether the bucket will overflow as all are in fractions, but fret not, we can just find the difference.

Volume of bucket - Volume of ice melted after 5 hours

= $\frac{9375\pi }{2} -\frac{35697\pi }{8 } \\=\frac{1803\pi }{8}cm^{3}$

from we can see the bucket can still hold more melted ice even after 5 hours therefore it will not overflow.

4 0

2 years ago

7. Given the following information, calculate, in order, the amount credited and the outstanding balance.

svp [43]

Answer:

C. $824.74, $175.26

Step-by-step explanation:

1) Amount Credited

The formula to calculate the amount credited =

Amount paid ÷ ( 100% - Discount)

Discount is given in the question as 3/10

Where 3 = Discount rate

Amount paid = $800

Amount credited = 800/( 100% - 3%)

= 800/ 97%

= 800/ 0.97

= $824.74

b) Outstanding balance = Invoice - Amount credited

Invoice = $1000

Amount credited = $824.74

Outstanding balance = $1000 - $824.74

= $175.26

5 0

3 years ago

Write an expression for the sequence of operations described below.

Triss [41]

Answer:

(h+g)=n➗2 is the answer. happy to help you

5 0

3 years ago

Please help! Correct answer only!

arlik [135]

Answer:

<em>Expected Payoff ⇒ $ 1.50 ; Type in 1.50</em>

Step-by-step explanation:

Considering that 1 out of the 100 tickets will have a probability of winning a 150 dollar prize, take a proportionality into account;

$100 - Number of Tickets,\\1 - Number of Tickets You Can Enter,\\\\1 / 100 - Probability of Winning,\\$ 150 - Money Won,\\\\Proportionality - 1 / 100 = x / 150, where x - " Expected Payoff "\\\\1 / 100 = x / 150,\\100 * x = 150,\\\\Conclusion ; x = 1.5 dollars$

<em>Thus, Solution ; Expected Payoff ⇒ $ 1.50</em>

5 0

2 years ago

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.

devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0

3 years ago