All categories

2 years ago

A pile of sand on the beach has the approximate shape of a cone with a diameter 28 in. and a height of 11.75 in.

Mathematics

2 answers:

DerKrebs [107]2 years ago

7 0

Answer:

A) Volume of the pile of sand is 2411.7 in³ .

Step-by-step explanation:

Given : A pile of sand on the beach has the approximate shape of a cone with a diameter 28 in. and a height of 11.75 in.

To find : volume of the pile of sand.

Solution : We have given that diameter of cone = 28 in.

height = 11.75 in.

Radius = $\frac{diameter}{2}$ .

Radius = 14 in.

Volume of cone = $\frac{pi(radius)^{2}*height }{3}$ .

Volume of cone = $\frac{3.14(14)^{2}*11.75 }{3}$ .

Volume of cone = 2410.47

Volume of cone ≈ 2411.7 in³.

Therefore, A) Volume of the pile of sand is 2411.7 in³ .

makvit [3.9K]2 years ago

6 0

Use the formula for volume of a cone.
V=pi × r2<span> × h /3
</span>V= 2411.7 in.^3

You might be interested in

Solve equation 3 - 8c = 35

NeTakaya

We would subtact 3 from both sides allowing the equation to simplify to -8c=32

We would then divide both sides by -8 to equal -4

The final answer C=-4

5 0

2 years ago

3/4 divides by 1/7 =

anyanavicka [17]

3/4 divides by 1/7 =

copy dot flip

3/4 * 7/1

21/4

4 goes into 21 5 times with 1 left over

5 1/4

4 0

2 years ago

Read 2 more answers

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300. The

zubka84 [21]

Answer:

the variance of the refund payment to the couple = 9463.394

Step-by-step explanation:

Given that :

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300.

It is possible that the couple won't be able to refund up 100 per day or more than 100 per day.

SO; let assume that the refund payment happens to be 0, 100,200, 300

Let X be the total refund payment on the seven day cruise.

We can say X = 0, if there is no rain on all 7 days.

$P(X = 0) = _nC_x * P^x * (1 - P)n-x$

$P(X = 0) = _7C_o * 0.2^0 * (1-0.2)^{7-0$

$P(X = 0) =1 * 1* (1-0.2)^{7$

$P(X = 0) =(0.8)^{7$

$P(X = 0) =0.2097152$

If it rains on any one day; then X = 100

$P(X = 100) = _nC_x * P^x * (1 - P)n-x$

$P(X = 100) = _7C_1 * 0.2^1 * (1-0.2)^{7-1$

$P(X = 0) =7 * 0.2* (1-0.2)^{6$

$P(X = 100) =7* 0.2* (0.8)^{6$

$P(X = 100) =0.3670016$

if it rains on any two day ; then X = 200

$P(X = 200) = _nC_x * P^x * (1 - P)n-x$

$P(X = 200) = _7C_2 * 0.2^2 * (1-0.2)^{7-2$

$P(X = 200) = 21 * 0.2^2 * (0.8)^{5$

$P(X = 200) = 0.2752512$

if it rains on any three day or more than that ; then X = 300

$P(X \ge 300) = 1 - P(X < 300) \\ \\ P(X \ge 300) = 1 - [P(X = 0) + P(X = 100) + P(X = 200)] \\ \\ P(X \ge 300) = 1 - [0.2097152 + 0.3670016 + 0.2752512] \\ \\ P(X \ge 300) = 0.148032$

Now; we have our probability distribution function as:

P(X = 0) = 0.2097152

P(X = 100) = 0.3670016

P(X = 200) = 0.2752512

P(X = 300) = 0.148032

In order to determine the variance of the refund payment to the couple; we use the formula:

variance of the refund payment to the couple $[Var X] =E [X^2] - (E [X])^2$

where;

$E[X^2] = \sum x^2 \times p \\ \\ E[X^2] = 0^2 * 0.2097152 + 100^2 * 0.3670016 + 200^2 * 0.2752512 + 300^2 * 0.148032 \\ \\ E[X^2] = 0 + 3670.016 + 11010.048+ 13322.88 \\ \\ E[X^2] =28002.944$

$(E [X]) = \sum x * p\\ \\ (E [X]) = 0 * 0.2097152 + 100 * 0.3670016 + 200 * 0.2752512 + 300 * 0.148032 \\ \\ (E [X]) = 0 + 36.70016 + 55.05024 + 44.4096\\ \\ (E [X]) = 136.16 \\ \\ (E [X])^2 = 136.16^2 \\ \\ (E [X])^2 = 18539.55$