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3 years ago

Find the volume of this Composite solid

Mathematics

1 answer:

damaskus [11]3 years ago

5 0

Answer:

D. 569.39 m^3

Step-by-step explanation:

First, we can split up the solid into two shapes: a cone and a cylinder.

Then we can find the volume of the cone

The formula is v= pi r^2 h/3

We know the radius is four and so is the height.

Plug it in and we get 67.02 m^3

Next is the cylinder

Formula is pi r^2 h

We know that the radius is 4 and the height is 10

Plus it in and we get 502.65 m^3

We add 67.02 and 502.65 together to get the total volume.

Answer is D. 569.39 m^3

Mine is a bit higher than D since I used a pi button instead of 3.14

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Is this a right triangle?

vesna_86 [32]

Answer:

yes!

Step-by-step explanation:

if it has a 90° angle it is a right triangle :)

8 0

3 years ago

Read 2 more answers

K=mv^2/2 solve for v

matrenka [14]

Answer: $v=\sqrt[]{\frac{2K}{m} }$

Step-by-step explanation:

$K=\frac{mv^2}{2}$

First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;

$2*K=\frac{mv^2}{2}*2$

$2K=mv^2$

Divide by m to isolate $v^2$ .

$\frac{2K}{m}=\frac{mv^2}{m}$

$\frac{2K}{m} =v^2$

To eliminate the square and isolate v, extract the square root.

$\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}$

$\sqrt[]{\frac{2K}{m} }=v$

let's rewrite it in a way that v is in the left side.

$v=\sqrt[]{\frac{2K}{m} }$

4 0

3 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$