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1 year ago

Find the volume of the pyramid to the nearest cubic unit. Use a calculator.

Mathematics

2 answers:

Goryan [66]1 year ago

4 0

Answer:

64 m³

Step-by-step explanation:

Formula

$\textsf{Volume of a square based pyramid}=\dfrac{1}{3}a^2h$

where:

a = side length
h = perpendicular height

Given:

a = 8 m
h = 3 m

Substituting the given values into the formula:

$\begin{aligned}\implies \textsf{Volume}& =\dfrac{1}{3} \cdot 8^2 \cdot 3\\\\& = \dfrac{3}{3} \cdot 64\\\\& = 1 \cdot 64\\\\& = 64\:\sf m^3\end{aligned}$

ryzh [129]1 year ago

4 0

Answer: 64 m^3

Step-by-step explanation:

V=1/3*a^2*h

V = 1/3*64*3

V=64 m^3

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

Please help me now plz I will mark you brainliest

EastWind [94]

Answer:

1229 km

Step-by-step explanation:

The missing angle is ...

180° -60° -75° = 45°

Knowing that, the Law of Sines can be used to find the side opposite the 75° angle.

?/sin(75°) = 900 km/sin(45°)

? = (900 km)sin(75°)/sin(45°) ≈ 1229.423 km

The UFO is about 1229 km from KA-12.

5 0

3 years ago

What is the area of this parallelogram? in or in2

Ket [755]

Answer:

A=30 in

Step-by-step explanation:

Solution

A=bh=5·6=30

6 0

2 years ago

Read 2 more answers

Worth a lot of points

Dima020 [189]

6. Is 6MPH because 12M / 2H
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2 years ago

If y varies inversely as x and y=3 when x=14, find x when y=6

alukav5142 [94]

X=7 What you can do is look at the first two values given, and make them x1 and y1. Then your next value here is y2.

So x1 is 14, and y1 is 3. y2 becomes 6. X2 is unknown.

Then make the formula: x1/y2 is equal to x2/y1
(You are setting two fractions equal to each other)

That makes 14/6=X/3. When we cross multiply, we find that x=7.

3 0

2 years ago