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3 years ago

What is 28 divided by 3.2 ?? Please include steps :(

Mathematics

1 answer:

expeople1 [14]3 years ago

8 0

thats the answer just look at the picture :)

for the work

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Simplify the expression.

sasho [114]

2^3z^9
A for question 1=8z^9

(-8ab)^2
(-8^2a^2b^2)
A for question 2=64a^2b^2

6 (x^3)^2
6 x^6
6x^6

Enjoy!=)

4 0

2 years ago

Read 2 more answers

You plan to charge $1 for each time a student plays, and the payout for a win is $5. According to your calculations, the probabi

galben [10]

Answer:

- $0.75

Step-by-step explanation:

For each game played the chance of winning is 0.05 and losing is 0.95.

If the student wins, they get $5 -$1 = $4, but they lose $1.

<u>Expected value is:</u>

4*0.05 - 1*0.95 = -0.75

It means you will make $0.75 per average game.

3 0

3 years ago

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Joey and his family are taking a road trip. On Monday they travel 68 miles. On Tuesday , they travel 25 miles. On Wednesday they

Stolb23 [73]

Answer:

42 miles

Step-by-step explanation:

so average is found by taking the sum of numbers of all the numbers given, and dividing by the number of numbers

68+25+33= 126

126/3=42

they drove an avg of 42 miles

hope this helps

6 0

2 years ago

When working with inequalities what do you do when its x squared > 16 and vice versa does the sign flip.

S_A_V [24]

Answer:

$x4$

$\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}$

Step-by-step explanation:

$x^2>16$

$x\sqrt{16}$

$x4$

The interval notation will be:

$\{x\in \mathbb{R}| \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right) \}$

The contrary,

$x^2$

is $-4$

6 0

3 years ago

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Evaluate the integral Find the volume of material remaining in a hemisphere of radius 2 after a cylindrical hole ofradius 1 is d

algol13

You want to find the volume inside the hemisphere $x^2+y^2+z^2=4$ (i.e. inside the sphere but above the plane $z=0$ ) and outside the cylinder $x^2+y^2=1$ . Call this region $R$ .

In cylindrical coordinates, we have

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_1^2\int_0^{\sqrt{4-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$

$\displaystyle=2\pi\int_1^2 r\sqrt{4-r^2}\,\mathrm dr$

$\displaystyle=-\pi\int_3^0\sqrt u\,\mathrm du$

(where $u=4-r^2$ )

$\displaystyle=\pi\int_0^3\sqrt u\,\mathrm du$

$=\dfrac{2\pi}3u^{3/2}\bigg|_0^3=2\sqrt3\,\pi$

7 0

3 years ago