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Leviafan [203]
3 years ago
11

A megabyte holds about 1,000,000 characters of data. A gigabyte holds about 1,000 times more datathan a megabyte. About how many

charaters of data does the gigabyte hold?
Mathematics
1 answer:
andrew-mc [135]3 years ago
6 0
Given:
A megabyte holds 1,000,000 = 10⁶ characters of data.

Because a gigabyte is 1,000 times more than a megabyte, it holds
10⁶*1,000 = 10⁶ * 10³ = 10⁹ characters of data.

Answer:
A gigabyte holds 10⁹ or 1,000,000,000 characters of data.
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Goshia [24]

Answer:

The answer is C

Step-by-step explanation:

An isosceles triangle has two angles that measure the same, so...

30 + 30 = 60

180 - 60 = 120

No

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180 - 90 = 90

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5 0
4 years ago
Find the surface area of the cube. 5 ft S.A. = [?] ft? 5 ft A Hint: A cube has 6 square faces. Find the area of one face, then m
atroni [7]

First, let's find the area of one face of the cube.

Area = base x height

A = 5 x 5 = 25

There are 6 faces on a cube, so we can multiply the area of one face by 6 to get the total surface area.

SA = 25 x 6 = 150 ft^2

Hope this helps!

4 0
3 years ago
585 divided by 83 what's the answer
trapecia [35]
7.05 is the correct answer
6 0
4 years ago
Read 2 more answers
Please help I'm dying rn​
amid [387]

Answer:

-3√5

Step-by-step explanation:

3√5-2√45

3√5-2•√9×√5

3√5-2•3√5

3√5-6√5

-3√5

6 0
3 years ago
Read 2 more answers
Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
kompoz [17]

a.

f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}

is a proper joint density function if, over its support, f is non-negative and the integral of f is 1. The first condition is easily met as long as C\ge0. To meet the second condition, we require

\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}

b. Find the marginal joint density of X and Y by integrating the joint density with respect to z:

f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz

\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}

Then

\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy

\approx\boxed{0.12886}

c. This probability can be found by simply integrating the joint density:

\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz

\approx\boxed{0.012262}

7 0
3 years ago
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