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3 years ago

A number cube is rolled 63 times. predict how many times the cube will land on a number that is a factor of 6 .

1 answer:

4 0

You said cube, so I assume a standard 6 sided die with numbers 1,2,3,4,5,6

how many times you said
we need to find how many fulfill the condition

how many numbers from 1,2,3,4,5,6 are a factor of 6?
1 is a factor
2 is a factor
3 is a factor
6 is a factor

so 4 numbers
there are 6 sides, so we would expect it to land of a factor of 6 a grand total of 4/6 or 2/3 times

so 2/3 of the time we will get a factor of 6

so find 2/3 of 63 to get number of times we get a factor of 6
2/3 of 63=2/3 times 63=42
so it will land about 42 times

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MatroZZZ [7]

The volume is 502.6

the volume of a cylinder is V= (pi)(radius squared)(height)

in this case it would be pi times 4^2 times 10, which is 502.6 :)

3 0

3 years ago

NOT RIGHT, PLEASE TRY AGAIN

mariarad [96]

Answer:

Ok I will

Step-by-step explanation:

5 0

3 years ago

How would I solve (-2-2i)(-4-3i)(7+8)

siniylev [52]

$(-2-2i)(-4-3i)(7+8)=\\
15(-2-2i)(-4-3i)=\\
15(8+6i+8i-6)=\\
15(2+14i)=\\
30+210i$

5 0

3 years ago

Out of 400 applicants for a job, 179 are female and 72 are female and have a graduate degree. Step 2 of 2 : If 118 of the applic

Dafna11 [192]

Answer:

36/59 or 0.610

Step-by-step explanation:

P(female and degree)/P(degree)

72/118

36/59

0r 0.610

7 0

3 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago