Answer:
A
Step-by-step explanation:
This explanation mostly depends on what you're learning right now. The first way would be to convert this matrix to a system of equations like this.
g + t + k = 90
g + 2t - k = 55
-g - t + 3k = 30
Then you solve using normal methods of substitution or elimination. It seems to me that elimination is the quickest method.
g + t + k = 90
-g - t + 3k = 30
____________
0 + 0 + 4k = 120
4k = 120
k = 30
No you can plug this into the first two equations
g + t + (30) = 90
g + t = 60
and
g + 2t - (30) = 55
g + 2t = 85
now use elimination again by multiplying the first equation by -1
g + 2t = 85
-g - t = -60
_________
0 + t = 25
t = 25
Now plug those both back into one of the equations. I'll just do the first one.
g + (25) + (30) = 90
g = 35
Therefore, we know that Ted spent the least amount of time on the computer.
The second method is using matrix reduction and getting the matrix in the row echelon form, therefore solving using the gauss jordan method. If you would like me to go through this instead, please leave a comment.
Answer:
<em>The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938
</em>
Step-by-step explanation:
<em>Given:</em>
<em>Here Section 1 students = 20
</em>
<em>
Section 2 students = 30
</em>
<em>
Here there are 15 graded exam papers.
</em>
<em>
(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070
</em>
<em>
(b) Here if x is the number of students copies of section 2 out of 15 exam papers.
</em>
<em> here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15
</em>
<em>Then,
</em>
<em>
Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798
</em>
<em>
(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)
</em>
<em>
so,
</em>
<em>
Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938
</em>
<em>
Note : Here the given distribution is Hyper-geometric distribution
</em>
<em>
where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.</em>
Answer:
A lumber yard is the answer to that riddle.
Step-by-step explanation:
Answer:
429.42 cm
Step-by-step explanation:
<u>Volume of a rectangle prism:</u>
V = whl
V = (5)(12)(7)
V = 420
<u>Volume of a cylinder:</u>
V = πr²h
V = (3.14)(1²)(3)
V = (3.14)(1)(3)
V = 9.42
<u>Combined:</u>
420 + 9.42
429.42
Answer:
1) 
2) ![\sqrt[3]{y^5}=y^{\frac{5}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D%3Dy%5E%7B%5Cfrac%7B5%7D%7B3%7D)
3) ![\sqrt[5]{a^{12}}=a^{\frac{12}{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D%3Da%5E%7B%5Cfrac%7B12%7D%7B5%7D%20%7D)
4) ![\sqrt[4]{z^{9}}=z^\frac{9}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D%3Dz%5E%5Cfrac%7B9%7D%7B4%7D)
Step-by-step explanation:
1) 
We know that 
So,
2) ![\sqrt[3]{y^5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D)
We know that ![\sqrt[3]{x}=x^{\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B3%7D)
So,
3) ![\sqrt[5]{a^{12}}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D)
We know that ![\sqrt[5]{x}=x^{\frac{1}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B5%7D)
So,
4) ![\sqrt[4]{z^{9}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D)
We know that ![\sqrt[4]{x}=x^{\frac{1}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B4%7D)
So,
