Answer:
1.19
Step-by-step explanation:
1+0.02+0.05+0.12 = 1.19
The correct answer is: [B]: " (2, 5) ".
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Given:
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-5x + y = -5 ;
-4x + 2y = 2 .
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Consider the first equation:
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-5x + y = -5 ; ↔ y + (-5x) = -5 ;
↔ y - 5x = -5 ; Add "5x" to each side of the equation; to isolate "y" on one side of the equation; and to solve in terms of "y".
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y - 5x + 5x = -5 + 5x
y = -5 + 5x ; ↔ y = 5x - 5 ;
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Now, take our second equation:
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-4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ; and solve for "x" :
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-4x + 2(5x - 5) = 2 ;
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Note, 2(5x - 5) = 2(5x) - 2(5) = 10x - 10 ;
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So: -4x + 10x - 10 = 2 ;
On the left-hand side of the equation, combine the "like terms" ;
-4x +10x = 6x ; and rewrite:
6x - 10 = 2 ;
Now, add "10" to each side of the equation:
6x - 10 + 10 = 2 + 10 ;
to get:
6x = 12 ; Now, divide EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;
6x/6 = 12 / 6 ;
x = 2 ;
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Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
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-5x + y = -5 ;
-5(2) + y = -5 ;
-10 + y = -5 ; ↔
y - 10 = -5 ;
Add "10" to each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;
y - 10 + 10 = -5 + 10 ;
y = 5 .
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So, we have, x = 2 ; and y = 5 .
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Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
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first equation:
-5x + y = -5 ;
-5(2) + 5 =? -5?
-10 + 5 =? -5 ? YES!
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second equation:
-4x + 2y = 2 ;
-4(2) + 2(5) =? 2 ?
-8 + 10 =? 2 ? Yes!
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So, the answer is:
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x = 2 , y = 5 ; or, "(2, 5)" ; which is: "Answer choice: [B] " .
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C. 6/5 * 4
6/5 = 1.20 where .20 represents the 20% increase in serving.
1.20 * 4 = 4.8 is the total cups of flour needed to increase serving by 20%
4 is the 100% ; the desired increase is 20%; therefore: 100% + 20% = 120% based on 4 cups. So, 120% ÷ 100% = 1.2 * 4 = 4.80 cups
Answer:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

Step-by-step explanation:
For this case we have the following distribution given:
X 0 1 2 3 4 5 6
P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02
For this case we need to find first the expected value given by:

And replacing we got:

Now we can find the second moment given by:

And replacing we got:

And the variance would be given by:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:
