Answer:
yes
Step-by-step explanation:
but really nice to the world I'm happy for you because you is so buetiful
Answer:
x = -5, and y = -6
Step-by-step explanation:
Suppose that we have two equations:
A = B
and
C = D
combining the equations means that we will do:
First we multiply both whole equations by constants:
k*(A = B) ---> k*A = k*B
j*(C = D) ----> j*C = j*D
And then we "add" them:
k*A + j*C = k*B + j*D
Now we have the equations:
-x - y = 11
4*x - 5*y = 10
We want to add them in a given form that one of the variables cancels, so we can solve it for the other variable.
Then we can take the first equation:
-x - y = 11
and multiply both sides by 4.
4*(-x - y = 11)
Then we get:
4*(-x - y) = 4*11
-4*x - 4*y = 44
Now we have the two equations:
-4*x - 4*y = 44
4*x - 5*y = 10
(here we can think that we multiplied the second equation by 1, then we have k = 4, and j = 1)
If we add them, we get:
(-4*x - 4*y) + (4*x - 5*y) = 10 + 44
-4*x - 4*y + 4*x - 5*y = 54
-9*y = 54
So we combined the equations and now ended with an equation that is really easy to solve for y.
y = 54/-9 = -6
Now that we know the value of y, we can simply replace it in one of the two equations to get the value of x.
-x - y = 11
-x - (-6) = 11
-x + 6 = 11
-x = 11 -6 = 5
-x = 5
x = -5
Then:
x = -5, and y = -6
Answer:
a) 0.3246
b) 0.0043
Step-by-step explanation:
- For player 1 ; Probability of winning = P(W) = 1/3
- Probability of loosing; P(winning) + P( Loosing) = 1
a) To find Find P(N <= 10) = P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)
= (1/3)^2 + (1/3)^2 x 2/3 + (1/3)^2 x (2/3)^2 + (1/3)^2x (2/3)^3 + (1/3)^2 x (2/3)^4
X (1/3)^2 x (2/3)^5 + (1/3)^2 x (2/3)^6 + (1/3)^2 x (2/3)^7 + (1/3)^2 x (2/3)^8
= 0.3246
b) Find P(N = 10) = (1/3)^2 x (2/3)^8 = 0.0043
32 × 2 = 68
32 × 3 = 96
32 × 4 = 128
So far we can conclude that the answer to your question lies somewhere between the numbers 3 and 4. To narrow down the answer some more, multiply 32 by 3.5 (a midway point between 3 and 4).
32 × 3.5 = 112
The number 112 tells us that the decimal we are looking for is higher than 3.5. (Because we need to get to 125, not 112.) Let's try some decimals between 3.5 and 4.
32 × 3.7 = 118.4
32 × 3.8 = 121.6
32 × 3.9 = 124.8
32 × 4 = 128
As we narrow down our answer, we can see that the number we are looking for lies between 3.9 and 4 on the number line. Now we need to start testing some decimals between 3.9 and 4.
32 × 3.905 = 124.96
Again, use the number five as a "midway" point to decide if you should use numbers that are higher or lower than 3.905. In this case, we need to use numbers higher than 3.905.
32 × 3.906 = 124.992
32 × 3.907 = 125.024
We are getting even closer to our number now that we know the decimal is somewhere between 3.906 and 3.907.
32 × 3.9065 = 125.008
With our midway point we can see that our number lies between 3.906 and 3.9065. Let's try a quarter point to see where our number lies from there.
32 × 3.90625 = 125
And BINGO! We have found the answer to the question. To be rephrased, our answer can be put like this:
= 3.90625
: 6
:
- i just plugged in every number in.
4 + 5 > 10
9 > 10? nope.
5+5 > 10
10 > 10 ? nope.
6+5 > 10
11 > 10? YES.