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Dmitry_Shevchenko [17]
3 years ago
14

−10x+3y=5 ASAP x=y−4 ​ x ? y?

Mathematics
1 answer:
nexus9112 [7]3 years ago
3 0

Answer:

x = 1 and y = 5

Step-by-step explanation:

Use substitution because you know that x = y - 4, and plug this into the first equation to get -10(y - 4) + 3y = 5, or -10y + 40 + 3y = 5. This is -7y = -35 so y = 5. Plug this into the 2nd equation to get that x = 1 and y = 5.

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You have no idea how much it would mean to me and make my day if you helped me with this question ! :)
rodikova [14]
So to do this, you would use the formula: 

Number of favorable outcomes
________________________

Total number of outcomes.


In the first question you are asked: Probability of an even number being spun. We can see that there are 4 even numbers, which is our favorable outcome and that over the number of outcomes, which is 8, would be 4/8 = 0.5 or 50%. Therefore the answer to #1 is 0.5 as a decimal or 50% as a percent. (Do it the way the directions tell you to).

In the second question you are asked the probability to spin a number greater than 3. This does not include 3, so you have 4, 5, 6, 7, and 8. That is 5 numbers. 5 numbers divided by the total number of outcomes is 5/8, which is equal to 0.625, or 62.5%. Therefore the answer to #2 is 0.625 as a decimal or 62.5% as a percent.

The third question asks for the probability that an odd number would be spun. We can see the odd numbers are: 1,3,5,7. This is 4 favorable outcomes divided by 8, the total number of outcomes. 4/8 is equal to 0.5 or 50%. Therefore the answer to #3 is 50% as a percent or 0.5 as decimal.

Hope this helps. Please rate, leave a thanks, and mark a brainliest answer. (Not necessarily mine). Thanks, it really helps!

8 0
3 years ago
Find the coordinates of the missing endpoint if B is the midpoint of AC. C(-5,4), B(-2,5) I need the steps and answer pls and ty
mylen [45]
\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ x}}\quad ,&{{ y}})\quad 
%  (c,d)
C&({{ -5}}\quad ,&{{ 4}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)

\bf \left( \cfrac{-5+x}{2}~,~\cfrac{4+y}{2} \right)=\stackrel{B}{(-2,5)}\implies 
\begin{cases}
\cfrac{-5+x}{2}=-2\\\\
-5+x=-4\\
\boxed{x=1}\\
----------\\
\cfrac{4+y}{2}=5\\\\
4+y=10\\
\boxed{y=6}
\end{cases}
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Let width be x

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ATQ

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