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weqwewe [10]
2 years ago
8

What is the value of y in the equation 5x + 2y =20 , when x = 0.3

Mathematics
2 answers:
Burka [1]2 years ago
8 0

To solve this equation, we must substitute in 0.3 for the value of x and then solve the equation for y.

5(0.3) + 2y = 20

To simplify, we begin by multiplying the two numbers on the left side of the equation.

1.5 + 2y = 20

Next, we should subtract 1.5 from both sides of the equation to get the variable term alone.

2y = 18.5

Finally, we should divide both sides of the equation by 2 in order to get the variable y completely alone on the left side of the equation.

y = 9.25

Therefore, the value of y is 9.25.

Hope this helps!

Alex_Xolod [135]2 years ago
8 0
5x + 2y = 20
5 X 0.3 + 2y = 20
1.5 + 2y = 20
2y = 20 - 1.5
2y = 18.5
y = 18.5/2
y = 9.25

The value of y is 9.25
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. A high school currently has a 30% dropout rate. They’ve been tasked to decrease that rate by 20%. Find the equivalent percenta
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The equivalent percentage point drop = 10%

Step-by-step explanation:

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The percentage change is the difference between the final and initial values

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What do the differences between the points (as shown on the graph) represent?
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They represent the rise and run of the graph.

Step-by-step explanation:

<em>The difference between the x-axis of the points represents the "run" of the graph (or how much you should run along x-axis to get to the next point.)</em>

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The ratio of rise to run is the slope of the graph, which tells us  how many steps should we take on the y-axis for every step we move forward on the x-axis.

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umka21 [38]

Answer:

0.4007

Step-by-step explanation:

Let's define the following events:

A: method A is used

B: method B is used

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R: the eye is recovered in a month

The probability that the eye has not recovered in a month is 0.002 if method A is used, i.e., P(NR|A) = 0.002, so P(R|A) = 0.998.

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If an eye is recovered in a month after surgery is done in the hospital, what is the probability that method A was performed? We are looking for P(A|R), then, by Bayes' Formula

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