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4 years ago

7

If

2 answers:

aleksandrvk [35]4 years ago

8 0

Hello !

5x² + 6x = 70
5x² - 6y = 10

We can reduce 5x² by decreasing the second ecuation so:
5x² + 6x - (5x² - 6y) = 70 - 10
   5x² + 6x - 5x² +6y = 60
     6x + 6y = 60

Now we can divide by 6 and eliminate it too :
              x+y = 10

Multiplying by 10,we'll have :
              10x+10y = 10*10 = 100

Svetach [21]4 years ago

7 0

To solve this problem, there is a very simple trick, once you see it. The first step is to subtract the second equation from the first equation. This gives:

6x + 6y = 60

Multiply by 5/3 to get 10x + 10y

10x + 10y = 60 * 5/3

10x + 10y = 100

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4 years ago

Read 2 more answers

Merta reports that 74% of its trains are on time. A check of 60 randomly selected trains shows that 38 of them arrived on time.

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Answer:

No, the on-time rate of 74% is not correct.

Solution:

As per the question:

Sample size, n = 60

The proportion of the population, P' = 74% = 0.74

q' = 1 - 0.74 = 0.26

We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.

Now,

The proportion of the given sample, p = $\frac{38}{60} = 0.634$

Therefore, the probability is given by:

$P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]$

P $(p\leq 0.634) = P[z\leq -1.87188]$

P $(p\leq 0.634) = P[z\leq -1.87] = 0.0298$

Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %

Thus the on-time rate of 74% is incorrect.

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3 years ago

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Hope it will help :)

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