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Greeley [361]
3 years ago
14

Step by step linear combination. 3x+2y=-4 4x-y=13

Mathematics
1 answer:
Paraphin [41]3 years ago
5 0

Answer:

The solution is the point (2,-5)

Step-by-step explanation:

we have

3x+2y=-4 ------> equation A

4x-y=13 ------> equation B

Multiply equation B by 2 both sides

2(4x-y)=2*13

8x-2y=26 ------> equation C

Adds equation A and equation C

3x+2y=-4\\8x-2y=26\\---------\\3x+8x=-4+26\\11x=22\\x=2

Find the value of y

substitute the value of x in equation A

3(2)+2y=-4

2y=-4-6

2y=-10  

y=-5  

The solution is the point (2,-5)

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Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

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4 years ago
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